The math of the pass line at craps

17 April 2016

By John Grochowski

QUESTION: Can you explain how house edges are derived? Maybe you could walk through the math for the house edge on the pass line as an example of how this is done.


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/.