Suppose the integers 1-5 are randomly assigned to five people (call them A, B, C

Question

Suppose the integers 1-5 are randomly assigned to five people (call them A, B, C, D, and E). Then they play a game in which A and B first compare their numbers and the person with the higher number wins the round, then that person proceeds to compare his or her number with C’s number. The game proceeds in this manner until the fourth comparison (involving E and the winner of the previous comparison) takes place. Let the random variable W represent the number of comparisons for which A is the winner. Determine the probability distribution of W.

Explanation / Answer

As per the problem, any one of the numbers 1-5 can be assigned to A, with equal probability.

When A is assigned the number 1, A cannot win in any of the 4 comparisons. Therefore, probability of success, p = 0
Similarly, when A is assigned the number 2, A can win only in one of the 4 comparisons, i.e. against the one assigned the number 1. Therefore, probability of success, p = 0.25

By the same logic,
We have the following probabilities of success for the various assignments to A:
Assignment = 1: p = 0
Assignment = 2: p = 0.25
Assignment = 3: p = 0.50
Assignment = 4: p = 0.75
Assignment = 5: p = 1

If W represents the number of comparisons for which A is the winner, for a given assignment, the probability distribution of W follows a binomial distribution with n = 4 (number of experiments) and p = probability of success for that particular assignment.

Since, the assignments to A are equally likely, the resulting probability distribution of W is an equally weighted mean of 5 binomial distributions,

P (W = k)
= 0.2 * nCk * pk * (1-p)4-k,where the distributions are summed over different values of p.

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