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Lets say I have a bunch of horses in a race, and I know that probabilities of them winning are:

  • $P(A) = 0.4$
  • $P(B) = 0.3$
  • $P(C) = 0.2$
  • $P(D) = 0.08$
  • $P(E) = 0.02$

QUESTION 1: How would I go about calculating the probability that the top 3 horses turn out to be arranged in this order:

  1. Horse $B$
  2. Horse $A$
  3. Horse $C$

QUESTION 2: How would I go about calculating the probability that Horses $A$, $B$, and $C$ are all in the top $3$, irrespective of order?

SIMILAR POSTS

I have tried searching for similar posts, but haven't found one that gives me a clear idea about how to answer my specific question.

Alessio K
  • 10,599
ronrest
  • 103
  • My answer to https://math.stackexchange.com/questions/625611/given-every-horses-chance-of-winning-a-race-what-is-the-probability-that-a-spe/4247021#4247021 addresses your question directly and in reasonable generality (arbitrary performance distributions). The answers below assume Luce' Axiom of choice without realizing it. – Peter Cotton Jan 01 '22 at 05:38

2 Answers2

1

The problem is that although you have given us the probability that each horse, we can't deduce the probability of a particular ranking without making further assumptions.

Here is a reasonable set of assumptions: let $S$ denote a subset $S \subset \{A,\dots,E\}$. Let $i$ denote an element of $S$. I assume that

  • The probability that $i$ comes first in a race among the elements of $S$ is $$ P = \frac{P(i)}{\sum_{j \in S} P(j)} = \frac{P(i)}{1 - \sum_{j \notin S} P(j)}. $$ That is, it is equal to the probability that $i$ comes first in a race of all horses given that the horses not in the race do not come first.

  • The probability of a ranking in which $i$ comes first is equal to the above probability multiplied by the overall probability of the ranking of $S \setminus \{i\}$. That is, the presence of the top horse does not affect the relative ranking of the remaining horses.

With these assumptions, we find that the overall probability of the order $B,A,C$ (the answer to question 1) is given by $$ P = P(B) \cdot \frac{P(A)}{1 - P(B)} \cdot \frac{P(C)}{1 - P(A) - P(B)} \approx 11.4 \%. $$

Ben Grossmann
  • 225,327
  • Thank you very much :) I appreciate that you provided the general formula for how to calculate it. – ronrest Sep 03 '20 at 22:53
1

Probability of B winning the race $= 0.3$

Now to find the probability of $A$ finishing 2nd given $B$ wins the race, you have to calculate the odd of $A$ winning against remaining horses

$= \dfrac{0.4}{0.7}$

Now to find the probability of $C$ finishing 3rd given $A$ and $B$ have finished in top 2, you have to calculate the odds of $C$ winning against remaining horses

$= \dfrac{0.2}{0.3}$

So Probability of $B A C$ being ranked $1, 2, 3 = \dfrac{0.3 \times 0.4 \times 0.2}{0.7 \times 0.3} = \dfrac{4}{35}$

This is obviously based on assumption that probabilities of winning the race can be extended.

Math Lover
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