一道英文概率题Consider the following version of the game of craps:T
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一道英文概率题
Consider the following version of the game of craps:The player rolls two dice.If the sum on the first roll is 7 or 11,the player wins the game immediately.If the sum on the first roll is 2,3 or12,the player loses the game immediately.However if the sum on the first roll is 4 5 6 8 9 or 10,then the two dice are rolled again and again until the sum is either 7 or 11 or the original value.If the original value is obtained a second time before either 7 or 11 is obtained,then the player wins.If either7 or 11 is obtained before the original value is obtained a second time,then the player loses.Determine the probability that the player will win this game.
Consider the following version of the game of craps:The player rolls two dice.If the sum on the first roll is 7 or 11,the player wins the game immediately.If the sum on the first roll is 2,3 or12,the player loses the game immediately.However if the sum on the first roll is 4 5 6 8 9 or 10,then the two dice are rolled again and again until the sum is either 7 or 11 or the original value.If the original value is obtained a second time before either 7 or 11 is obtained,then the player wins.If either7 or 11 is obtained before the original value is obtained a second time,then the player loses.Determine the probability that the player will win this game.
2 3 4 5 6 7 8 9 10 11 12
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
7 or 11 8/36=2/9 win on first roll
2 3 or 12 4/36=1/9 lose on 1st roll
else 24/36 =2/3 continue from first roll
p of obtaining original value from second roll=1/36
p of contuinuning the game from second roll and after=(2/3-1/36)=23/36
(2/9)+(2/3)(1/36)+(2/3)*(23/36)*(1/36)+(2/3)*(23/36)^2(1/36)+.
=2/9+(2/3)(1/36)(1+23/36+(23/36)^2+(23/36)^3.)
=2/9+(2/3)(1/36)/(1-23/36)
=2/9+(2/3)(1/36)(36/13)
=2/9+2/39
=26/117+6/117
=32/117
1/36 2/36 3/36 4/36 5/36 6/36 5/36 4/36 3/36 2/36 1/36
7 or 11 8/36=2/9 win on first roll
2 3 or 12 4/36=1/9 lose on 1st roll
else 24/36 =2/3 continue from first roll
p of obtaining original value from second roll=1/36
p of contuinuning the game from second roll and after=(2/3-1/36)=23/36
(2/9)+(2/3)(1/36)+(2/3)*(23/36)*(1/36)+(2/3)*(23/36)^2(1/36)+.
=2/9+(2/3)(1/36)(1+23/36+(23/36)^2+(23/36)^3.)
=2/9+(2/3)(1/36)/(1-23/36)
=2/9+(2/3)(1/36)(36/13)
=2/9+2/39
=26/117+6/117
=32/117
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