Question

In: Statistics and Probability

Peter and Paul bet $1 on each game. Both players start with $10. Peter is extremely...

Peter and Paul bet $1 on each game. Both players start with $10. Peter is extremely nervous on the first game, and the probability of him winning that game is only 1/4. After that, his probability of winning any given game is back to 5/8. Under these circumstances, what is the probability that Peter bankrupts Paul?

Expert Solution

Let Ai be the event that Peter wins the ith game

and Bi be the even that Paul wins the ith game, i = 1, 2, ......

Since the bet is $1 for each game and both players start with $10, so in order to Peter bankrupt Paul, Peter needs to win 10 more games than Paul.

Since, only either of Peter or Paul can win the ith game therefore, Ai and Bi are mutually exclusive and exhaustive.

i.e. P(Ai) + P(Bi) = 1

We have, P(A1) = 1/4

Therefore, P(B1) = 1 - 1/4 = 3/4

And P(Ai) = 5/8

Therefore P(Bi) = 1 - 5/8 = 3/8, for all i = 2, 3, .....

Since the probability of Peter wins the first game is different from that of the following games, there can be two mutually exclusive cases:

Case I - Peter wins the first game

The probability that Peter bankrupts Paul

= (Probability that Peter wins first 10 games) + (probability that Peter wins 11 games and Paul wins 1 game) + (probability that Peter wins 12 games and Paul wins 2 game) + ... + (probability that Peter wins 19 games and Paul wins 9 games) [Paul cannot win more than 9 games because if he wins 10 games, then Peter will bankrupt and the game will end]

= (1/4)(5/8)9 + (1/4)(5/8)10(3/8) + (1/4)(5/8)11(3/8)2 + ... + (1/4)(5/8)18(3/8)9

= (1/4)(5/8)9[1 + (5/8)(3/8) + (5/8)2(3/8)2 + ... + (5/8)9(3/8)9]

= (1/4)(5/8)9[1 - (5/8)10(3/8)10]/[1 - (5/8)(3/8)]

= 0.00475

Case II - Peter loses the first game

The probability that Peter bankrupts Paul

= (Probability that Peter wins next 11 games) + (probability that Peter wins next 12 games and Paul wins 1 game) + (probability that Peter wins next 13 games and Paul wins 2 games) + ... + (probability that Peter wins 19 games and Paul wins 8 games) [Paul cannot win more than 9 games because if he wins 10 games, then Peter will bankrupt and the game will end]

= (3/4)(5/8)11 + (3/4)(5/8)12(3/8) + (3/4)(5/8)13(3/8)2 + ... + (3/4)(5/8)19(3/8)8

= (3/4)(5/8)11[1 + (5/8)(3/8) + (5/8)2(3/8)2 + ... + (5/8)8(3/8)8]

= (3/4)(5/8)11[1 - (5/8)9(3/8)9]/[1 - (5/8)(3/8)]

= 0.00557

Therefore, the probability that Peter bankrupts Paul = 0.00475 + 0.00557 = 0.01032

Answer: The probability that Peter bankrupts Paul is 0.01032.

orchestra answered 2 years ago

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