integral with many variables

[fredhoyle]

The problem is this:

We have a number N of runners each known to cover the distance at a mean time Ti (i=1 to N) and the standard deviation around the mean time is the same, say ó.
What are the probablitities of runner i winning the race ?


Typical cases:

100 meters sprint: mean values are of the order of 10'' and ó is of the order of 0.1''

1000 meters horse race: mean values are of the order of 60'' and ó is of the order of 0.5''


This can be computed fairly easily with the use of normal pseudo-random numbers.
But I am looking for a closed solution or approximation.

[alkatran]

Let me get this straight.. you have N amount of runners.. which are represented as a group (no individual stats)...

Looks like the chances of winning are 1/N to me? ... But that's at a glance...

[sql_lall]

No, Ti (i = 1 to N) means they have different times....


Interesting question...
I guess it'd have to be:
P(#i wins) = P(#i beats #1) AND P(#i beats #2) AND P(#i beats #3) ....

Which is just:
P(#i wins) = P(#i beats #1) x P(#i beats #2) x P(#i beats #3) ....

To find P(#i beats #1)...if dist(i) and dist(1) are the distributions for i's and 1's times, let dist(Z) = dist(i) - dist(1), and find P(Z < 0)

[fredhoyle]

No that's wrong
Suppose N = 3 and suppose the pair probabilities are:

P(1 beats 2) = 0.60 , P(1 beats 3) = 0.90
P(2 beats 1) = 0.40 , P(2 beats 3) = 0.60
P(3 beats 1) = 0.10 , P(3 beats 2) = 0.40

Then, with your formula:

P(1 wins) = 0.60 x 0.90 = 0.54
P(2 wins) = 0.40 x 0.60 = 0.24
P(3 wins) = 0.10 x 0.40 = 0.04

and P(1) + P(2) + P(3) = 0.54 + 0.24 + 0.04 = 0.82 = error

It does n't work like that !

Monte Carlo approach is:

Generate first a pair of uniformly distributed random numbers in the interval (0,1) (that's what the rnd function does).
Then using those numbers (r1 and t2) compute a value x from

x = sqr(-2 ln(r1)) . cos (2 . ð . r2)

This is a normally distributed random number with zero mean and unit variance and can be adjusted to mean m and standard deviation ó by

x1 = ó . x + m

If you do this, the numerical value of the integral (*) is easy to compute. It may be however a bit of a slow process as you may require a few thousand samples.

I wonder if some kind of approximation exists.
Does it have anything to do with the "extreme value distribution" or not ?



(*) that is, P(t1 = least) , P(t2 = least) a.s.o.