Odds of Winning in Dice Game of CrapsDate: 02/20/2007 at 08:47:58 From: Ekal Subject: Independent repeated trials The dice game Craps is played as follows. The player throws two dice, and if the sum is 7 or 11, then he wins. If the sum is 2, 3 or 12, then he loses. If the sum is anything else, then he continues throwing until he either throws that number again (in which case he wins) or he throws a 7 (in which case he loses). Calculate the probability that the player wins. Isn't the process infinite? Then how to find the answer? Date: 02/20/2007 at 12:12:12 From: Doctor Anthony Subject: Re: Independent repeated trials Hi Ekal - 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 which as you would expect will be less than 1/2 so that over the long run you will lose to the bank. - Doctor Anthony, The Math Forum http://mathforum.org/dr.math/ |
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