I have a heavily simplified horse race problem and I'm curious if anyone knows how to solve it:
I have a set of n horses and a dice with n*1000 faces. Each face is assigned randomly to a horse and for some number of throws, I throw the dice. Whichever horse is assigned to the face that lands up takes a single step, always of the same size, forward. All of the other horses stand still.
A horse wins the race when two conditions are met: 1: it is the furthest forward 2: the difference between its position and the position of the next furthest forward horse crosses a certain threshold.
Example: If the threshold is 10 and at time 2000 horse A is at position 1000 and horse B is at position 997, neither horse has won. But, if at time 2000, horse A is at position 11 and horse B is at position 0, horse A wins.
I want the probability that a horse wins at each time.
I have solved this problem for only two horses. I need it for 3 or more.
In response to comments: The point of having 1000n faces is that a horse can have any number of faces assigned to them. Sometimes, when I’ve posed this question, people make the mistake of thinking that we’re not dealing with independent variables. We are, but they are not identically distributed.
So, at every time t, each horse j has a probability p_j of taking a single step forward. Only one horse steps forward at a time.
The solution for two is non-trivial, but essentially we set the difference between position A and position B to be a markov chain that moves along the number line one step at a time. We then set absorbing states for at the threshold and stop the race at time of absorption.
If we take the same approach, we can set up (n-1)^2 Markov chains that represent the difference between the positions of each ordered pair of horses. The trouble is that we can’t use a simple absorbing state set.
Thanks all