Awnik repeatedly plays a game that has a probability of winning of 31β. The outcomes are independent. What is the expected value of the number of games he will play until he has both won and lost at least once?
Answer Choices:
A. 25β
B. 3
C. 516β
D. 27β
E. 415β
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After the first game, there is a 31β chance he has won a game and a 32β chance he has lost a game. If he has won, then he plays until he loses a game, so in expectation, he will play 2/31β=23β more games. If he has lost, then he plays until he wins a game, so in expectation, he will play 1/31β=3 more games. Hence, the expected value of the number of additional games he will play is given by:
E[additional games]β=P(win 1st game)β
E[additional gamesβ£win 1st game]+P(lose 1st game)β
E[additional gamesβ£lose 1st game]=31ββ
23β+32ββ
3=25ββ
Hence, including the first gmae, Awnik will play 27β games on average, which corresponds to (D) 27ββ.
The problems on this page are the property of the MAA's American Mathematics Competitions