ExcelRaceRatings
New Member
- Joined
- Jan 9, 2016
- Messages
- 25
- Office Version
- 2016
- Platform
- Windows
Hi All,
I am looking for assistance for a mathematical formula to be able to turn my assessed win betting prices (win %) into place betting probabilities. The below should explain the equation a little further.
[TABLE="width: 256"]
<tbody>[TR]
[TD="width: 64"][/TD]
[TD="width: 64"]W%[/TD]
[TD="width: 64"]2nd %[/TD]
[TD="width: 64"]3rd %[/TD]
[/TR]
[TR]
[TD]Horse 1[/TD]
[TD="align: right"]35[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 2[/TD]
[TD="align: right"]15[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 3[/TD]
[TD="align: right"]13[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 4[/TD]
[TD="align: right"]10[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 5[/TD]
[TD="align: right"]9[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 6[/TD]
[TD="align: right"]7[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 7[/TD]
[TD="align: right"]5[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 8[/TD]
[TD="align: right"]3[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 9[/TD]
[TD="align: right"]2[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 10[/TD]
[TD="align: right"]1[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
</tbody>[/TABLE]
As previously mentioned, the probability of a horse finishing in a place equals the sum of the probabilities of the horse coming first, second and third. To determine these probabilities, let's suppose the horse we wish to determine the place odds for is horse X and the probability of this horse winning is P(X). Furthermore, we will assume the number of horses in the race equals N.
The probability of horse X coming first is simply the win probability, namely P(X). The probability of horse X coming second is more complicated. It equals the sum of the following probabilities:
P(Horse 1 wins AND Horse X comes second)
P(Horse 2 wins AND Horse X comes second)
P(Horse 3 wins AND Horse X comes second)
...
P(Horse N wins AND Horse X comes second)
As Dedman notes, the probability that Horse 1 wins and Horse X comes second is the same as the probability that Horse 1 wins times the probability that Horse X wins if Horse 1 is scratched. This is given by P(1)*P(X)/(1-P(1)) where P(1) is the probability that Horse 1 wins (and '*' is computer notation for multiply). Similarly for Horse 2 winning with Horse X coming second and so on.
A similar process is used to determine the probability of Horse X coming third but this time there are many more probabilities to add together as there are many more ways in which the remaining horses can fill the first two places. This will become apparent when you recognise that the probability of horse X coming third is equal to the sum of the following probabilities:
P(Horse 1 wins AND Horse 2 comes 2nd AND Horse X comes third)
P(Horse 1 wins AND Horse 3 comes 2nd AND Horse X comes third)
P(Horse 1 wins AND Horse 4 comes 2nd AND Horse X comes third)
...
P(Horse 1 wins AND Horse N comes 2nd AND Horse X comes third)
Plus the sum of:
P(Horse 2 wins AND Horse 1 comes 2nd AND Horse X comes third)
P(Horse 2 wins AND Horse 3 comes 2nd AND Horse X comes third)
P(Horse 2 wins AND Horse 4 comes 2nd AND Horse X comes third)
...
P(Horse 2 wins AND Horse N comes 2nd AND Horse X comes third)
Plus the sum of:
P(Horse 3 wins AND Horse 1 comes 2nd AND Horse X comes third)
P(Horse 3 wins AND Horse 2 comes 2nd AND Horse X comes third)
P(Horse 3 wins AND Horse 4 comes 2nd AND Horse X comes third)
...
P(Horse 3 wins AND Horse N comes 2nd AND Horse X comes third)
And so on right through to and including the sum of:
P(Horse N wins AND Horse 1 comes 2nd AND Horse X comes third)
P(Horse N wins AND Horse 2 comes 2nd AND Horse X comes third)
P(Horse N wins AND Horse 3 comes 2nd AND Horse X comes third)
...
P(Horse N wins AND Horse N-1 comes 2nd AND Horse X comes third)
Now the probability that Horse 1 wins and Horse 2 comes second and Horse X comes third first involves working out the probability that Horse X wins if both Horses 1 and 2 are scratched. This is equal to P(X)/{(1-P(1))*(1-P(2))} where P(1) is the probability that Horse 1 wins and where P(2) is the probability that Horse 2 wins. This is then multiplied by the probability that Horse 1 wins and Horse 2 comes second (the latter being determined as outlined on the previous page).
Thank you for those who offer some advice!
I am looking for assistance for a mathematical formula to be able to turn my assessed win betting prices (win %) into place betting probabilities. The below should explain the equation a little further.
[TABLE="width: 256"]
<tbody>[TR]
[TD="width: 64"][/TD]
[TD="width: 64"]W%[/TD]
[TD="width: 64"]2nd %[/TD]
[TD="width: 64"]3rd %[/TD]
[/TR]
[TR]
[TD]Horse 1[/TD]
[TD="align: right"]35[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 2[/TD]
[TD="align: right"]15[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 3[/TD]
[TD="align: right"]13[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 4[/TD]
[TD="align: right"]10[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 5[/TD]
[TD="align: right"]9[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 6[/TD]
[TD="align: right"]7[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 7[/TD]
[TD="align: right"]5[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 8[/TD]
[TD="align: right"]3[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 9[/TD]
[TD="align: right"]2[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
[TR]
[TD]Horse 10[/TD]
[TD="align: right"]1[/TD]
[TD][/TD]
[TD][/TD]
[/TR]
</tbody>[/TABLE]
As previously mentioned, the probability of a horse finishing in a place equals the sum of the probabilities of the horse coming first, second and third. To determine these probabilities, let's suppose the horse we wish to determine the place odds for is horse X and the probability of this horse winning is P(X). Furthermore, we will assume the number of horses in the race equals N.
The probability of horse X coming first is simply the win probability, namely P(X). The probability of horse X coming second is more complicated. It equals the sum of the following probabilities:
P(Horse 1 wins AND Horse X comes second)
P(Horse 2 wins AND Horse X comes second)
P(Horse 3 wins AND Horse X comes second)
...
P(Horse N wins AND Horse X comes second)
As Dedman notes, the probability that Horse 1 wins and Horse X comes second is the same as the probability that Horse 1 wins times the probability that Horse X wins if Horse 1 is scratched. This is given by P(1)*P(X)/(1-P(1)) where P(1) is the probability that Horse 1 wins (and '*' is computer notation for multiply). Similarly for Horse 2 winning with Horse X coming second and so on.
A similar process is used to determine the probability of Horse X coming third but this time there are many more probabilities to add together as there are many more ways in which the remaining horses can fill the first two places. This will become apparent when you recognise that the probability of horse X coming third is equal to the sum of the following probabilities:
P(Horse 1 wins AND Horse 2 comes 2nd AND Horse X comes third)
P(Horse 1 wins AND Horse 3 comes 2nd AND Horse X comes third)
P(Horse 1 wins AND Horse 4 comes 2nd AND Horse X comes third)
...
P(Horse 1 wins AND Horse N comes 2nd AND Horse X comes third)
Plus the sum of:
P(Horse 2 wins AND Horse 1 comes 2nd AND Horse X comes third)
P(Horse 2 wins AND Horse 3 comes 2nd AND Horse X comes third)
P(Horse 2 wins AND Horse 4 comes 2nd AND Horse X comes third)
...
P(Horse 2 wins AND Horse N comes 2nd AND Horse X comes third)
Plus the sum of:
P(Horse 3 wins AND Horse 1 comes 2nd AND Horse X comes third)
P(Horse 3 wins AND Horse 2 comes 2nd AND Horse X comes third)
P(Horse 3 wins AND Horse 4 comes 2nd AND Horse X comes third)
...
P(Horse 3 wins AND Horse N comes 2nd AND Horse X comes third)
And so on right through to and including the sum of:
P(Horse N wins AND Horse 1 comes 2nd AND Horse X comes third)
P(Horse N wins AND Horse 2 comes 2nd AND Horse X comes third)
P(Horse N wins AND Horse 3 comes 2nd AND Horse X comes third)
...
P(Horse N wins AND Horse N-1 comes 2nd AND Horse X comes third)
Now the probability that Horse 1 wins and Horse 2 comes second and Horse X comes third first involves working out the probability that Horse X wins if both Horses 1 and 2 are scratched. This is equal to P(X)/{(1-P(1))*(1-P(2))} where P(1) is the probability that Horse 1 wins and where P(2) is the probability that Horse 2 wins. This is then multiplied by the probability that Horse 1 wins and Horse 2 comes second (the latter being determined as outlined on the previous page).
Thank you for those who offer some advice!