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The math of the pass line at craps17 April 2016
ANSWER: I’m going to get more math-y than I usually get in this column, because you’ve picked one of the more involved calculations in craps, because the pass line is a multipart bet. You have to account for both come-out roll wins on 7 and 11 and losses on 2, 3 and 12, and post-come-out wins and losses on the point numbers.
For the first part of the bet, you need to add the probability of rolling 7 to the probability of rolling an 11. Since there are 36 possible two-dice combinations, with six totaling 7 and two totaling 11, we’re adding 6/36 to 2/36, giving us 8/36. Then we set that aside – we’ll need that total later.
For the second part of the bet, the formula is pr(4)×pr(4 before 7) + pr(5)×pr(5 before 7) + pr(6)×pr(6 before 7) + pr(8)×pr(8 before 7) + pr(9)×pr(9 before 7) + pr(10)×pr(10 before 7). The “pr” stands for “probability,” so the long form is “the probability of rolling a 4 times the probability of rolling a 4 before rolling a 7,” and so on.
There are three ways to roll a 4, so the probability of rolling a 4 is 3/36. For the probability of rolling 4 before 7, we’re confining ourselves to the three ways to roll 4 and six ways to roll 7, and ignoring all other numbers. That means the probability of rolling 4 before 7 is 3/9.
We do the same thing for all other point numbers. The probability of rolling 5 is 4/36, and the probability of rolling 5 before 7 is 4/10. The probability of rolling 6 is 5/36, and the probability of rolling 6 before 7 is 5/11.
The probabilities when the point is 8 are the same as on 6, those on 9 are the same as on 5, and those on 10 are the same as on 4.
When we plug all those numbers into the formula, we get (3/36)×(3/9) + (4/36)×(4/10) + (5/36)×(5/11) + (5/36)×(5/11) + (4/36)×(4/10) + (3/36)×(3/9). We can reduce that to (2/36) × (9/9 + 16/10 + 25/11).
In order to work with common denominators with the three fractions on the right, we multiply 9/9 by 110/110, 16/10 by 99/99 and 25/11 by 90/90. That’s the same as multiplying each by 1, so the values don’t change, but now they all have 990 as a denominator.
That takes us to (2/36) × (990/990 + 1584/990 + 2250/990), and that comes to 9648/35640. That’s the probability of winning in the second portion of the bet.
For our total probability of winning, we need to add the 8/36 from the first portion to the 9648/35640 of the second. We need a common denominator there, too, so we multiply 8/36 by 990/990 and get 7920/35640. So our chance of winning is 7920/45640 + 9648/35640, or 17568/35640, which can be reduced to 244/495.
If we win 244 times per 495 decisions then the house wins 251 times – seven more wins than the players. Divide those 7 extra wins by 495 trials, then multiply by 100 to convert to percent, and you get the house edge of 1.41% on the pass line.
You’re not going to go wrong by just reading craps house house edges off a chart. But if you want to try it for yourself, Michael Shackleford has formulas at the Wizard of Odds site, http://wizardofodds.com/games/craps/appendix/1/.
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