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Worked solution

Sum the chances of winning on successive rounds

Three people A,B,C roll a fair die for a prize. The first person to roll 6 wins. Each round proceeds A, B, C and repeats until someone wins. Find the probability that (i) A wins on the first throw, (ii) C wins on the first attempt, (iii) no one wins in a round, (iv) A wins.

CJC · Lecture · Example 11

(i) A succeeds immediately

P(A wins on first throw)=1/6P(\text{A wins on first throw})=1/6.

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(ii) C succeeds on the first opportunity

A and B must both fail before C succeeds. By independent rolls, (5/6)(5/6)(1/6)=25/216(5/6)(5/6)(1/6)=25/216.

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(iii) All three fail in one round

q=P(no winner in a round)=(5/6)3=125/216q=P(\text{no winner in a round})=(5/6)^3=125/216.

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(iv) List the first three and general winning cases

The printed “A won on the nth throw” refers to A’s own nth attempt, in round n, not the nth throw of the whole game.

CaseDescriptionProbability
1A wins on own first attempt1/61/6
2A wins on own second attempt(125/216)(1/6)(125/216)(1/6)
3A wins on own third attempt(125/216)2(1/6)(125/216)^2(1/6)
………
nA wins on own nth attempt(125/216)n−1(1/6)(125/216)^{n-1}(1/6)

Each case requires all earlier rounds to have no winner, then A to roll 6.

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(iv) Sum the infinite geometric progression

These winning cases are disjoint. Their sum is (1/6)+(125/216)(1/6)+(125/216)2(1/6)+⋯(1/6)+(125/216)(1/6)+(125/216)^2(1/6)+\cdots. The source identifies first term a=1/6, common ratio r=125/216 and infinitely many terms. Since 0≤r<1, P(A wins)=(1/6)/(1−125/216)=36/91P(\text{A wins})=(1/6)/(1-125/216)=36/91. This incorporates the Teacher’s series evaluation.

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Common mistakes

Do not count A’s later attempts unless B and C also failed in every previous round. Rolls are independent. “Nth throw” in the source case table is the player’s nth turn; the game’s corresponding throw is 3 n−2.

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