(a) Find the joint PMF of X, Y, Z.

(b) Find the probability that the game is decisive. Simplify your answer (it should not involve a sum of many terms).

(c) What is the probability that the game is decisive for n = 5? What is the limiting probability that a game is decisive as n ! 1? Explain briefly why your answer makes sense.

Solution:
(a) This is a Multinomial(n, (1/3, 1/3, 1/3)) distribution.
(b) The game is decisive if and only if exactly one of X, Y, Z is 0. These cases are disjoint so by symmetry, the probability is 3 times the probability that X is zero and Y and Z are nonzero. Note that if X = 0 and Y = k, then Z = n−k. This gives P (decisive) = ([2^n] - 2) / (3^[n-1]). As a check, when n = 2 this reduces to 2/3, which makes sense since for 2 players, the game is decisive if and only if the two players do not pick the same choice.
(c) The probability is (2^5-2)/3^4 = 30/81 = 0.37 approximately for n = 5. The limiting probability is 0 as n goes to infinity, which make sense since if the number of players is very large, it is very likely that there will be at least one of each of Rock, Paper, and Scissors.

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