The home of intelligent horse racing discussion
The home of intelligent horse racing discussion

Stats Help Rqd

Home Forums Archive Topics Stats Help Rqd

Viewing 15 posts - 1 through 15 (of 17 total)
  • Author
  • #4301
    • Total Posts 8798

    Any stats gurus out there? I need help to prod me in the direction of a method for calculating the following :-

    Operating a particular system I get x winners from y selections. For the next group of y selections (assuming all other things are equal, which I know they are not!) I want to ascertain the probabilities regarding the number of winners I am likely to see in the next group of y selections.

    So, for example, if i get 20 winners from 100 selections, what are the statistical probabilities regarding the number of winners in the next 100. I would assume 20 is the most likley answer but that there will obviously be a chance of it being higher or lower, say, for arguments sake, a 95% chance of between 17 and 23 winners etc. How do I calculate the actual numbers/% probabilities. Am I right in assuming that the answer lies in calculating and applying standard deviations?

    Anybody help or guide me to someone/where who can?


    Nick Hatton
    • Total Posts 399

    Pretty simple to do this …. you just need a factorial (!) button on your calculator. I’m not too hot on the theory but this is how it would work in practise :

    You have achieved 20 wins from 100 bets.

    The %chance of getting 20 wins from the next 100 bets would be ((100!/(20!*80!))*100)*(0.2 to the power 20)*(0.8 to the power 80)

    The %chance of getting 19 wins from the next 100 bets would be ((100!/(19!*81!))*100)*(0.2 to the power 19)*(0.8 to the power 81)

    I use this method practically everyday and unfortunately do not have the mathematical knowledge to explain it fully but I swear that it works. It was shown to me last year by a maths teacher who is a family friend.

    Good luck !


    (Edited by Nick Hatton at 9:01 pm on Mar. 18, 2003)

    william winalot
    • Total Posts 102

    Hello Nick Hatton.

    I don’t know if this is the right approach.<br>Let me know if it is.

    Try this.



    Or rounded off 17 winners in the next 100

    Divide the number of bets by the number of winners plus one.

    Then divide 100 by the above answer.


    6 bets,4 winners=6/4+1.00=2.5

    100/2.5=40 or 40% probability of winning.

    Is this the right approach.

    Hope it helps.

    Nick Hatton
    • Total Posts 399

    Hi William …..

    I’m a bit confused by your post ….

    You lost me when you divided 100 by 6 to give 16.66% This is obviously the %age chance of 5/1 shots (or decimal odds of 6).

    I don’t really understand how this is relevant to working out the future probablity of 20% chances which are 4/1 shots or decimal odds of 5 ?!?!

    I promise you that to the best of my knowledge the method in my original post works, but as I stated I haven’t really got a good theoritical knowledge of the maths involved. For this reason I don’t really want to elaborate much further as I don’t want to accidentally mislead anyone.

    Maybe someone else on the forum would like to add to this thread ……

    All the best <br>Nick

    (Edited by Nick Hatton at 12:44 am on Mar. 19, 2003)

    • Total Posts 8798

    Nick – thanks for the reply, your answer seems to make sense and seems to be exactly what I’m looking for so I’ll utilise the formula. It’s really to help try and establish staking levels for applying to a system I operate during the flat turf season as, once I establish the probabilities of number of winners in future based on past performance it allows me to establish more accurate and effective staking levels. Thanks again Nick.

    William – thanks for the post but you lost me also. I’m afraid I couldn’t follow the logic of what you were trying to illustrate and what (in the last line) the ‘40% chance of winning’ actually meant. It’s probably me so if you want to clarify it’d be great.


    Nick Hatton
    • Total Posts 399

    Nice to know that was some help cormack

    Exactly the same reason as I wanted to know the formula last year, namely to determine correct staking levels. I put it into practise this Winter for the a/w season and it definitely helped increase profits. The only word of caution (which I’m sure you’ve already considered) is that your original set of results must be a large sample otherwise the assumed %age chance of winning may not be accurate enough.

    Hope you win more from now on !!  :biggrin: <br>

    (Edited by Nick Hatton at 5:10 pm on Mar. 19, 2003)

    william winalot
    • Total Posts 102


    Maybe I don’t understand what you were looking for.

    I miss typed in my Example.Should have read.6/4 winners=6/4+1.00=2.5,sorry about that.

    You say that you wan’t to know the probabilities of your WINNERS against Selections.

    My understanding would be as follows.

    If you had 1 WINNER from 2 Selections,then the probability of getting the next selection right,would be 50%.

    If you had 20 WINNERS from 100 Selections,then the probality of getting another 20 Winners from the next set of 100 selections would be the same,namely 20%.

    But the Law of Probality must come into your calculations.


    A 6/4 chance has a 40% chance of winning,according to the odds.


    It doe’s not matter how many times there is a 6/4 chance in a race,the chance is always 40% against the rest of the field,so if you have 20 winners from 100 selections,the chance of getting 20 from the next 100 is always the same 20%,providing that you can guarantee 20 winners from 100.

    You must base your calculations around 100 percent.<br>This is how a betting book is made up.

    Am I right to assume that this is what to were looking for.

    (Edited by william winalot at 4:39 pm on Mar. 19, 2003)

    • Total Posts 8798

    William – I think we’re at cross purposes. Nick’s formula seems to be the one I was looking for and seems to make sense. I think I wasn’t very clear with my original query but I do appreciate your response which I think answered a different question than the one I intended.

    william winalot
    • Total Posts 102


    I can’t see how Nick Hattons formula works.<br>I have tryed to work it out.<br>The solution I get is as follows.

    <br>20 wins from 100 bets.


    How doe’s this  calculation tell you the chance of your next probability.

    As I said if you get 20 wins from 100 selections,then your next sequence is the same.

    Example:Each year my age increases by 1 year,it doe’s not matter how long I live,my age only increases by 1 year.

    If you pick 20 winners from 100 selections,your next 100 selections SHOULD be the same,or to put it another way,see below.

         2 winners from     10 selections<br>    20 winners from    100 selections<br>  200 winners from   1000 selections<br>2000 winners from 10000 selections<br>So on.ETC:

    It doe’s not matter how many selections there are,the percentage will always be the same.

    If I am wrong please enlighten me.<br>

    • Total Posts 8798

    William –

    As I read it Nick’s formula returns 9.93% so you may want to revisit your calculation.
    What this is saying is that if I have 20 winners from 100 selections then, all other things being equal, there will be a 9.93% chance that I will have 20 winners from the next 100 selections.
    You state that I SHOULD get 20 winners from the next 100 and you are right, in that it is the most likely outcome. However, statistically, other outcomes are possible also. This is because each individual selection stands a 1 in 5 chance of being a winner (or a 20% chance). Each event outcome is independent of another so there is in fact a possibility that there may be more or less than 20 winners in the next 100. It is most likely that there will be 20 winners but there is  also a probability that there will be 10 or 22 or 28, for example. From my limited understanding of statistics such probabilities fall on various distribution curves and the values on these curves (as with most curves) can be calculated mathematically and that is what Nick’s formula does. So, given the results from a sample of my slections I can predict the probabilities of obtaining any given number of winners during the next ‘lot’ of selections.
    The best way to rationalise this is by thinking of a toin coss. 2 possible outcomes, head or tail, but if we toss the coin 100 times we do not always get 50 of each. This is because of statistical probability. Each individual outcome is not dependent on another. It also stands to reason that the more times we toss the coin the greater the likelihood, statistically, that the number of heads and tails will gravitate towards their true probabilities, 50% each.
    The interesting thing for me is that, if I can calculate the probabilities for each outcome, I can then say, for example, that, having had 20 winners from 100 selections say, there is a 95% chance (and I haven’t calculated this so don’t hold me to ransom on the numbers – these are to illustrate my point only) that I might have between 12 and 28 winners during my next ‘lot’ of 100 selections. This does 2 things, it allows me to calculate the likely overall losses which may ensue and also calculate the statistical likelihood of the length of losing runs. For example with 20 winners the longest losing run theoretically possible during my 100 selections is 80 and the shortest is 4. The most likely losing run lies somewhere between these two numbers and I can calculate the probability for each losing run, therefore enabling me to set my betting bank to maximise profit while eliminating a lot of the risk of losing my bank.
    What I think you are saying is that, given I have 20 winners from 100 selections, then the most likely outcome is that I’ll have 20 from the next 100 and this is indeed correct. However it does not stand, as you state, that the second sequence will be the SAME as the first. It will tend to be the same but, as explained by mathematical laws of probability, it is possible that there will be variation and these can be calculated, in this case using the formula proposed by Nick.
    Phew! Clear as mud I s’pose!!

    (Edited by cormack15 at 9:21 pm on Mar. 19, 2003)

    Nick Hatton
    • Total Posts 399

    Great post cormack

    I Get 9.93% too so you’re definitely reading it exactly as  intended  :biggrin:

    William …. If you’re trying to use the formula I believe the slight mistake that you’re making is that you are not pressing the ‘factorial’ button on your calculator. This button looks like an exclamation mark.

    I can make it look just a little less complicated than the original post. Try this ……..<br>(100! / (20!*80!))*100*(0.2^20)*(0.8^80)

    It might be easiest to do the 20!*80! bit first and put the answer into the memory of the calculator and then take it from there

    Hopefully you’ll get 9.93 as the answer

    (Edited by Nick Hatton at 12:01 am on Mar. 20, 2003)

    <br>(Edited by Nick Hatton at 12:02 am on Mar. 20, 2003)<br>

    (Edited by Nick Hatton at 12:07 am on Mar. 20, 2003)

    william winalot
    • Total Posts 102

    Ok Lads.

    I agree with you both.

    But I still say that if you consistently get 20 winners from 100 selections,then the winners should’nt vary to much from this calculation.It is possible that you will get 0 winners from 100 selections,where doe’s your formula calculate for this.If your selection method holds to 20 winners over any given period,then I am sure that your percentage will always be 20%,it can not vary from this.<br>The law of percentages is always based around 100,so how do you get 9.93%,mathematics is a proven formula.Your selections are random,your selections can vary one way or the other,UP or Down.The formula that you have shown me can only work if your selections are constant,that is,they never vary from 20 out of 100.If they vary even by one,then no formula can give you a true percentage.

    If I was trying to work out a probable percentage for your selections,then I would  use the method below.

    I work with GWBASIC to do this program.It should work with any BASIC PROGRAM.

    This program will tell you how many times a given number will come up out of 100 selections or more.

    This might not  be what you are trying to achieve.

    If you do not have GWBASIC,then if you wish I could E-Mail the program to you.Here is the program I would use.

    1 DIM C(10000)<br>10 ClS:KEY OFF<br>20 RANDOMIZE TIMER<br>30 A=INT(RND*100)+1<br>35 RANDOMIZE TIMER<br>40 B=INT(RND*100)+1<br>50 IF A=B THEN C(A)=C(A)+1:D=D+1<br>60 IF D=100 THEN GOTO 80<br>70 GOTO 20<br>80 FOR E=1 TO 200<br>90 IF C(E)>=1 THEN F=C(E)+F:PRINT E;"=";C(E);"{";F;"}"<br>100 NEXT

    • Total Posts 8798

    William – Whilst I appreciate where you are coming from your logic is wayward on this occasion.

    Putting it as simply as possible. If I toss a coin 100 times and it comes up heads 50 times this gives an indication, statistically, about what is likely to happen the next 100 throws. However it does not guarantee that we will get exactly 50 heads next time. In practice (try it and see) you will often get more or less than 50 heads (or 50%). This is because with random chance events we get statistical variation. In essence the true probability of getting a head is 50% but that in itself does not guarantee we will get 50 heads each time we toss a coin 100 times. We do know, however,  that if we repeat it often enough then the statistical variation will even itself out and we will get 50%.<br>What this formula does is indicate the likelihood of getting a certain number of heads. Most likely will be 50 then next likely will be 49 or 51, then next likely 48 or 52 and so on, with the possibilities diminishing each time until we get to 0 and 100 where the possibilities are extremely low indeed, but still theoretically possible. <br>It is the same with using historical data about the number of winners to runners. It gives an indication of the future probabilities and no more. However using mathematical formula we can calculate the statistical probabilities of various outcomes.<br>Can’t make it any more clear than that William and I don’t want to keep on at you about it so I’ll let it rest after this post. Suffice to say that Nick’s formula gives me the info I was looking for.<br>Interesting stuff though!  

    dave jay
    • Total Posts 3386

    This calculation is known as ‘binomial distribution probability’ ….

    If you want to use this in Excel the function you are looking for is ‘BINOMDIST’ … to set this up input data in columns for each probability …. hope this helps ….

    william winalot
    • Total Posts 102

    Ok Lads.

    Just trying to help.

    Must have gotten mixed up somewhere.

    Try out the small program I posted,see what you think.

    Good Luck.

Viewing 15 posts - 1 through 15 (of 17 total)
  • You must be logged in to reply to this topic.