The dice game craps is played as follows. The player throws two dice and if the...
The dice game craps is played as follows. The player throws two dice and if the sum is seven or eleven, the player wins. If the sum is two, three, or twelve the player loses. If the sum is anything else, the player continues throwing until either that same number is thrown again (in which case the player wins) or the player throws seven (in which case the player loses). Calculate the probability the player wins within 0.01 (hint - simulate this in Excel).
Homework Answers
We start by finding the probability of winning at the first roll, that is obtaining 7 or 11: Pr(7) 1 6 6/36 = 1/6 probability of getting 7 2 5 3 4 (use fractions rather than decimals) 4 3 5 2 6 1 Pr(11) 5 6 2/36 = 1/18 probability of getting 11 6 5 So the probability of winning on first roll = 1/6 + 1/18 = 2/9 Now we find the probability of losing at the first roll (i.e. getting 2, 3, 12): Pr(2) Pr(3) Pr(12) 1 1 = 1/36 2 1 = 2/36 6 6 = 1/36 1 2 The probability of losing on first roll = 1/36 + 2/36 + 1/36 = 1/9 Then the probability of making a point will be 1 - 2/9 - 1/9 = 2/3 The individual probabilities for the numbers 4, 5, 6, 8, 9, 10 are shown below. Pr(4)(3/36 = 1/12) Pr(5)(4/36 = 1/9) Pr(6)( = 5/36) 3 1 4 1 5 1 2 2 3 2 4 2 1 3 2 3 3 3 1 4 2 4 1 5 Pr(8)( = 5/36) Pr(9)(4/36 = 1/9) Pr(10) (3/36 = 1/12) 6 2 6 3 6 4 5 3 5 4 5 5 4 4 4 5 4 6 3 5 3 6 2 6 Now ADD these probabilities. P(4u5u6u8u9u10) = 3/36 + 4/36 + 5/36 + 5/36 + 4/36 + 3/36 = 24/36 = 2/3 (as we have already shown above) This is the probability of making a point. We now consider the individual probabilities depending on which 'point' has been obtained. (E.G., probability of getting 4 as the 'point' = 1/12) I shall work through this problem in detail and the result can then be applied to the various 'point' probabilities. The probability of gaining 4 as the 'point' = 1/12 " getting 7 = 1/6 " game continuing = 1 - 1/12 - 1/6 = 3/4 So the probability of winning in this case is (1/12) + (3/4)(1/12) + (3/4)^2(1/12) + ..... to infinity = (1/12)[1 + (3/4) + (3/4)^2 + ..... to infinity] = (1/12)[1/(1 - 3/4)] = (1/12)(4) = 1/3 So the total probability of gaining the 'point' 4 and then winning = (1/12) (1/3) This will be the same probability if the initial 'point' was 10. If the 'point' was 5 or 9 the initial probability of 5 or 9 is 1/9 and the probability of continuing thereafter is 1 - 1/9 - 1/6 = 13/18 and by the same argument as above the probability of winning is (1/9)[1/(1 - 13/18)] = (1/9)(18/5) = 2/5 and the total probability is (1/9)(2/5) If the 'point' is 6 or 8 the probability of continuing is 1 - 5/36 - 1/6 = 25/36 and the probability of winning is (5/36)[1/(1 - 25/36)] = (5/36)(36/11) = 5/11 and the total probability is (5/36)(5/11) We can now find the overall total probability of winning = 2/9 + 2[(1/12)(1/3) + (1/9)(2/5) + (5/36)(5/11)] = 244/495 = 0.493
So the Answer is .493
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