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Please help my rational, I am trying to work out the likelihood of a $4$ horses passing the finish line in a particular order for every permutation of horses. I know the odds of each horse. Below I will attempt to explain what I have done so far and where I am getting stuck.

I have a list of race horses and the odds that each horse will win eg :

Horse Number | Odds of winning
------------------------------
           1 : 1 to 1
           2 : 5 to 1
           3 : 80 to 1
           4 : 100 to 1

For simplicity sake in this example the horse number corresponds to the likelihood of a win. So the first thing I do is assign a weight to each horse (smaller being more likely to win, and larger less likely)

Horse Number | Weight
------------------------------
           1 : 1
           2 : 2
           3 : 3
           4 : 4

(if, in a real world example, $2$ horses share the same odds the weight will be the same for both of them.)

Now, the most likely combination of horses passing the finish line would be $1,2,3,4$ in that order, the less likely combo would be $4,3,2,1$.

I have generated a list of all possible combinations of the horses, $1,2,3,4\,$ - $\,2,3,4,1\,$ - $\,4,1,3,2$ etc..

and I would like to order them by probability of passing the finish line in that order.

So my thinking was to use the weight assigned above and the position of the horse in the finishing order to assign another weight to the combination.

for example

Positon        [1][2][3][4]
Horse Weight   [1][2][3][4]

If I subtract the position of the horse with its weight (if the result is a negative I change it to a positive) I then get a total that is smaller if its a likely combination and larger if its a unlikely combination.

So the above example :

combo : 1,2,3,4 : total weight = 0
combo : 4,3,2,1 : total weight = 8

I don't know if I am on the right path here, I should probably be considering the neighbouring horses into the equation, eg if horse $1,2$ are first or $1,3$ are first they will have the same total weight but $1,2$ should be more likely to finish than $1,3$.

Any help with my thinking would be greatly appreciated.

Edit: I am programming this in java if it matters.

1 Answers 1

0

If we interpret the weight $w_i$ as

Horse $i$ is $\frac{w_i}{w_j}$ times more likely to reach the finish line before horse $w_j$

we can construct roulette.

For example probability, that among horses $a,b,c,d$ probablity that horses $c$ reach the first line is $$\frac{w_c}{w_a+w_b+w_c+w_d}$$ then we want horse $d$ on the finish line, with probablity $$\frac{w_d}{w_a+w_b+w_d}$$ and so on.

Probability, that the finish line will be reached in sequence $(i,j,k,l)$ is $$P(i,j,k ,l)=\frac{w_i}{w_i+w_j+w_k+w_l}\frac{w_j}{w_j+w_k+w_l}\frac{w_k}{w_k+w_l}$$

You can order your sequences using this probablity formula.