Probability in Craps GameDate: 6/27/96 at 22:44:45 From: Anonymous Subject: Probability in Craps Game Dr. Math, In the game of craps, the shooter rolls two dice and wins if the sum of the dice is 7 or 11 ("natural"); he loses if the sum is 2,3, or 12 (craps). If the sum is 4,5,6,8,9, or 10, then the result is not yet decided. He must roll the dice again and again, as often as is necessary until the initial sum, be it 4,5,6,8,9, or 10 is repeated (shooter wins) or until the sum is 7 (shooter loses). Find the probability that the shooter will win in a game of craps. Richard Date: 6/28/96 at 18:14:38 From: Doctor Anthony Subject: Re: Probability in Craps Game A good way to start is to draw up a square grid with margins 1 to 6 vertically and horizontally, and the sum of the marginal values filling up the grid. There are 36 grid points, and the probability of a sum of 5 (say) is to count the number of 5's in the grid and divide this number (=4) by 36, to give a probability of 4/36 = 1/9 We have to enumerate the ways that the shooter could win. P(7)+P(11) = 1/6 + 1/18 = 2/9 Now consider throwing a total of 4, and needing to repeat this to win. Probability of a 4 is 1/12. Now we have to continue until another 4 occurs. This gives rise to an infinite sequence with probability at each throw of 1/12 to win, 1/6 to lose (by throwing a 7) and 3/4 to have a further go. We could win after 1, 2, 3, ....... infinity number of throws. The probability sequence looks like this 1/12 + (3/4)(1/12) + (3/4)^2(1/12) + (3/4)^3(1/12) + .... (1/12){1 + (3/4) + (3/4)^2 + ...} = (1/12)(1/(1-3/4) = (1/12)(1/(1/4)) = 4/12 = 1/3 It's not really necessary to sum the series. Since rerolls don't affect whether you win or lose, you can ignore the 3/4 probability and say that your probability of winning with 4's is P(win) / P(win + lose) = (1/12) / (1/6 + 1/12) which of course gives the same answer of 1/3 as the series sum. So complete probability of winning with 4's = (1/12)*(1/3) = 1/36 Since probability of 10 is also 1/12, this too gives an overall probability of 1/36 We now consider probability of winning with two 5's or two 9's. These have probability 1/9 of occurring and by a repeat of the argument given above, the total probability of winning with 5's is (1/9)*(2/5) = 2/45. The same applies for winning with two 9's. Finally we consider winning with two 6's or two 8's. The probability of a 6 is 5/36, and a repeat of above argument with infinite series, gives a probability of (5/36)*(5/11) = 25/396 and the same applies for two 8's. We now add up all the probabilities of winning = (2/9) + 2(1/36) + 2(2/45) + 2(25/396) = (2/9) + (1/18) + (4/45) + (25/198) = 244/495 (=0.492929...) -Doctor Anthony, The Math Forum Check out our web site! http://mathforum.org/dr.math/ |
Search the Dr. Math Library: |
[Privacy Policy] [Terms of Use]
Ask Dr. Math^{TM}
© 1994- The Math Forum at NCTM. All rights reserved.
http://mathforum.org/dr.math/