Probability of top $3$ Horses ranked in a certain order

Question

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.

• How to find top 3
• Calculating probabilities in horse racing!
• Given every horse's chance of winning a race, what is the probability that a specific horse will finish in nth place?

Answer 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 $$ P = P(B) \cdot \frac{P(A)}{1 - P(B)} \cdot \frac{P(C)}{1 - P(A) - P(B)} \approx 11.4 \%. $$

Answer 2

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.