Alice, Bob and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice, Carol always follows Bob, and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is 1/6, independent of the outcome of any other toss.)
Alice, Bob and Carol repeatedly take turns tossing a die. Alice begins; Bob always follows Alice, Carol always follows Bob, and Alice always follows Carol. Find the probability that Carol will be the first one to toss a six. (The probability of obtaining a six on any toss is 1/6, independent of the outcome of any other toss.)
The correct option is D 2591
If Carol wins
in the first round, then she must have rolled a six after two non-sixes have
occurred. This happens with the probability
(56)(56)(16)=(16)(56)2
If Carol wins in
the second round, then five non-sixes preceded her six. This happens with
probability
(56)(56)(56)(56)(56)(16)=(16)(56)5
This pattern
continues. Carol wins in the third round with probability (16)(56)8
and in the fourth with probability (16)(56)11.
It is possible
(though unlikely) that the game could continue forever. The probability that
Carol wins the game is equal to the sum of the probabilities that she wins in
any given round:
=(16)(56)2+(16)(56)5+(16)(56)8+....=(16)(56)2[1+(56)3+(56)6+.....]
=(16)(56)2∞∑n=0(56)3n=(16)(56)2∞∑n=0(5363)n
=(16)(56)2⎡⎢
⎢
⎢
⎢⎣11−5363⎤⎥
⎥
⎥
⎥⎦
=2591
If Carol wins in the first round, then she must have rolled a six after two non-sixes have occurred. This happens with the probability
(56)(56)(16)=(16)(56)2
If Carol wins in the second round, then five non-sixes preceded her six. This happens with probability
(56)(56)(56)(56)(56)(16)=(16)(56)5
This pattern continues. Carol wins in the third round with probability (16)(56)8 and in the fourth with probability (16)(56)11.
It is possible (though unlikely) that the game could continue forever. The probability that Carol wins the game is equal to the sum of the probabilities that she wins in any given round:
=(16)(56)2+(16)(56)5+(16)(56)8+....=(16)(56)2[1+(56)3+(56)6+.....]
=(16)(56)2∞∑n=0(56)3n=(16)(56)2∞∑n=0(5363)n
=(16)(56)2⎡⎢ ⎢ ⎢ ⎢⎣11−5363⎤⎥ ⎥ ⎥ ⎥⎦
=2591
