# Runners in a race, probability paradox

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• cosmicminer
In summary, the double normal distribution has a property where if one runner has a distribution with a small σ then the other runner has a higher probability of winning. However, if there are more than two runners, the order in which they are faster does not always follow the order predicted by the integral.
cosmicminer
There are a number n of runners in a race.
We know their expected times from start to finish μ(i) and the corresponding standard deviations σ(i).
The probability of runner 0 to finish first is given by this integral:

It's from here:

https://www.untruth.org/~josh/math/normal-min.pdf

The 0 is one of the i's really but is suffixed as 0 in the above image of the formula.
I would write i instead of "0" and then in the product j ≠ i rather.

This can be computed easily using Simpson's rule and the approximation for erfc from Abramowitz-Stegun perhaps.

The strange thing is this:
If I choose n = 2 and any values for μ and σ then the following holds true:

if μ(1) < μ(2) then always P(1) > P(2) irrespective of the σ's ................. (1)

This is a property of the double normal distribution.
Thus if runner 1 has a delta function for a distribution (limiting normal with σ = 0) and runner b is close second but with big σ then the 1 has higher probability irrespective.

But if n > 2 the law (1) may or may not hold - depending on the sigmas.
So for n > 2 it is possible that one of the theoretically faster runners has lower probability than a slower runner with bigger σ.
How is this possible ?

It's not just about being faster than each rival in a pairwise comparison, but about being faster than all of them simultaneously. Thus if (because of the sigmas) there are several with a significant probability of beating the one with lowest μ, that one may not be the most likely to finish first.

Consider an extension of your example: 3 runners, A, B and C. A has a delta function, while B and C have identical (but independent) distributions with μ and σ such that the probability of B finishing after A is 55%.
The probability of A winning is the probability that both B and C finish later, i.e. 0.552 ≈ 0.3.
The probability that at least one of B and C finishes before A is 0.7.
By the symmetry of the situation, P(B) = P(C) = 0.35.
So A is more likely to beat B than B is to beat A, likewise with A and C. But B and C are each more likely to finish first of the 3 than A.

mjc123 said:
It's not just about being faster than each rival in a pairwise comparison, but about being faster than all of them simultaneously. Thus if (because of the sigmas) there are several with a significant probability of beating the one with lowest μ, that one may not be the most likely to finish first.

Consider an extension of your example: 3 runners, A, B and C. A has a delta function, while B and C have identical (but independent) distributions with μ and σ such that the probability of B finishing after A is 55%.
The probability of A winning is the probability that both B and C finish later, i.e. 0.552 ≈ 0.3.
The probability that at least one of B and C finishes before A is 0.7.
By the symmetry of the situation, P(B) = P(C) = 0.35.
So A is more likely to beat B than B is to beat A, likewise with A and C. But B and C are each more likely to finish first of the 3 than A.

I 'm not sure of what you say.
Doing this with Monte Carlo random numbers (Box-Muller) does n't seem to help.
With n = 2 and 100,000 samples it even finds errors, P(2) > P(1) and we know that for n=2, P(2) < P(1).

The proof for n = 2 exists somewhere but I don't have the proof that the order 1 > 2 is strictly valid only for n = 2. But if they integral says so then it is so unless some error is introduced by the Simpson rule approximation.

I try

μ1 = 60, σ1 = 0.001
μ2 = 60.05, σ2 = 3

Integral says ok, P1 = 0.512884, P2 = 0.487116

Monte Carlo with 100,000 Box-Muller samples finds "error":

P1 = 0.48982, P2 = 0.51018

I add a μ3 = 60.05, σ3 = 3, so it's 3-way contest.
Integral finds P1 = 0.2598426, P2 = 0.3700787, P3 = 0.3700787
Monte Carlo with 100,000 Box-Muller samples again finds:
P1 = 0.27264, P2 = 0.35933, P3 = 0.36803

So it seems Monte Carlo finds it difficult even with 100,000 samples, while the integral says that the strict ordering is for n = 2 only.
27% to 36% looks like big difference to be caused by integration errors.

However when I increase the steps of integration from 200 to 1000 I find:

for n = 2, P(1)= 0.50519, P(2) = 0.4948101
for n = 3, P(1) = 0.2559455, P(2) = 0.3720273, P(3) = 0.3720273

So 200 to 1000 affects the second decimal digit.

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