First review the probability of throwing any given number on any given throw:

- 2: 1/36
- 3: 2/36
- 4: 3/36
- 5: 4/36
- 6: 5/36
- 7: 6/36
- 8: 5/36
- 9: 4/36
- 10: 3/36
- 11: 2/36
- 12: 1/36

From the above the probalility of rolling 7 is 6/36 = 1/6 and the probability of rolling an 11 is 2/36 = 1/18. These are the numbers that win on the first throw.

Next lets assume the point thrown on the first roll is a 4, what is the probability of throwing it again before a 7?

Let pr(x) stand for the probability of event x happening on any given roll. The answer is:

pr(4) +

pr(anything other than 4 and 7) * pr(4) +

pr(anything other than 4 and 7)^{2} * pr(4) +

pr(anything other than 4 and 7)^{3} * pr(4) +

pr(anything other than 4 and 7)^{4} * pr(4) +

+ ...

Pr(4) = 3/36 = 1/12, pr(anything other than 4 and 7) = 1-3/36-6/36 = 27/36 = 3/4.

pr(rolling a 4 before a 7)

= 1/12 + (3/4 * 1/12) + ((3/4)^{2} * 1/12) + ((3/4)^{3} * 1/12) + ...

= 1/12 * sum for i = 0 to infinity of (3/4)^{i}

= 1/12 * (1/(1-3/4))

= 1/12 * 4
= 1/3.

The probability of rolling a 10 is the same as the probability of rolling a 4 so pr(rolling a 10 before a 7) also equals 1/3.

Next assume the point thrown is a 5.

Pr(5) = 4/36 = 1/9, pr(anything other than 5 and 7) = 1-4/36-6/36 = 26/36 = 13/18.

pr(rolling a 5 before a 7)

= 1/9 + (13/18 * 1/9) + ((13/18)^{2} * 1/9) + ((13/18)^{3} * 1/9) + ...

= 1/9 * sum for i = 0 to infinity of (13/18)^{i}

= 1/9 * (1/(1-13/18))

= 1/9 * 18/5
= 2/5.

The probability of rolling a 9 is the same as the probability of rolling a 5 so pr(rolling a 9 before a 7) also equals 2/5.

Next assume the point thrown is a 6.

Pr(6) = 5/36, pr(anything other than 6 and 7) = 1-5/36-6/36 = 25/36.

pr(rolling a 6 before a 7)

= 5/36 + (25/36 * 5/36) + ((25/36)^{2} * 5/36) + ((25/36)^{3} * 5/36) + ...

= 5/36 * sum for i = 0 to infinity of (25/36)^{i}

= 5/36 * (1/(1-25/36))

= 5/36 * 36/11
= 5/11.

The probability of rolling an 8 is the same as the probability of rolling a 6 so pr(rolling a 8 before a 7) also equals 5/11.

Now we're ready to add all this together...

The probability of winning the pass line bet is:

pr(7) + pr(11) +
pr(4)*pr(4 before a 7) +
pr(5)*pr(5 before a 7) +
pr(6)*pr(6 before a 7) +
pr(8)*pr(8 before a 7) +
pr(9)*pr(9 before a 7) +
pr(10)*pr(10 before a 7)

= 6/36 + 1/18 + (1/12 * 1/3) + (1/9 * 2/5) + (5/36 * 5/11) +
(5/36 * 5/11) + (1/9 * 2/5) + (1/12 * 1/3)

= 6/36 + 1/18 + 1/36 + 2/45 + 25/396 + 25/396 + 2/45 + 1/36

= 330/1980 + 110/1980 + 55/1980 + 88/1980 + 125/1980 + 125/1980 + 88/1980 + 55/1980

= 976/1980 = 244/495.

The expected return on this wager is the product of the probability of winning and the ratio of what you keep (including your original wager) to the original bet if you do win (in this case 2). Thus the expected return is 244/495 * 2 = 488/495 =~ 98.59%.

The house advantage is what the casino gets to keep, on average, which is 1 minus the expected return, which equals 1-(488/495) = 7/495 =~ 1.41%

In the appendix to this problem I show the house advantage on the don't pass bet.

Michael Shackleford, A.S.A., 9/15/1998