Craps Probability?

Question

Compute the house edge of the don't pass (Bar 12) line bet in craps.

I know I am to find the probability that the shooter loses and the probability that the shooter wins (which I have already found to be .493), and I can disregard if I get a 12, because that pushes. I know that the house edge is 1.36%, but I just need help calculating the probability of the shooter losing.

I have the beginning of the problem started: The probability of getting a 2 - (1/36), a 3 - (2/36), and this is where I get stuck.

Answer 1

(1) You already know the probability the pass line bet will win is 0.493.

(2) The don't pass only loses when the pass line wins, so the probability the don't pass line will lose is 0.493.

(3) The probability that the don't pass line will push because a 12 is rolled is 1/36 or 0.0278

(4) If the don't pass line does not lose or push, it wins. That makes the probabilty of the don't pass line winning

1 - (0.493 + 0.0278) = 0.479

(5) If you want a win vs. loss probability, disregarding the push, you will have to adjust the two previously calculated probabilities by eliminating the 1 roll in 36 that will push, so you will need to multiply them by

36 / 35 = 1.0286.

That makes the probability of winning

0.479 x 1.0286 = 0.493

and the probability of losing

0.493 x 1.0286 = 0.507

(6) By the way, the 1.36% house edge is only accurate if you do not disregard pushes. If you disregard pushes the house edge is 1.4% since

0.493 - 0.507 = -0.014