Converging Factor Racing Formulas

Converging Factor Racing Formulas

CONVERGING FACTORS Chapter One The bookmaker lives on figures, and the answer to most problems is in statistics. Yes, but only the USE we make of them. They are as available to the stay-at-home backer as to the man who attends a meeting every day of racing. And in fact, they are the only way that the stay-at-home man can win. The important thing about mathematical and statistical methods is that, quite simply put, they bring METHOD into the business. And that is all-important. You must bring that in, and it must be founded on sound facts. The point I want to make here is that it does not matter if those facts disagree with your previous conceptions and prejudices. Are they facts? Can they be used to our advantage? If so, why worry about our farmer ideas? I’ve often been laughed and sneered at in the past because I have used unorthodox methods of approach. But the methods got results. Isn’t that the important thing? For example, I have suggested using the fact that experts will get more losers than winners, and turning it to our advantage in order to get WINNERS. When I first mooted it, it was called “crazy,” and much worse. But it is a fact, and if so, why not turn it to advantage? Even on Nap Selections, they will always give more losers than winners. One winner of the Sporting Life Challenge Cup had 58 Winners, 131 Losers, and 14 Non-runners. He was the best of the lot, don’t forget. Out of 189 actual runners, he had 58 winners, and that is “good going.” Cannot we use that fact? But if he had 58 winners, he had 131 losers. He tops the list because he showed the biggest profit on level stakes. The bottom of the class, on this basis, gave 45 winners, and 149 losers, from 194 actual runners. There was nearly K100 difference between them in cash on a f1 level stake basis. One was plus f40-odd, and the other nearly minus f.60. But the actual winners were: – 5845 Winners and 189194 Runners Not a great difference! The first main difference becomes clear if we look at the column headed Consecutive Losers. The winner had ......... 0 The wooden spoonist had ...... 14 That is one item that knocked the profits for six. But what if we can evolve a system of selecting winners from the fact that there is a definite percentage of losers? Then the wooden spoonist would immediately become more useful to us than the winner of the Cup! He would obviously be considered the expert that is the least successful, by the ordinary backer. But as far as we are concerned, he would be the other way about! In actual fact, what we are looking for is consistency; it does not matter if it is in losing. 1t will do just as well. And if all the experts are consistent in anything, it is

in the simple fact that they give more losers than winners – that goes on from season to season. The difference in actual winning nap percentage is not much as between winner and wooden spoonist. There is not a yawning gulf at all there. When you consider all the experts on the list, you will not see a tremendous difference. There is considerable similarity! They all give a lot more losers. For convenience later in this book, I’ll give the list as I want to make use of it. In order of success on level staking. Total Runners Winners Losers 189 58 131 194 80 114 197 75 122 194 76 118 193 61 132 196 66 130 197 51 146 194 71 123 191 62 129 191 56 135 197 65 132 192 73 119 194 66 128 196 62 134 196 49 147 187 50 137 195 62 133 194 57 137 185 42 143 193 52 141 198 46 148 183 38 145 194 45 149 There they are, the twenty-three. Quite a lot of difference in the amount won or lost to a level £1 stake. And that is how they are judged. But I am not concerned with passing an opinion about them, I am only concerned with using them. Surely that is what matters? There are indeed many ways in which we can do this. We can take a pair of experts, or three, or indeed, any number. We can assume an expert from a pair to fail, or two from three, or, again, indeed, any different combination. What I want to do from this angle, is to examine the consistent statistical trends, and suggest ways in which they can be utilised. And the fact that experts give less winners than they do losers, must obviously be regarded as without any shadow of a doubt, one of the more consistent trends! But the method of using the experts must depend on the data we are dealing with, and its own particular consistency in any one given direction. It can win often, it can lose often, it does not matter, as long as it shows a definite trend. Our usage of the experts can do the rest for us.

Chapter Two I am always intrigued by the betting forecast. For this is also a statistical source that can be USED to our advantage: it does keep up a steady consistency. Couple it with other proven ideas, and you will get somewhere. For example, have a look at the forecast for 21 consecutive 2 - Y 0 Favourites system races. Forecast Starting Price Result 11/4 3/1 W 11/4 10/11 L 13/2 8/1 L 7/4 5/4 2 9/4 9/2 3 9/4 3/1 W 4/1 10/1 L 6/4 6/4 3 5/2 13/8 W 4/1 4/1 L 4/1 9/2 3 Ev. 4/5 W 6/4 4/9 W 5/2 10/1 W 3/1 15/8 3 3/1 100/30 L 9/4 5/2 2 13/8 3/1 W 5/4 4/9 L 5/4 8/15 W 9/4 7/1 2 The first point to note is that they are CONSECUTIVE races. They are not weeded out to prove a point. Following the system, you would have found them in this particular order. That is very important, in view of the ultimate uses we shall put them to. Row what are the next things to note? 1 –Note that in two instances, the starting price was exactly the same as the forecast. 2 –ln nineteen instances, it differed. 3 –Of the nineteen, the bookmaker gave a better price in nine, and not so favourable to the backer, in ten. 4 –In the 21 races there were 8 winners, 7 placed, and 6 losers. 5 –Of the 8 winners, 5 were in races where the Starting Price was not so good as

the Forecast. And 3, of course, were in the races where the S.P. was better, for the two races where Forecast and S.P. agreed, did not produce a winner for the system. 6 –There was a winner in every three consecutive races, at least. 7 –Of the placed horses from the system, 4 were in races where the S.P. was not so good as the Forecast, 2 were in the races where it was better, and one in the race where it agreed with the Forecast. 8 –But note that although here was a winner in every three consecutive races, apart from one instance, there was no more than one winner. In other words, there was exactly one winner in every three races, as they came. I give them again, in groups of three races: WLL 23W L3W L3W WW3 L2W LW2 There are several interesting things to note as well as the numbered points given above. But I shall not go into them in any detail at the moment. One is that we almost seem to have nearly an even chance of getting a better starting price than the Forecast one. But it is safer, if we want winners, to bet in the races where the Starting Price is not so good as the Forecast thinks it will be. Then, when the Forecast really takes a turn for the better, and the S.F. shows a wide plus gap, we cannot really expect many winners. All these, and many more, are there if we look for them. And we can use all of them. Let us assume, for example, that we can find the winner of a race in the first three horses in the betting forecast, or in the first three horses in the betting. If we can in any way, reduce the possibilities of the winning to three things. We cannot always, but there are many occasions, especially in races of this type, when we can. We call the runners 1, 2, 3. If the forecast favourite wins one race in three, as our figures show, then we could find the winners of three consecutive races in... 111123232323 232311112233 223322331111 TwelveTrebles And, what’s more, we should have done it in six of seven consecutive groups of three consecutive races!! I talked in the first chapter, of experts failing to give winners more often than not! I’ll go into that in new detail later. But for the moment, assume that we have two experts for the first group of three races, and they give . . 2 2 1 and 3 1 2. If we assume one of them to fail, then we shall not bother with... 12233 21132 22311 Four of our twelve trebles, they are the ones that agree with both the experts, in one race or another. 12233 21132

22311 have all got something of both... 23 21 1 2. And we are now left with seven trebles... 3322111 1132233 2 3 1 1 3 2 3. That is only one very simple method of using the statistics we gave for the 21 races. There are many useful things indeed in all that information. Many things that we can use to our advantage. What is more, they can be used in conjunction with one another. Or, if you prefer it, they can be used to cover each other. And those are the lines we shall try and follow in this volume.

Chapter Three I want at this point to introduce some of the technicalities of the methods I shall employ, and start with the very simplest, so that you should find no difficulty in using them. What is even more important, if you master them thoroughly, you will be able to use them with data of your choosing, and with your own particular systems. I have already mentioned two sources of information, the experts and the betting forecast. Suppose we deal first of all with the experts in a general way, and then come down to the particular. If we were dealing with two races, and we have reduced the “possibles” to three horses, the first three in the betting, or what have you, then we should have nine possible doubles for the two races... 111222333 123123123 It is possible, of course, to arrange to cover as many as you want, and we can have four possibles in two races, for example, in... 1234123412341234 1 1 1 1 2 2 2 2 3 3 3 3 4 4 4 4. But at the moment, let us start with the simpler, the three-horses-to-contain-thewinner suggestion. If we have two experts, A and B, and they fancy as follows: AB 23 31 then if both are assumed to have a winner, the only doubles we shall need are: 32 21

Surely that is quite evident? All you have to do is to look along the nine possible doubles, and all of them bar these two, will not agree with BOTH experts. To make it clear, let’s try... 1 1 will not agree with A. 1 2 will not agree with B. 1 3 will not agree with A or B! 2 2 will not agree with B. 2 3 will not agree with B. 3 1 will not agree with A. 3 3 will not agree with A. Therefore, if we assume that both experts will get a winner, the only doubles are: 32 21 But if you assume that EITHER A or B, or BOTH, will have one winner at least, that is, both could have a winner, or one could have the two winners, or at least one, then you are playing very safe, and the only rejected double from the possible nine will be: 1 3 It says that neither will get anything! And its only value will be a NEGATIVE one. In other words, if... One of the two experts gets one winner at least, then if the first race is won by 1, the second race will not be won by 3. But let’s try something a little more ambitious. Three experts – A ,B, C. ABC 213 1 2 3. If A and B both give a winner, then we reject: 3332211 1 2 3 3 1 3 2. But let’s cover, by assuming so, OR B and C to give a winner.

Then all of the seven rejected above, we only reject: 12233 2 1 3 3 1. Again cover, by assuming that we are not even safe there. What we really want is: A or B to give a winner, OR B and C to give a winner, OR A and C to give a winner. Of the five above, that only rejects: 132 2 3 1. In other words. assuming any two of the three experts to give at least one winner, will reject their original forecasts. That simplifies matters considerably. So that we do not have to work out at all, we just reject the three original forecasts of the experts, the three doubles that A, B, and C in turn fancy! Let me look at it from the other way round. A or B to have a winner will retain... They give 21 12 keeps 12 1 2. But B or C to have a winner, will keep, from the 7 rejected by the above, and they give 13 23 keeps 13 32 And A or C to have a winner, will keep from the 5 now rejected, and they give 23 13 keeps 32 13 And it still brings us back the three doubles thrown input if we assume any two of the three experts to have at least one winner. They are: 132 2 3 1 .… the original fancies of the three experts! So, to assume any two of the three experts to give us at least one winner, simply reject their selections, and keep the rest. I have shown work-outs from two different angles, but in actual fact, that’s all it means.

Chapter Four

I will pursue the same idea into the area of the Treble, and you will see that the same simple rule applies. If we have the winners in three horses – 1, 2, 3 – then the possible trebles will obviously be 123123123123123123123123123 111222333111222333111222333 111111111222222222333333333 Have three experts – A, B, C. If their fancies for the three races were: ABC 123 213 123 and we assumed that each would get a winner, at least, then we should only keep six trebles, the six that will agree with all three experts: 112233 132321 323121 But that would probably be a little too ambitious! I am not keen on assuming that all three would get a winner, even in three races! My experience of experts and racing tells me to fight shy of it. And if we are going to play safe, as in the case of the doubles, we shall only get a negative indication. But that is still worth getting, for it tells us what to avoid, and that can be of the utmost value in any betting system. But the point I shall make here is that something that looks very complicated indeed, can be resolved in quite a simple manner, as in the case of the doubles. I shall again go through the work-out, but prove that all you want is a very simple formula. ABC 123 213 123 First of all, assume A or B to have a winner, and you reject 333333322221111 333221133113322 123313232233131

15 trebles

But we don’t want to finally reject that number. We assume A and B, OR B and C – all right. If B and C have a winner, then we would reject, from the 15 above: 333321111 223312233 9 trebles 1 3 3 1 2 3 1 1 3. These nine are the rejection of assuming EITHER A and B, OR B and C have a winner.

But we want to assume A and B, OR B and C, OR A and C to have a winner, in other words, any two experts from the three pairs. A and C would only reject from the 9 above: 123 213 123 So that if we assume A and B, OR B and C, OR A and C, to have a winner, i.e. any two experts, or any two pairs really, from the three pairs, then we reject finally: 123 213 123 or, strangely enough, the original selections of the experts. Or the other way round... A and B to have a winner keeps 332222211111 211322211123 211121331222

12 trebles

B and C to have a winner, keeps 222333 133311 6 trebles 323223 And A and C to have a winner 111333 233322 6 trebles 331131 A total kept of 24 trebles, therefore, a total rejected, of three trebles. And the three rejected are the original selections of the experts. I have shown two different “work-outs,” but the ending is the same, and, indeed, can be arrived at without any work-out at all. To know what to reject if any one pair from any two pairs have a winner, all you do is to reject their original selections. So simple isn’t it? With just three experts, it only has a negative value. You will know what to do in the event of any result after the first race. If 1 wins the first race, avoid 2 1 as a double for the next two. If 2 wins, then avoid 1 2 as the next double If 3 wins, then avoid 3 3 as the next double. But the point I want to make is that you can use several groups of experts, in threes.

You can have, say, 15 experts! Or ideas. In five groups of three each group. Any pair from each group of three possible pair combination, to have a winner, and you automatically know without any complicated workout what trebled to exclude. ABC 123 123 213 213 1 2 3 excludes just 1 2 3 their own selections. DEF 213 213 321 321 1 3 2 excludes just 1 3 2 their own selections. And so on. But the point I really want to get at, is this. All the runners in any race can be arranged in such a way that something on these lines could be done to eliminate unwanted selections. For you can’t tell me that we are asking a lot when we ask that any pair from any two pairs possible from three experts will give a winner – one winner in three races! It is a very modest asking! And my idea in this volume is to provide vital statistics that have stood the tests of many years, and then show how to use them in such a way that will get the best out of them and evade the worst. In other words, to avoid the errors. And to provide simple formulas that are seen at one quick glance. I show you the round about workouts, just the best method of proving the formulas, but all you need worry about are the ideas and the formulas themselves.

Chapter Five In this volume, the two main things that matter are: 1 –Statistical facts beyond any doubt. 2 –Mathematical methods of using them to our advantage. I have already indicated two statistical factors – the betting forecast, and the expert selections. From one point of view, it cannot be said that either is “beyond doubt,” but their trends over the years are certainly very clear, and can therefore be contained in that category! I will leave them for the time being, as I want to introduce some techniques of usage. I have suggested ideas for assuming that experts will have so many winners. And that so many of the experts will have them. But one of the safest ways that I know of allowing for some experts to fail, when obviously we do not know which ones will, or if any at all will, is to take it that in a group of experts’ selections, so many CONSECUTIVE experts will not fail. For example, any two consecutive in four to get a winner, can mean: One and Two. Two and Three. Three and Four.

And that gives them a fair chance. Remember, any consecutive two experts to get at least one winner. The first essential, as previously shown, is to arrange all the runners “that matter” into groups that can be contained in any type of permutation, For the moment, l shall stick to the conventional 1, 2, 3. But I shall come to more ambitious ones later on, and the same basic ideas will hold good then as well. The 1, 2, 3 can contain one horse each, or two each, or more. They can be formed from any basis you like. The order of betting forecasts, the majority expert choice, or what have you. At the moment, it’s the technique that matters. If we assume the winners of two races to be in 1, 2, 3, then the possible doubles are obviously in 123123123 111222333 and have our experts who give: 1213 1132 And we assume any consecutive two to have a winner, then these doubles are out: 123 322 For these three will not have any part of any two consecutive lines. You can test them and see. So if we can expect that no two consecutive experts will fail completely, then the winning double will not be one of these three. Suppose we call the two experts of one paper A and B, and the experts of another paper C and D, and place them thus: ABCD 1211 1132 complete failure by any two consecutive is very unlikely. And the lessons learned may just be negative ones, but they are most useful just the same. In this case, if 2 or 3 won the first race, then we should not back 2 in the second race. 1f 1 won the first race, then don’t back 3 in the second. It will be obvious that the number of doubles rejected will vary according to the relationship of one expert selection to another. For example: ABCD 2132 1213 assuming any two consecutive to have a winner, will reject: 2113 3321 2 winning the first race, and we would not have 3 in the second. 1 winning the first race, and we refuse both 3 and 2 for the second. 3 winning the first race, and we have no faith in 1 for the second!

And another four: ABCD 3212 2113 would reject: 3321 2 3 3 2. And the same type of remarks will apply, but if we had taken the three groups of four experts (twelve experts) and assumed that two consecutive in each four would get a winner, then we reject all these, and some are duplicated, of course... 12321133321 32233212332 And what have we left?... 12 11 Only two doubles. Is it too much to ask two consecutive experts to get a winner in two races? It may be, but in more than two races, in three, and especially four, they are more likely to do so, and that is what I want to drive at. The more races, the better the chances of getting at least one winner. So after showing you the basic idea in two races, I shall move on to better propositions.

Chapter Six Just a few words on the technique of using consecutive experts in Trebles, before I move on to the data from which we can actually put these techniques into betting operation. I have dealt with it on Doubles. We assumed that two consecutive experts from four would get a winner in two races. The chances of two consecutive experts from four getting a winner in three races is naturally much better, and remember, it’s any two consecutive from four. All they have to do is to get one winner. If we are to have the winners in 1, 2, 3, then for a treble there are 27 possibles: 123123123123123123123123123 111222333111222333111222333 111111111222222222333333333 Finding out what would be rejected when we assume any two consecutive to score on doubles, is (quite simple. For it can be SEEN at a glance that certain columns do not agree with any two consecutive doubles. Perhaps it’s not quite so simple on a treble? All right. Let me try and show that it can be done. Four experts, and their choice. ABCD 1231 2121

3232 The method is to take A and B first. Note the trebles rejected if we assume that they both have a winner. Finally, of course, we do not rely oil them, we rely on any two consecutive, a rather safer thing. But for a start, we take their rejects. And they will be 15 trebles, that is, 15 that do not agree with both experts A and B. They are: 333333322221111 3 3 3 2 2 1 1 1 1 3 3 3 3 2 2 15 trebles 312132112121313 They are the rejects of A and B, NOT the final rejects. Now take B and C, the next consecutive pair, and from the above 15 trebles, reject any that do not agree with both B and C. They will be: 333322221111 332233113322 12 trebles 133121213131 Now take C and D, the final consecutive pair, and from the above 12 trebles. Reject any that do not agree with them both. They will be: 122223333 3 1 1 3 3 2 2 3 3 9 trebles 112121313 So that 9 trebles are rejected by assuming that either A and B, or B and C, or C and D, will have at least one winner. Any one consecutive pair to have a winner, and the steps to prove the method are really quite simple. Just deal with the rejects only in every step taken. Worked in the same way: EFGH 2121 3131 1321 would be dealt with: E F, F G, G H And their final rejections would be, assuming any two expert consecutive, to score 333332211 322213212 223132233 And if we had: I J KL 1232 3212 2132

Their 113 313 333

final throw-outs would be 321 123 312

Taking all these groups: A B C D, E F G H, I J K L and assuming that in each group there would be two consecutive trebles with at least one winner, then we should reject all the trebles as dealt with in turn, above. And we should have left over: 332221111 112312211 213332121 There would be duplications in the groups’ rejections, but they are easily seen, and the trebles that matter, those left behind, are the rest of the original 27. You may think that this process is not easy. It may take a little effort for you to master, but once seen, there’s no trouble. Just follow the examples given. To get winners from definite data is possible, and practical, but only if we decide to master the techniques of doing so.

Chapter Seven Let me deal, first of all, with one of the simplest possible examples of using both experts’ and betting forecast. I shall later on proceed to more complex examples, some that will afford us the maximum error-allowance possible. But you will not be able to grasp those unless you are first of all made familiar with the more simple techniques. The simpler the technique, as often as not, the less allowance we have for error. But we still have to walk before we run, in any sphere, and very much so in the rather tricky one we are dealing with at the moment. In this example, I shall make some elementary assumptions. Later on, I shall show methods of covering against some failure among them, But at the moment, let’s make it simple. The Daily Express gives me the race card for Hamilton Park. I quote actual figures and facts from that paper. We deal with the first two races – 2.30 and 3 o’clock. I assume: (a) That the winners of the two races will be in the first three horses in the Express betting forecast. (b) That the three experts – The Scout, Sir Harry, and Form will each give a winner in one of the two races. The assumptions may be ambitious, or not, at the moment I am not concerned with that. I merely want to show how we would deal with the two races, on those assumptions.. I will later on show you how to deal with races when you are making what might be termed less ambitious conditions. That is another story. At the moment we are dealing with A and B as probabilities. First of all, then, the data, as given in the paper: Betting forecast – first three for each race:

2.30 – Lilianje, Valdivia, Norcreet. 3.00 – Gallant Turk, Trim Curry, Happy Wanderer. If we term them 1, 2, 3 in order of the forecast, we have nine possibilities for the two races: 123123123 111222333 For convenience, I shall take the initials of each horse. L V N L V N L V N GT GT GT TC TC TC HW HW HW Nine possible doubles. Next, the experts’ selections for the two races: 2.30 - Lilianje (The Scout), Valdivia (Sir Harry), Norcrest (Form) 3.00 – Gaunt Turk (The Scout), Trim Curry (Sir Harry), Gallant Turk (Form) Or, in terms of our figures: 123 121 So far, so good! We have all the data. What is the procedure? Well, we have taken assumption A, that the winner would be in the first three horses in the betting forecast, and we have only considered those horses. We have got the selections into terms of our figures instead of names of horses, for convenience. Remember, the possibles, in those terms, are: 123123123 111222333 If The Scout is: to give a winner, and he has, 1 1, then, from the nine possible doubles, we shall not require these: 2323 2233 The reason is clear, they do not agree with The Scout at all. If Sir Harry is to have a winner, with 2 2, then we shall also get rid of: 113 3 1 1, for they don’t agree with Sir Harry. And if Form is to give a winner, with 3 1, then this is out: 1 2, for it does not agree with Form. And we have only one double left: 2

1 In terms of horses, Valdivia Gallant Turk. In fact, the actual winning double. Now this winning double was achieved quite simply. Because our two main assumptions came up. Remember? (a) The winners to be in the first three in the betting forecast. (b) The three experts to get at least one winner. I have made simple assumptions -because I want to put over the method. You can make more elaborate ones, assumptions that cover more against failure. In fact, we can assume that the experts will not all get winners in two races, and we would be more right than wrong in most races. What I am aiming at is to show how to take advantage of conditions that prevail more often than not, and to make allowance for their occasional failure. And this chapter marks the first step towards understanding the techniques of using simplified maths to make winning bets.

Chapter Eight In the last chapter, I took the data from the Daily Express, And made two assumptions about two races at Hamilton Park. (a)The winners would be in the first three horses in the Express betting forecast. (b)Three of the papers’ experts would each give a winner in the two races. Both assumptions came up, so we could find the winning double in one column! But things are not always so. Sometimes we should like to cover against failure on the part of the experts. How can we do it? Assuming the winners to be in the first three, then the possible doubles are: 123123123 111222333 And in terms of figures, their order in the betting forecast (fell names, etc., are given in the last chapter), the experts gave: 123 121 Now instead of assuming that all three experts will give a winner, let us hedge a little. Assume that any two consecutive selections will give a winner. This covers you against one of the experts failing altogether. How do we find out the betting doubles? Well, proceed in this way... If we assume any two to have a winner, then first of all, deal with experts one and two: 12 12

If the first one has a winner, we reject: 2323 2233 As in the last chapter. And if the second one has a winner, then we reject 113 311 Again, as the last chapter. Altogether, we reject: 2323113 2233311 retaining 12 2 1. But we do not reject finally at this stage. For we said any two consecutive to give a winner. And the next two consecutive are: 23 21 What will 2 2 reject from the seven above – in other words, what is still thrown out by this consecutive pair as well? 3113 3311 And 3 1 will reject: 22 32 Both together will reject: 311322 3 3 1 1 3 2, and retain 3 2 only. We now have found out what is finally retained by assuming any two consecutive selections to give a winner in the two races. They are: 213 1 2 2, with the winning double: 2 1 among them.

We are left with three doubles, not one, but we have allowed for one expert to fail. In order to do this, we have to make more than one rejection. And the golden rule is to keep every retained column in each rejection process. But I also had a look at another paper's selections. And I want to take two of those experts, making five in all and allow for two of the five experts to fail. By assuming any three consecutive. To give a winner. How can I do that? The same procedure, but I take three experts at a time, and not two. With the three already given, they gave, and we call them A, B, C, D, E, for short. ABCDE 12313 12133 From the nine possibles, A would reject: 2323 2233 B would reject: 311 113 C would reject: 1 2 And: 2 1 is retained, and, remember, finally kept. So we have 8 columns to deal with again: 23233111 22331132 From them, B would reject 3113 1133 C would reject 122 2 2 3. D would reject 3 2 Between them, 31131223 11332232 In fact, all the columns. Nothing retained. So we deal with the same columns for the last, three consecutives: C, D, E. We have already taken A, B, C, and we found they only retained one double. And we see that B, C, D do not keep anything at all. From the eight doubles: C will reject

2211 2323 D will reject 33 12 E will reject 1 1 And this time, one will be retained: 3 3 So we have two doubles to back: 23 13 2 1 or Valdivia Gallant Turk being the winning double. We took five experts. We assumed the winners of the two races to be in the first three horses in the betting forecast of the Daily Express. But we assumed any three consecutive of the experts to give a winner. We could have had two complete failures, and still have had the winning double in two attempts. I think you’ll agree that it is better cover than the assumption that all the experts you use will get a winner. Obviously, the main assumption here, really, is the one that the winners will be found among the first three in the betting forecast. As I have shown you ways of allowing for the experts to fail, so it can be shown that the winners can come from a rather more democratic area! The net can be wider. Although l would expect quite a lot of winners from the first three in the forecast, with no more than two experts from five failing to give any winner at all! We have already dealt with distinct possibilities!

Chapter Nine I have already dealt with two races when we assume: 1 – That the winners will be found in the three first horses in the betting forecast of the morning paper. 2 – Where we assume that three experts will all have a winner. I have also dealt with the method of assuming that any two consecutive experts will have a winner. This acts as a cover against one of the experts failing to get a winner. I have also suggested that it would in fact be more of a practical proposition to assume that one expert or more in any group you take will fail to

get a winner. We have to deal with things as they are, and it is quite a common occurrence for an expert, or even for more than one, from a group, to fail to get a winner, especially in two races. So we want to assume failure on the part of one or more experts. But we also want to take advantage of the fact that one expert, at least, in a group, is likely to give a winner. In other words, we should take the facts as they come. Some experts will fail, but not all experts will fail. It has always happened, and is more than likely to happen again. Some will fail, and some will score. Which ones will do what? No one knows, but we can take advantage of the facts without knowing anything about that! We can make the necessary allowance, because we are dealing with mathematical methods, and they do not worry about personalities or fancies. They can arrive at an unprejudiced answer. 1%at is their strength. Looking at the card for the York meeting, I note the betting forecast in the Daily Express for the two races 3.0 and 3.30. Let us assume the winners to be in the first three horses in the forecast. Number the horses in the forecast order 1, 2, 3. 3.00. Prairie Emblem (1), Antonamy (2), Arietta (3). 3.30. Cannebiere (1), Sweden (2), Joyeaux II (3). All the possible doubles, then, assuming the winners to be in the above: 123123123 111222333 Or, in abbreviated form: P.E. A . Ar. P.E. A. Ar. P.E. A. Ar. C. C. C. S. S. S. J. J. J. The paper gave three different expert opinions: A B C Prairie E. Prairie E. Arietta Joyeaux II Sweden Cannebiere Or in terms of our numbers: ABC 113 321 So far, so good. We want two different things. And both are very likely to happen. That is the important thing. 1 – We want to allow for one expert in two to fail to give a winner 2 – We also want to assume that at least one of the three will give at least one winner, in the two races Both are very likely things to happen. It is logical to expect them. The 9 possibles then, are. 123123123 111222333 If we assume that A or B will NOT give a winner then we reject: 111 321 Far they all agree with both A and B – that is clear enough. If either B or C is to fail, then we also reject: 3

2 For that particular double agrees with both. So does 1 1 but that has already been thrown out in the first move. And if A or C will not both give a winner, then we reject: 3 3 For that agrees with both A and C. What have we now left? 2223 3211 But we don’t think that all three expert will fail in both races, surely? So we assume... A or B to have a winner, and that rejects: 32 11 For the two columns will not have it that either A or B will have a winner at all. And if we also assume that either B or C will have a winner, then we reject: 2 3 What does that leave us with? 2 2 One double. In terms of horses: Antonomy Sweden There is nothing magical about it. The first essential, in this instance, was for the winners to be in the first three horses in the betting forecast. Then we really took the averages as they generally prevail. Some of the experts to fail, and one or two to get a winner. Nothing unusual, in fact the ordinary run of day-to-day racing. We simply took advantage of facts, and we used mathematics, in very simple form. Some expert will fail, and some expert will give a winner, every day that comes along. It happens the run of the experts of any one particular newspaper as well. Although we can take as many papers as we wish. We can play the experts of one paper against the experts of another one. It does not matter, the techniques can be many, what matters is the idea, and the facts, as they come, day by day.

Chapter Ten There is another ingenious method of taking the “best of both worlds,” as far as the experts, are concerned. I saw a “perfect” example at Sandown the other day, using the Daily Express and its racing experts. The races were the 2.30 and 3.0. The assumption that the winners wauld be in the first three in the paper’s betting

forecast. The forecast named these: 2.30. Durham Ranger (1), Trackabu (2), Runantic (3). 3.00. U.10 (1), Shirastrin (2), Predominate (3). The experts gave these: 2.30. Durham Ranger (The Scout), Trackabu (Course Corr.), Durham Ranger (Form). 3.00. Predominate (The Scout), U.10 (Course Corr.), U.10 (Form). Or in terms of position in the betting forecast: 121 311 Instead of looking at the experts in the normal way, and taking them to fail, or to give winners, let me suggest an alternative. If the winners are in the first three in the betting forecast, then the “possible” doubles must be these: 123123123 111222333 Let us take the doubles that are outside the ones suggested by the experts. In other words, the alternatives to the experts. For 1 3 they will be 23 12 For 2 1 they will be 13 23 For 1 1 they will be 23 23 So that we have six columns: 231323 122323 They are the “reverse” of the experts. And in two races, we must not expect the experts to give winners in “profusion,” we must get some failure, but we must also get some success. So that we can use the “reverse” of the experts and still expect some of those to fail. What if we assume that there will be at least one column in every consecutive three of these “ reverse ” ones without a winner? The reverses should be two-thirds effective to get at least one winner. That is reasonable, if the experts don’t do too well! Prom the 9 possible doubles, then, If

231 122 is to have at least one column that fails, we reject 2 2 For, as you can see, it agrees with all three. If 313 223 is to have a failure column, we reject 3 2 If 132 232 is to have one column to fail, we reject Nil – for there is not one left to agree with all three. If 323 323 is to have a failure, we reject 2 3. And we have left: 111332 3 1 2 1 3 1. We have assumed that every three consecutive would not have a winner in at least one column. But let us look at the other side. Assume that every two consecutive will have one winner in the two races. Remember, we are dealing with the reverse of the experts. If 23 12 is to have a winner between them, we keep 3 1. For of the six left above, it is the only column to agree with both. If 31 22 are to have a winner, we keep 1 2. If 13 23 is to have a winner, we keep

1 3. 32 32 will not keep us anything, and neither will 23 23 So all we keep are three doubles: 113 321 Or Durham Ranger Predominate

Durham Ranger Shirastrin

Runantic U10

...with the last double, Runantic and U10, being the winning one. S.P. of 8-1 and 4-1. A very nice double indeed. Note that we reversed the expert selections, and made more allowance for the reverses to score, than not. Quite reasonable. But we also knew that the reverses would not be 100 per cent. The experts are expected to score at least one winner in two races, between all three of them. We rely a little on the reverses to lose, and a little more on them to win. All the eggs are not in one basket, and that is sound reasoning, as you’ll agree.

Chapter Eleven Still talking about the betting forecast in the morning paper, you will find that the winners are nearly always in the first six in the order given. On a bad day for the forecast, that is, a good day for the better-priced horses, you will find that in two consecutive races, you will get winners from OUTSIDE the first three, very often in 4, 5, 6 of the betting forecast. On a good day for the forecast, you are apt to get the winners in the 1, 2, 3 in the forecast. “It never rains but it pours.” There is no need to do any more than just give you that thought. So, if we had the first six, in two groups, we should have many winners indeed. Something like this for two consecutive races: 123123123 111222333 and 456456456 444555666 Have a look at the Daily Express betting forecast, and expert selections for two consecutive races at Kempton. The 2.0 Betting forecast order: Henley Brianna Straight Flight 1 2 3

Lago Last Port Fighting Fox 4 5 6 The 2.30: First Light Nautch Petra 1 2 3 Parched Bright SilkChou Chin Chow 4 5 6 The experts: The Scout Course Corr. Form 121 423 Now, in case any of the experts get winners, or in case more than one will do so, we want to keep all doubles that agree with them in any way. The two groups of doubles are, again. 123123123456456456 111222333444555666 So let us keep all doubles ending with 4.2.3., the experts’ choice for the second race. They will be: 456123123 4 4 4 2 2 2 3 3 3. Nine doubles. The Scout is the only expert to have a horse outside the first three in the betting forecast, in other words, to have a horse in the 4.5.6. group. So we have to keep all three doubles ending in 4, as we have no other expert to “compare” him with. So we keep: 456 4 4 4. In terms of horses: Lago Last Port Fighting Fox Parched Parched Parched We could go over to other newspaper experts in order to get a comparison, but at the moment we are taking all our information from one paper, so we keep the three doubles. But we have two experts with their selections in the other group. In other words, they think that the winners will be in 1.2.3 in the betting forecast. They say: 21 23 If we assume that they will not both get a winner, then we must discard, from the 1.2.3 group, the following doubles: 21 32 For they agree with both experts. So we are left with: 2313 2 2 3 3, of the 1.2.3 group, and with: 456

445 of the 4.5.6 group. Seven doubles altogether: Lago Last Port Fighting Fox Brianna Straight Flight Parched Parched Parched Nautch Nautch and Henley Straight Flight Petra Petra The winning double being: 6 Fighting Fox (100-8) 4 Parched (5-1) The prices, as the winners were outside the 1.2.3 group, were exceptionally good, and for seven doubles, we have the first six horses in both races running for us. More often than not, the winners will be found among the first six horses in the betting forecast. As they should be! You will note that in all our racing calculations, we are assuming that while the experts will not all give winners, they will give some, and we are on very safe ground indeed when we assume that. The rest is up to us!

Chapter Twelve In any system of betting there must always be a waiting period. The most simple idea of all, perhaps, the idea of following an expert every day, on the assumption that he must give a winner, or winners, if he himself is to “ survive,” is a case in point. It then becomes a matter of just how long you can carry on, and the price of the winner when it does come. I have met several men who tell me that getting winners in racing is the very simplest of affairs. You simply follow one particular expert until he gets a winner. They say that it is rather unlikely for an expert to go one complete meeting without a winner, and if he does, watch him in the next one! It might well be possible to make money in that way, if we had an inexhaustible supply of cash and patience, and if the bookmaker would accept the tremendous bets that must accrue. But I don’t notice the people who tell wallowing in any luxuries. So they have not got the cash it would need to carry on, for the stake would have to be progressively raised as they went along. But if the price of the winner, when it came, was a good one, then it might well be a practical system. Unfortunately, the expert is more likely to break his losing spell with a favourite, than with an outsider! But still the winning systems are really

waiting ones. The basic idea that I have been discussing, the assumption that the winner will ’be found in the first three horses in the betting forecast, for example. You can wait for that, as well. Or you can include more than three horses. Have the first six in the betting forecast. In that case, your betting will have to be heavier. But you are more likely to get nearly all the winners in your initial calculations. It becomes, then, a case of, either: A.Having more than three horses in your initial columns, OR B.Devising a waiting method that will give you the winners taken on this first three basis, when they do come. They will come, as indeed they must. They have a better chance of scoring than any one expert’s selections in any one day. First of all, they are three, and not one horse. And they are easily built into a sequence. Suppose we think of a waiting method, then. There are six races in the meeting. And it is odds on that at least one consecutive pair of those races will be won by horses that are in the first three in the morning paper’s betting forecast. How can we divide the six races in doubles of two consecutive races? For example: The 2.0-2.30, 2.30-3.0, 3.0-3.30, 3.30-4.0, 4.0-4.30 You will find that one double, at least, and very often more then one, will be won by horses from the first three in the betting forecast. So that if you covered all the possible pairs of races, you would be concerned with five doubles. As I would reduce my bets by making assumptions on the expert selections, I would be able to bring my bets down to, perhaps, two doubles for every possible two consecutive race pair. So that I would have ten doubles in all, and if the winning double, or doubles, more than covered that, then I make a profit. For, do remember that we are bound to get winning doubles. There will be races won by the first three horses in the betting forecast. All we have to watch is that our reduction methods are sound. The thing to note about expert selections is, that while they do not all give winners, some of them always do give at least one. By that, I mean, that of a group of experts, one at least will be very likely to give a winner. But not all the experts in the group will have a winner. A does well today, and tomorrow is another day for B. C gets the winners here, and D gets them there. All we have to do is to remember these facts, and use them, and we must not depend on any individual expert to fail or win, but on a group to do the “average” thing. If we can arrange our doubles to cost us less than the gain from the winning one or ones, then we are “in”. A suggestion can be made here. We can discriminate, and use only the races where the experts available all agree that the winners should come from 1.2.3 horses. I shall follow that idea later on. Suppose we have four experts, who all give horses that are in the 1.2.3 group. And their selections: ABCD 1213 2312 We can make two sensible assumptions. A. That one expert in every two consecutive ones, will NOT get a winner. In other

words, that: A or B will fail, B or C will fail, C or D will fail Or both in any pair, of course. There are nine possible doubles, given three horses in each race: 123123123 111222333 If A or B must fail, then we delete: 21 2 3. For they agree with both. If B or C must fail, then we delete: 2 1. If C or D must fail, then we reject: 31 1 2. But we can assume that someone must have a winner. For the winners we want, and will eventually get, are in 1.2.3 And the expert selections are also there: So, if we assumed, say, that either A or C would get one winner at least, then we get rid of: 32 33 And if we thought that either B or D would get a winner, then we get rid of: 1 1. And we have only one double left over: 3 2. It will not always work down to one double. But it isn’t necessary. The winning double will often more than pay for several losing doubles. We are assuming a minority of one double in. the five possible to have horses winning from the first three in the betting forecast. Very often you will get two and three winning doubles. I will proceed to study the effect of various reducing ideas in practice, and also the method of not betting where ALL the experts have not got selections from the first three horses in the forecast. It will be, I am sure you will agree, very interesting.

Chapter Thirteen There are many other angles to the business of taking expert selections to team up with the betting forecast. Instead of having several experts, and making assumptions about them, let me just show a way of using one. Taking a day’s racing at Leicester, and the Form selections from the Daily Express. I assume that the expert will get two winners in the six races, at least. Thus assuring me of a winning double. I am interested in all races where the expert selections are also from the first three in the morning’s betting forecast. If all his selections are in that category, then I have 15 doubles on my hands. So I must find a method of

reducing a little. I will take all the races that have the oper expert selectors giving horses from the first three in the betting forecast. So I have two conditions. The selections must be from the first three horses in the betting forecast. If they are all so in the case of the “main” expert, then we reduce to less than the 15 full possible doubles, by eliminating all races in which the other experts go outside the first three in the forecast. That’s clear enough. First of all, let’s have a look at the “main” expert, the Form column in the paper. He gives: Race one: Canastra two: Diesel three: Racing Clouds four: Thorney Hill five: Mainswitch six: Taphius That, of course, is one selection for each race, and if we are to consider all of them, we have any two from six, equals 15 doubles. But we shall only consider those that are 1.2.3 in the forecast. The next obvious step is to have a look at the betting forecast. I give them in order of 1.2.3. Race one: King Bomba, Canastra, Huddersfield two: Diesel, Downtown, Greek Consort three: Racing Clouds, Colarani, Clermont four: Thorney Hill, Richie, Prince Silver five: Mainswitch, Ellwood, Racing Clouds six: Taphius, Mode, Honey Harvest All Form column selections are in the 1.2.3 horses, as you can see by comparing Form, and the above Forecast list. So that, in a column, his selections are: Race one:2 two:1 three:1 four:1 five: 1 six: 1 If we look no further, than, we have 15 doubles, for we expect him to give us at least one winning double. Calling them F, after Form, we have: Race one: F F F F F two: F F F F F three: F F F F F four: F F F F F five: F F F F F six: F F F F F Fifteen doubles. But we want to have a smaller number, if we can. So we have a look at the other experts. And we find that there are only races one, four, five, and six, that make the other experts choose horses from the first three in the betting forecast. So we shall only use Form selections for those races. And our rearranged columns will look thus:

Race one: F F F two: three: four: F F F five: F F F six: F F F Six Doubles In terms of horses, they will be: Canastra Canastra Canastra Thorney Hill Thorney Hill Mainswitch Taphius Mainswitch Thorney Hill Mainswitch Taphius Taphius The six winners were, in the betting forecast: 253113 Form, then, had winners in races one, four and five: 2 1 1 The six doubles yielded three winning ones: Races One and Four – at 3-1 and 8-11 Races One and Five – at 3-1 and 5-4 Races Four and Five – at 8-11 and 5-4 The winning doubles were: Canastra and Thorney Hill Canastra and Mainswitch Thorney Hill and Mainswitch The rules are simple. We take the selections of a “main” expert in so far as they agree with 1.2.3 in the betting forecast. If they all agree, as in the example given, then we deduct from them all races for which the other experts in the paper do not select from 1.2.3 horses. If we took all the possible doubles for six races, any two from six, we have fifteen. But the other experts disagreed with 1.2.3 of the betting forecast in two of the six races. So we have any two from four left, and that equals six. Of the six possible doubles, three came up. The prices were not so good as they might have been, but they would have shown a fair enough profit. This, obviously, is only a very simple version of using the experts and the betting forecast, but it will give you some idea of the possibilities there are. When we go on to couple this with other covering systems, then we may well find it one of the most effective means of all of getting our account on the credit side of our ledger!

Chapter Fourteen Another very simple, and efficient, method of utilising the betting forecast and the experts, and one that should score very often. The main points are: We select a “main” expert, and we expect that he will give us at least one double in the races selected. The races chosen wi1l be those in which our first expert, and another expert, will have their fancies from the first three horses in the betting forecast. The meeting is Beverley, and the first four races come into this

category. First of all, the betting forecast: 2.30 Gallant Turk Trim Curry Lavendon 1 2 3 3.0 Ocarillus Dignified Gledhill Boy 1 2 3 3.30 Jacmil Harding Lauder 1 2 3 4.0 Pearl Band Reine Galante Kessaway 1 2 3 The selections of the Scout and another expert, given in number order in the betting forecast: The Scout Expert A 2 2 2 3 3 2 1 2 They both, for these four races, give selections that are in the first three horses in the paper’s betting forecast. So that we can consider the possibility of taking all possible doubles. But we are really basically concerned with the selections of the “main” expert, in this case, The Scout. If we take all the possible doubles on his selections, they can be six; any two from four. Races: 1 with 2 1 with 3 1 with 4 2 with 3 2 with 4 3 with 4 But I always assume that no one expert, however good, will have it all his own way. So I will take only the races in which he disagrees with the other expert, as far as doubles are in question. There will, then, be only the doubles: Races: 2 and 3 2 and 4 3 and 4. For they both agree on race 1. As it happened in this particular meeting, we lost winning doubles by doing this, but we have to follow an idea. In the event, we would have got one winning double, on races 3 and 4, Lauder and Pearl Band, at 2-1 and 7-2; still a paying proposition with the expenditure of three doubles. The winners of the four races being: Race one: Trim Curry at 2-1 two: Gledhill Boy at evens three: Lauder at 2-1 four: Pearl Band at 7-2

One winning double in three, perhaps, is not bad going. Have you got the rules? Take all races at a meeting where the two experts will have their selections from the first three in the betting forecast. Have one expert, that we can call the “main” one, and assume that he is going to give us at least one winning double among his selections. So that if we have three races to consider, we have three “possible” doubles: 1 with 2, 1 with 3, and 2 with 3 Four races give us six possibles: 111223 234344 Five races give us ten possibles: 1111222334 2345345455 Six races give us fifteen possibles: 111112222333445 234563456456566 But the second rule will reduce the number of doubles. We only use the “main” expert's selections in the races where he disagrees with his fellow experts. The assumption is that as they do disagree on most races, we shall have less doubles to back, and those that do come up may be better priced. We shall miss some winners that way, but we shall also get some, and no expert will have it all his own way. So that we can expect expert the second to fail in at least some races. Remember that our main expert is the one we look to for the winners, at least on any one given day. We can choose the main expert on the grounds that he has had two very bad days on the run, if you like. That would be a sensible basis. For he cannot go on not getting any winners at all, and when they do come, they come in more than one! In other words, in two or three doubles!!

Chapter Fifteen

There is another simple method of covering against failure oil the consecutive double idea. We need four experts. Call them A, B, C, D. Let A be your main expert. Let D be the expert on whom you depend next. Consider only the races in which expert A's selections are 1.2.3 in the betting forecast. Consider only consecutive doubles. A and D can be your choice for your own reasons. They can have had a couple of bad days, if you like. Suppose in the six races, we have the following information: Expert A selections in betting forecast 1 2 2 5 2 1 We are, therefore, concerned with consecutive races as under: Race One and Two Race Two and Three Race Five and Six For the expert does not have a 1.2.3 selection in race 4, and that cancels 3 and 4, and 4 and 5. Suppose we also cover the expert against failure in our doubles. For races One and Two, instead of his 1 2 we cover for any failure within the 1.2.3 base: 11123 1 3 2 2 2. Five doubles. And we have experts B, C, D giving: 232 123 If we assume that either B and C will get a winner each, OR, failing that, D to get at least a winner, then we shall get rid of these from our five doubles: 113 122 And we are left with: 21 23 How? Well, if 2 1 is to have a winner, we shan’t keep: 113 322 And if 3

2 is to have a winner, we shan’t keep: 1 1 So we’ve rejected: 1131 3 2 2 1. And kept 2 2 But we cover against failure by B and C, by only rejecting those that D will reject as well. Or, if you prefer it, by keeping those that D wants. Of the four thrown out above, three are also thrown out by D, they do not agree with him either. They are: 131 221 And we ARE left with: 2 1 23 So we have two doubles instead of five. Then the next consecutive pair of races used, races 2 and 3. The expert A gives: 2 2.So we cover with: 2 2 2 1 3 1 3 2 2 2. Then B C D 123 311 lf we use the same method, and assume either B and C, OR D to have a winner, we reject: 12 1 2, and we keep: 232 3 2 1. Three doubles for five. Then the other pair of races that we can use: races 5 and 6. Expert A gives: 2 1. So we have: 22213 2 3 1 1 1. Experts: BCD 131 211 Using them as previously shown, we reject: 22 32

and we are left with: 123 1 1 1. The actual winners of the six races, were, and they are taken from a day’s racing at Catterick: 1 3 2 6 2 1. Taking, of course, their positions in the Express betting forecast. We have reduced the five doubles needed for cover to quite a few less. All our remaining doubles will be: Races: **Winners marked 121 * 223 __________ * 2232 * 3321 __________ * 5123 * 6111 We would actually have to get eight doubles. And they would have given three winning ones. The rules again: Have four experts. They must give selections in the 1.2.3 ff the betting forecast. Expert A we follow, but we allow him up to one failure in every double that we take from him. The failure to be covered by one of the 1.2.3 horses. That makes it five doubles based on each of his doubles. Then we have experts B, C, D. Assume that if B and C do not get a winner, that D will or vice versa. That makes the reductions. When we assume things like this about the experts, we are on safe ground. We need not be afraid of it, for some will fail, and some will give winners, but they won’t all fail, and they will not all give winners. We only use the races where expert A agrees with one of 1.2.3 horses. Those are all the rules. Quite simple. And they are based on observation of the trends of races and experts, and the betting forecast!!

Chapter Sixteen

I would like to offer something that I think has most interesting possibilities, and I want to emphasise that it is, as yet, a basis. I shall show you methods of reduction that will cut down the number of bets, and still retain the good points inherent in the idea. Any useful and logical reduction method would bring the number of bets down to make it very profitable indeed, and I would like to follow some of those in future chapters. At the moment, however, I would like to give you the basis. And may I say that, as it stands, it has often proved to be most profitable, so that any efficient reduction method would make it a really excellent proposition. The idea is to take one expert, after a bad day, if you like, or after a bad run. Then we are looking to him for some winners. The course is Wolverhampton, and we’ll take The Scout, one of my favourite experts, in any case! First of all, we’ll identify his selections in terms of the Daily Express betting forecast. Races Scout selections in betting forecast One 1 Two 6 Three 1 Four 1 Five 2 Six 1 Note that all except one are in the 1.2.3 of the betting forecast. First of all, we take all possible doubles that are in the 1.2.3 of the betting forecast. They will be TEN doubles, taking all The Scout selections INCLUDED in the paper’s betting forecast 1.2.3. For we shall not include race two, for which his selection is number 6 in the paper’s betting forecast. The doubles, then, are in terms of The Scout selections from the 1.2.3 horses: Race One with Three One with Four One with Five One with SixT en doubles Three with Four Three with Five Three with Six Four with Five Four with Six Five with Six In terms of horses, they would be: Free Gift with Entente Cordiale, with Inveroy, with Astrentia, and with Marly Knowe. Entente Cordiale with Inveroy, with Astrentia, and with MarlyKnowe. Inveroy with Astrentia, and with Marly Knowe. Astrentia with Marly Knowe. The winners were: Race No. in betting forecast 14 2 6* 3 1* 4 1* 54

6 1* Those given by The Scout are marked with an asterisk (*) Leaving out the second race, we have three winning doubles from the ten above. Race Three with Four, and with Six. Race Four with Six. 4-1 and 9-4. 4-1 and 5-4. 9-4 and 5-4. In terms of the betting forecast 1.2.3 we have now ten doubles: 1111111112 1121121211 The first four are race one with races three, four, five, and six. The next three are race three with races four, five, and six. The next two are race four with races five and six. The last one is race five with race six. Three of them were winning doubles. That is part of our basic plan. Now you will have noticed that it either is a good day for the forecast short-priced horses, or it is a bad day for them! There are not many in-betweens. So the second part of the basic plan is to take any race or races where The Scout’s choice is not in the 1.2.3 and double them with horses that are 4.5.6 in the betting forecast in all other races. They will be 6 in the betting forecast, in race two, with 4,5.6 in all other races. That will be 15 doubles: *6 6 6 *4 5 6 in race one I have marked the winning doubles with an asterisk 666 456 in race three The prices, however, were quite good, as it might be expected. 11-2 and 9-4, 9-4 and 20-1 666 456 in race four: *6 6 6 *4 5 6 in race five 666 456 in race six The basic idea, then, is in two parts: A.The Scout selections in all races where he agrees with 1.2.3 in the betting forecast. B.The Scout’s selection that is number 6 in the forecast, i.e. is outside the 1.2.3, with 4.5.6 in all the other races. Altogether there are 25 doubles. lt would be the basis. We have to find satisfactory ways of getting this down to fewer doubles. But that is the basis, 25 doubles. It so happened that we should have done quite well out of it, even at 25 doubles, but it should prove to be a very good idea if we can reduce a little. In the: example given, we got five winning doubles: 4-1 4-1 9-4 11-2 9-4 9-4 5-4 5-4 9-4 20-1

With a £1 stake on each double, you would have got a very good day’s wages indeed! But it would be very possible to have less doubles, and still get the winning ones, provided we have an efficient reducing medium.

Chapter Seventeen In the last chapter I gave you a basis for doubles, that, although involving quite a number of doubles, would yet make a really good profit for itself on many occasions indeed. I emphasised that the idea is to show you the basis, and then proceed to give you methods of reduction that will give it a chance to be very profitable indeed. In this chapter, I will again give you a basis for betting, and then I shall again give you methods of reduction. It is really surprising how many of these basics will persist in showing a profit even with a very large number of bets! But I want to provide the method of reduction to suit each particular idea later on, for we may strike a particularly short-priced number of winners. A day at Alexandra Park, and I first of all get the horses sorted into their positions in the betting forecast. I will go as far as six, and I call them, in the order of the forecast, 1.2.3.4.5.6. I have taken a day’s racing that does not qualify to be at all a good day for the idea. That is the fairest way to introduce it. And. the basis will involve a little matter of NINETY DOUBLES! Do remember that it is a basis, and later on methods will be shown of making considerable reductions. But, even so, well, I shall let you discover what happened for yourself! The secret of the idea is that we deal with consecutive pairs of races. In six races, there are five of them: 12 23 34 45 56 We have six horses in every race to consider: 1.2.3.4.5.6 in the betting forecast. First of all, take each consecutive pair of races, with this ordor of choosing your doubles: Races 1 and 2: 123123123 444555666 Races 2 and 3: 123123123 444555666 Races 3 and 4: 123123123 444555666 Races 4 and 5: 123123123 444555666 Races 5 and 6: 123123123

444555666 That is quite a number of doubles: 5 x 9=45. But there are more to follow! Take again the same order of races, but placing your doubles thus: Races 1 and 2: 456456456 111222333 Races 2 and 3: 456456456 111222333 Races 3 and 4: 456456456 111222333 Races 4 and 5: 456456456 111222333 Races 5 and 6: 456456456 111222333 Another 45 doubles! And that makes 90 doubles altogether. Of course, it’s only a basis, and we can move on from there. The winners, in terms of position in the betting forecast, were: Race one: 4 two: 6 three: 1 four: 1 five: 4 six: 1 Although they were all in the first six in the betting forecast, they could have been much better placed as far as our idea is concerned! So that only three were winning doubles from all the ninety! They were: Races four and five with 1 4 Races two and three, with 6 1. Races five and six, with 4 1. The odds for the first double from races four and five were: 9-4 and 20-1. The second double, from races two and three, were: 9-4 and 4-1. I he third double, from races five and six, were 20-1 and 5-4 With only three winning doubles from ninety, what happens? How much money do we lose, and how urgent is the need for a system that will reduce the number of doubles from ninety to about ten? And that could be done. But let me try and

get the answer to the main question first. The first double: 9-4 and 20-1. 1 x 9/4 = 2.25 plus 1=3.25. 3.25 x 20 = 65 plus 3.25 = 68.25 to come. The second double: 9-4 and 4-1 = 1 x 9/4 = 2.25 plus 1 = 3.25. 3.25 x 4 = 13 plus 3.25 = 16.25. The.third double: 20-1 and 5-4. 1 x 20 = 20 plus 1 = 21. 21 x 5/4 =26.25 plus 21 = 47.25 So 68.25 + 16.25 + 47.25 = 132.75 And even our 90 doubles do not look so bad! Obviously, we shall have to find reduction methods, but I hope that I have done enough to show you that by taking consecutive races, and covering one method against another, that it is possible to get good results And even on ninety doubles, show a little gain! And if we can devise a reducing method that gives good results, then we reality are in with a chance!

Chapter Eighteen It must be obvious that if we could cover the first six in the betting forecast, that we would get nearly all the winners! How we can reduce this number and still get most of the actual winners is one of the most interesting things to try and solve in the whole annals of betting. I think that it could be done, and quite often. We need to localise some information that will enable us to make the necessary reductions. For example, it will be readily admitted that all the experts will not get a winner in any two consecutive races. Neither will all the experts fail to get a winner in any two consecutive races. There will also be a percentage of winners from the first choice of the betting forecast. That will be obvious to anyone who follows racing at all. That is happening every day. To show you what I mean by a very simple reduction, let me give you some data from a day’s racing at Ripon. The betting forecast is from the Daily Express. I give the order of the betting forecast, the first six for each race, and the winner of the race. 2.0 Legal Jargon, Spring Day, Plaint, Tarantus, Toaci, Tovary Tovar. Winner: Legal Jargon, 6-5 2.30 Rock Council, Farrington, Hoar Frost, Defender, Lightkeeper, Bedouin.Winner: Lightkeeper 15-8 3.0 Pool Green, Prince Conkera, Irish Cheer, Moss Rose, Tudor Playboy, Miss Slipper. Winner: Moss Rose 7-4 3.30 Sunshade, Broken Ranks, Tom Cobbler, Sansu, Buffalo, All others no forecast Winner: Broken Ranks 15-8 4.0 Naval Fighter, English Wonder, Stratocruiser, Sleepy Session, Round the Clock, Hushabye Lu. Winner: Sleepy Session 8-1

4.30 Ihe Saint, Woodworm, Happy Clacton, Panshanger, Flying Snob, Cranford Cross. Winner: lee Saint 11-10 You will see then, that the order of the winners in the betting forecast, was: 1 5 4 2 4 1 And if we wanted to cover each pair of races, taking l and 2, with 3 and 4, and 5 and 6, we bet on the: 2.0 and 2.30, 3.0 and 3.30, 4.0 and 4.30 If we wanted to cover the first six horses, we could do it: 123123123 444555666 and 456456456 111222333 for each pair of races. That means: 18 doubles for the 2.0 and 2.30, 18 doubles for the 3.0 and 3.30, 18 doubles for the 4.0 and 4.30 Note that we would be assuming a “mixture” of winners from the 1.2.3 and from the 4.5.6 in the forecast. And on most occasions, we would be making a correct guess. But one more simple, but very logical assumption, and we make a very efficient reduction in the number of doubles. Why not assume that one of every pair of races would be won by the horse first in the betting forecast? In other terms, by number 1. That would reduce our expenditure quite a lot, but would not reduce the number of winners by anything like the same ratio. How, then, would our betting look for the three pairs of races given in our example? 2.0 1 1* 1 4 5 6 2.30 4 5* 6 1 1 1 3.01 1 4 5 6 3.30 4 5 1 1 1 Five doubles, because the forecast gives no sixth horse for the 3.30 race. 4.01 1 1 4* 5 6 4.30 4 5 6 1* 1 1 We would then have 23 doubles, with only two winning doubles. But that need only be the initial reduction, the assumption that we are going to get a winner in every two races from the horse first in the betting forecast. There are several methods of still further cutting down the number of bets. But the point I want to make here is that a mixture of the 1.2.3 horses with the 4.5.6 horses, from the betting forecast, would be capable of getting many winning

doubles indeed. That is the basis. On that basis we can build again, and make some very good bets in a reasonable number of doubles. And it is a good basis for trebles. Such as: 111456 456111 111564 This particular combination of six trebles would have given quite a number of winning ones in taking, as it does, the horses from the first six in the betting forecast. Perhaps I can pursue this combination of the first three with the second group of three in the betting forecast a little further. It definitely serves as a sensible basis for your doubles and trebles, assuming as it does that most of the winners will be in the first six, and that most of those, but not all, will be in the first three.

Chapter Nineteen This is an actual example of betting on three races, taken from a day’s racing at Ascot. In actual fact, Ascot is not one of the best places for betting on the horses that are foremost in the betting forecast, but nevertheless it is an actual example. It would be a relatively simple matter to evolve a winning system on doubles and trebles, if most of the winners came from the first three horses in the morning paper’s forecast. It would be quite easy to reduce the number of bets, and still keep the winning one, from an analysis of the expert selections. You must surely agree that the experts behave after a pattern. One day it will be A of the Daily -- that will have a very good day. The next day it will be B of the Daily -, his chief rival! That’s how it goes. There is not much doubt about that. We cannot, unfortunately, tell which one, A or B, is in the ascendant today. But we can surely evolve a system that will win if either is. Or if neither gets among the winners, if C of the Daily – does? The only essential thing is to reduce the number of horses that should contain the winner, to a reasonable number’. One way is to take the first three in the betting forecast. We can take more, but three is a very convenient number. My example is an actual one from Ascot. I have three experts, A, B and C. And I take three races. Assuming the winners to be among the first three in the betting forecast, 1.2.3, there would be 27 possibles for the treble: 123123123123123123123123123 111222333111222333111222333 111111111222222222333333333 That will be quite clear. And we need a reduction. I find three experts, call them A, B, C. They were all giving horses in the 1.2.3 groups. Their selections:

ABC 312 323 232 What is the first step? You will agree that almost every day is a good day for some expert or other. Well, I am going to assume that one of the three experts will get two or more winners, from the three races. It does not matter which expert does that. Mind you, and I should like to deal with it later, it would be possible to allow for any expert in a group to get two or more winners from a complete day’s racing. But at the moment, two or more winners three races. And that would be a very distinct possibility if we take all possible groups of three consecutive races in the meeting: 123 234 345 456 But for this example, two or more winners for any one expert, from the chosen three races. Look at the 27 possible trebles: If A is to have two or three winners, then we keep: 3333321 3332133 3212222 for they all agree at least twice with A: 3 3 2 If B is to have two or three winners, then we keep: 2311111 2232221 3332313 for the same reasons as far as B is concerned. And if C is to have two or more winners, then we keep: 2222 2133 2231 again for the same reasons as far as C is concerned. We thus have quite a number of trebles: 333332123111112222 333213322322212133 321222233323132231 But if we have said that some expert or other is always having a good day, it is also true that one or more of them can also have a bad day. Of the three given, if we assume that any one of them will FAIL to get a winner, in the three races, then we shall dispose of: 3322111 2332332 2332322 For they all agree, in one or more races, with ALL three experts. And we are left, then, with: 1 1 1 3 3 3 3 2 2 2 2*

1 2 2 2 1 3 3 1 3 3 2* 3 1 3 3 2 1 2 2 1 2 3* The winning treble marked with an asterisk (*).In an actual fact: Chantelsey – 100-30 Light Harvest – 100-6 Borghetto – 13-8 Quite a return for eleven trebles. And we could have got it to quite a few trebles if we assumed that either two of the first in the forecast, or two of the second in the forecast, would win. As in the example, it is taken from the full possibles of 27 trebles. And you will find that if you took all possible consecutive three races, you would do it almost daily! With all new knowledge it is the author’s task to explain his theories so clearly that the reader becomes eager to put them into immediate use. I hope I have done that.

Zapping into betting with a method that links winners CONVERGING CLOUT (Part 1) There are plenty of approaches to betting on horses. Most of them are well worn. But my basic idea will travel one that is not so worn. In fact, it begets entirely new methods. Some will say that nothing matters except form. Statistics are hurled at us in profusion. Then there are certain races to avoid, horses for courses, and horses for distances. Many of these things have some basis in fact. I am all for reasonably intelligent and logical factors but we have to handle them in a reasonably intelligent and logical manner as well. My view is that it's essential to consider factors in groups. The 'law of averages' has been decried time and time again, and it's pointed out that something cannot happen because something else has! If we are going to make statements of that kind they must be within certain conditions. Something may have to happen, provided something else HAS, if certain selected conditions prevail. In certain conditions, the IF must mature! Perhaps history not repeating itself can be the result of an individual factor, of which we are not aware. But a group's history is always repeating itself. Prophecies concerning groups can be very accurate. Given similar conditions, we may assume that the behaviour of a group will tend to repeat itself. It

may concern a larger or smaller part of the group than it did last time, but the trend is almost always very apparent. And there is always a sound reason for the trend, if we look closely enough. Okay, why do favourites win about 30 per cent of races? And second favs account for 25 per cent? Favourites in wfa races may even have a higher percentage. These things are very consistent. They come about every season, pretty near the same. The figures are constant and that gives us a clue. If the figures are so constant, and if groups are so fond of repeating, cannot we stick to figures and groups? There is, I believe, salvation in a group. A particular 2yo favourite will fail but will a group of 2yo favourites all fail? I do not think so. The best newspaper expert of the lot can tip a loser and indeed often does but will a group of experts, whose selections vary all fail for a complete meeting? I don't think so. What we must do, then, is to find a method of turning those group trends to our advantage in a given race, or races. We must ally the best of one group to the best of another. We may assume that while some may fail, all will not do so. Suppose we deal first of all with numbers. We assume that it is possible for a sequence of three factors to have two possibilities each, so that the full permutation would be 2 x 2 x 2 = 8. If the two possibilities are represented by ciphers 1 and 2, then the written out permutation would be as follows: 12121212 11221122 11112222 But if we can assume, on reasonable grounds, that the correct or desired combination will have BOTH a 1 and a 2 in it, then we can safely eliminate these: 111 and 222 as both are comprised of the same numbers. We are then left with only six combinations (211, 121, 221, 112, 212,122). If we also are given any reasonable grounds to assume that there will be a pair of the ciphers 1 in the correct or desired combination, we are able to dispense with three more: 221, 212 and 122. That leaves us with 211, 121 and 112. If we could again assume that cipher 2 will not appear in the second or third factors, then we will have only one column left and that will be: 211. This is the one that fulfils all the conditions or assumptions made. Note that our first assumption did not venture to prophesy any final combination; it merely said that there should be a 1 and a 2 in the final reckoning. It did not rashly guess in what order. There will, of course, be reasons for the assumptions. And the soundness of those reasons will depend on whether our final column is the correct one or not. If they are sound, then we must be correct. We have, in fact, synchronised varying factors in a bid to bring the full possible permutation down to one single correct column. What we are after is a definite indication for a given race, or races, from a number of factors that are generally apt to provide winners. If we stick to the group technique we can benefit by the best of them, whilst evading the worst. In the simple example on numbers, three conditions were specified. If they were all obtained, we would find the correct column. If one or more of our conditions let us down we would fail. In actual betting, this has not GOT to be so. One or more of our factors can fail and we can still get a winner. We can allow for errors and failures. Successful betting, then, is:

  

Dealing in groups Synchronisingvarioussoundfactors Allowance for failure, and more allowance for failure. We can take trebles, for example. If we pick 2 horses in each leg, that means a total combination of 2 x 2 x 2 equalling 8 combinations. Let's say we bet the first 2 favourites in each leg but we only expect the TOP favourite to win two of the legs and a 2nd fav to win only 2 legs. That means we eliminate some of the combinations. In all, the linkups would be (1 equalling the fav, 2 the second fav):

1st LEG 2nd LEG 3rd LEG 1 1 1 1 1 2 1 2 1 1 2 2 2 1 1 2 1 2 2 2 1 2 2 2 We are going to bet that there will only be 2 favs in any one treble, and only two 2nd favs in any treble. So we can eliminate the 1-11 combination and the 2-2-2 combination. Instead of betting 8 combinations, we bet only 6. That means we can ADD to the amount we place on the bettable combinations to boost our profits. Let's take another example and assume that we expect a second favourite to figure only once in any treble. That means we would bet only the following combinations: 1-1-2 1-2-1 2-1-1 Now we are betting only three combinations. If we can get 2 winning favs up and a 2nd fav wins the other leg, whichever one that happens to be, then we have landed the treble for just a 3 unit outlay. This is the beauty of making the stats work for you. In next month's article I intend to take this 'converging factors' idea a stage further. I have stressed that if we make statements to the effect that something must happen because something else has, they need to be statements made in certain definite conditions. We can be as 'certain' as we like, provided the appropriate conditions obtain. Example: If there are a lot of socks in a drawer and they are of two different colours, black and white, and you are looking for a matched pair in total darkness, you need take out only three socks to know you have a pair of the same colour. You MIGHT take three of the same colour out in three attempts but you MUST take two of the same colour. Given the conditions, then: 1. 2.

A mixture of black and white socks The taking out of three socks at random, and a pair of similar socks must ensue! The fact that you take three out will guarantee a matched pair. If you took only two out, you might very well succeed in getting a pair to match but there is NO guarantee. Suppose you had socks in three different drawers and in each drawer there was 75 per cent white socks and 25 per cent black socks. If you took two socks from each drawer - six socks in all - you could get three of each, you might even get more black than white, but it is very unlikely.

You are more likely to get at least two pairs of white socks, if not more, from the six socks taken at random. You might take a black pair from one drawer but it is most unlikely that you would do so from two drawers and highly unlikely that you would take a black pair from all three drawers. That is obvious common sense. The chances of getting at least one pair of white socks from the three drawers are excellent. There are MORE white socks available. The majority of the white socks is an undeniable and existent factor. And we cannot ignore factors with such credentials. If there were three horses running in the first race and three horses running in the second, then the two races, or the double, must be won by one of these combinations: 1-1, 1-2, 1-3, 2-1, 2-2, 2-3, 3-1, 3-2 or 3-3. But we can eliminate a lot of unwanted bets by making certain rational assumptions. The assumptions would be based on normally sound factors. Suppose the favourite will win ONE of the two races, then we would not need the combinations of 2-2, 2-3, 3-2 or 3-3 and we would be left with 1-2, 1-3, 1-1, 2-1 and 3-1. It is obvious that you can save a lot of 'useless' staking cash by applying such an approach. But the approach, of course, has to be based on sound principles. The bets you cut out have to be correctly 'denounced', so to speak.

Converging factors NUMBER CRUNCH (Part 2) Last issue, I wrote about the idea of 'converging factors' in planning your betting. In this second article, I am taking the idea a stage further, stressing again the importance of dealing in groups. The basic idea behind this series utilises the capacity of different factors to indicate winners, even if they do so only now and then. But the application depends on keepingthings in groups. By 'factors', I mean anything that tends to give reasonable winner indications. They can be newspaper experts, systems arising from form, time tests, or what-have-you! They must all be capable of reasonable percentages of winners. They must then be synchronised so that we can profit from their inherent capacities for winnerfinding. Suppose for a moment there are 2 races with only 2 runners in each race. Call them 1 and 2. The winning double, then, must be one of these: 1-1, 2-1, 1-2 or 2-2. You may query what is new about that? Wait a moment. Given the first double (1-1) winning, there are three columns with at least one winner in the four doubles. One will have two winners, and the other two will have a winner each (1-1, 2-1, 1-2) and the fourth double will not have a winner (22). We see that three of the four doubles must have at least one winner. If the second double (2-1) wins, then the position is the same. There are still three doubles that will give at least one winner. They are (1-1, 2-1, 2-2) and only one (1-2) is a complete failure.

If the third double (1-2) wins, there are also three columns with at least one winner (1-1, 1-2, 2-2) and the other one (2-1) would fail. Finally, if the fourth double (2-2) won, then three doubles would give a winner at least (2-1,1-2, 2-2) and the failure would be 1-1. If the winners of the two races are in the columns given (there are four in number) then there are three chances from the four that we shall get one or two winners with the doubles specified. In other words, the odds are 3/1 that we shall find at least one winner. Whatever the number of possible combinations we employ, the odds are WITH us that we shall get a winner, or part of the column correct. If we had two horses per race for three consecutive races, the full cover would be: 1-1-1, 2-1-1, 12-1, 2-2-1, 1-1-2, 2-1-2, 1-2-2, 2-2-2. So whatever the winners of the three races, there must be seven trebles that would contain at least one winner. For example: If the winners were 1-1-1, then the following trebles (1-1-1, 2-1-1, 1-2-1, 2-2-1, 11-2, 2-1-2 and 1-2-2) all have one or more winners. The chances are 7/1 that for any winning treble among them that there will be a winner or winners in any one column. The only one to fail altogether would be 2-2-2. In other words, for any one given result to the three races, there will be only one in eight that will fail to give a single winner. There will be only one to give three winners, but there will be seven that will give one or more winners. So we must obviously find a method that can turn this factor to profit. It is so much easier to get a percentage of winners that we must do something about it! And synchronisation does just that. I have given in the example here a constant factor: We do know that there will be seven trebles with at least one winner, given the conditions arranged. Don't be worried by the fact that I have two horses a race. I am merely giving an example and, in any case, you can group horses together as you want to. You would be interested if I told you that it is possible to get the winner of a race provided a couple of experts gave a winner each in four races! Well, it can be done. Everything in betting can be pinned on a reasonable percentage of winners being obtained. That is all it really amounts to. And by reasonable I mean the usual quotas that have regularly been turned up over the years by well-known factors. It is simply a matter of using them in the right way. The normal punter follows one idea for a while, gets no winners and then deserts the idea, and goes in for another one. By this time, the old idea is starting to provide winners again! That is always happening. What we must do is to get the best from a lot of ideas at the same time. That is what synchronisation means. It is more reasonable to expect three experts to give a winner each in a day than it is to expect one expert to give three winners. But it is still possible to get the three winners if the experts just give one each. It's easier, that's the point. I will develop this idea further in next month's January 1997 issue of the magazine. I am confident the ideas will give you plenty to think about.

Converging factors POWER FIGS (Part 3) Sixty per cent of all races are won by well-backed horses. I don't think anyone will disagree with this claim. If it is, indeed, correct, and I have no reason to doubt it, there does not seem to be much point in backing outsiders. So, if we are interested in, say, the first six horses in the pre-race betting, we can start to think seriously about how to use converging factors to ensure ourselves of a winner should one of them win. There is a tendency for the winner to be in the first two in the pre-race market. This happens some 50 per cent of the time. Suppose, then, we take two races and assume the following: - The winners will be in the first six in the pre-race market (assuming more than six runners). - Split them into three groups: A=first and 2nd in the market B=3rd and 4th in the market C=5th and 6th in the market * That one of the two races will be won by a horse either 1st or 2nd in the betting market - that is, in Group A. In that case, the winners of the two races would be contained in the following: A AA B C B C A AA in effect, then, we are backing five 'sets' of doubles. A into B is actually four doubles because each group contains 2 horses. The same goes for the rest. So we have a total of 20 doubles. If wenumberthem, theywould go as follows:

    

1 1 1 3 5

and and and and and

2 2 2 4 6

with with with with with

3 5 1 1 1

and and and and and

4 6 2 2 2

You'll note that we do not have to bother with somelinkings because we have assumed that at least ONE race will be won by a horse in the A group. Thus we don't need a pairing of 3 and 4 with 5 and 6, or 5 and 6 with 3 and 4. Why waste our money on these combinations when we are saying that horses in these groups cannot win both races? They might win one but not two. We also do not need to back 3 and 4 with 3 and 4, or 5 and 6 with 5 and 6. The same rationale applies. We don't think they can win both races, but we expect them to have a chance of winning one. So, instead of 36 doubles we have cut back the bet to 20 doubles. That allows us to either (a) save money by betting $20 and not $36 on the 6x6 combination, or (b) have more on each combination. With the 'saved' $16 we could raise the stake on each double to $1.80 (assuming such a sum was possible). Instead, we could add another $4, making the bet $40, and have $2 on each double.

The point I am making with this 'converging factor' idea is that you can make certain assumptions about selections and then save yourself money if the assumptions are correct. You might be backing jockeys. You look at a couple of races and decide Shane Dye can win at least ONE of them but probably not TWO. Therefore, if you were linking his mounts in doubles, you could eliminate some. Let's say you had the following combinations: 1st Leg: DYE, COOKSLEY, MOSES 2nd Leg: DYE, PELLING, MARSHALL. Your doubles would be: DYE into PELLING-MARSHALL COOKSLEY-MOSES into DYE In the first double you expect Dye to win so you can leave him out of the second leg (remember you expect him to win only one race), and back only those horses ridden by Pelling and Marshall in the second leg. In the second double, you assume he will not win the first leg, but now you are expecting him to win the second leg, so you need to link Cooksley and Moses into Dye alone. Thus you have cropped a potential 3x3 linkup down to only four doubles. That's a big saving and if things work to plan you will have achieved a nice 'edge' on the game. Once again, the saving allows you to double your bet on each double. What I now suggest is that you look back over my previous two articles in this series. Try to understand how you make the savings on your bets through reasoned assessments. If your assessments are backed with correct assumptions, then the converging factors approach will usually work out well over a period of time. It provides for much cost-cutting on bets. What you attempt to do is to say, with some authority that certain horses, or groups of horses, will only win certain races in any given group. Thus you can provide for backing them in the right places and to save on those races where they need not be bet. It's really a matter of synchronisation. In the next issue of the magazine I will extend the matter further by talking about the selections of newspaper experts and how you can make assumptions about their performances. Once you grasp the concept of the converging factors it becomes simply a matter of careful synchronisation with your bets to take fullest advantage of the likely outcomes.

Converging factors TIPSTER TRICKS (Part 4) Let me deal, firstly, with one of the simplest possible examples of using both tipsters and the prepost betting market. This is a relatively simple technique and it will probably allow you to move on to more complex techniques in regards to 'converging factors', the subject about which I have been writing these last four months. We all have to walk before we can run and very much so in the rather tricky area I am dealing with at the moment. The example that I am about to give will act as an introduction on how to set about applying converging factors to your betting benefit.

In this example, I shall make some elementary assumptions. Let's say we are dealing with the daily double at a major meeting (though it could be any meeting). I assume that: (A) The winners of the 2 races will be in the first 3 horses in the pre-post betting market. (B) That 3 chosen experts, say Scout, Harry and Form, will each give a winner in one of the two races. The assumptions may be ambitious, or not, but I merely want to show you how we could deal with the two races, using the above assumptions. At the moment we are dealing, then, with A and B as probabilities. So let's take a theoretical race. First 3 in the pre-post betting: FIRST LEG: Lilianje, Valdivia, Norcrest. SECOND LEG: Gallant Turk, Trim Curry, Happy Wanderer. If we term them 1, 2, 3 in order of betting, we have nine possibilities for the two races: 123123123 111222333 Next, the tipsters' selections for the two races:

SCOUT Lilianje G.Turk

HARRY Valdivia Trim Curry

FORM Norcrest G.Turk

So far, so good. We have all the data. What is the procedure? Well, we have taken assumption A that the winners will come from the first 3 horses in the betting, and we have considered only those horses. We have got the selections into terms of our figures instead of names of horses and we have considered only those horses. Remember, the possibles, in those terms, are: 123123123 111222333 Now, if Scout is to give one winner, and he has 1-1, then, from the nine possible doubles, we shall not require the following combinations: 2-2, 3-2, 2-3, 3-3. The reason is clear: They do not agree with Scout at all. Now, if Harry is to have a winner, with 22, then we shall also get rid of 1-3, 11, 3-1, because they don't agree with Harry. And if Form is to give a winner, with 3-1, then out goes 1-2 because it does not agree with Form. We have only one double left: 2-1. In terms of horses, Valdivia and Gallant Turk. Valdivia was Harry's top selection in the first leg while Gallant Turk was the top pick in the second leg of both Scout and Form. Remember that this approach is all about making assumptions -and then reducing your overall bets to accommodate the assumptions coming true. Using the tipster tricks a bit further, let's select a 'main' expert and assume that he will give us at least one double in the races selected. The races chosen will be those in which our expert, and another expert, will have their selections in the first 3 in the prepost betting market.

Let's assume a theoretical meeting at Sandown, with 4 races covered: The selections of our main tipster (Scout) and the other expert (Oracle) are as follows, numbered according to betting position: SCOUT 2-2-3-1 ORACLE: 2-3-2-2 They both, for the 4 races, give selections that are in the first 3 in the betting. So we can consider taking all combinations. But we are basically concerned with the picks of the Scout. If he takes all possible doubles on his selections, there can be six (races 1-2, 1-3, 1-4, 2-3, 2-4, 3-4). But I always assume that no one expert will have it all his own way. So I will take only the races in which he disagrees with the other expert, so far as doubles are concerned. There will, then, be only these doubles: Races 2 and 3, 2 and 4 and 3 and 4. They both agree on race one. This is the essence of the approach. We only use the main tipster's selections in races where he disagrees with his fellow expert/s. Remember, too, we are backing only the selections of the main tipster, no-one else's.