1.41% formula
February 9th, 2010 at 4:47:09 PM
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What is the formula for calculating the passline edge of 1.41% in craps?
February 9th, 2010 at 5:44:08 PM
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This is from the Wizard's Craps Appendix 1
Quote:The probability of winning on the come out roll is pr(7)+pr(11) = 6/36 + 2/36 = 8/36.
The probability of establishing a point and then winning 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) =
(3/36)×(3/9) + (4/36)×(4/10) + (5/36)×(5/11) + (5/36)×(5/11) + (4/36)×(4/10) + (3/36)×(3/9) =
(2/36) × (9/9 + 16/10 + 25/11) =
(2/36) × (990/990 + 1584/990 + 2250/990) =
(2/36) × (4824/990) = 9648/35640
The overall probability of winning is 8/36 + 9648/35640 = 17568/35640 = 244/495
The probability of losing is obviously 1-(244/495) = 251/495
The player's edge is thus (244/495)×(+1) + (251/495)×(-1) = -7/495 =~ -1.414%.
February 9th, 2010 at 10:36:42 PM
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I don't know if it helps if I retell this with more words.
(A) There are 6 ways to roll a seven and 2 ways roll an eleven on come out roll. So you will your come out bet 8 times out of 36 or (6+2)/36.
(B) There are 3 ways to roll a four or 3/36. At this point the only rolls that you care about are a seven or four or (6+3) rolls. On three of them you will complete your point so 3/(6+3) is the probability that you will complete the point. The odds for landing on a 4 and then completing the point require you to multiply the two probabilities to get 3/36 * 3/(6+3)
(C) To use a similar argument for five you get 4/36 * 4/(6+4)
(D) To use a similar argument for six you get 5/36 * 5/(6+5)
(E) You must double the point solutions to include eight, nine and ten,
The odds of winning
= (A) + ( (B) + (C) + (D) ) * (E)
= +(6+2)/36 + ( 3/36*3/(6+3) +4/36*4/(6+4) + 5/36*5/(6+5) ) * 2 -> you can put directly in Excel
= (+8 + 18/9 + 32/10 + 50/11 )/36
= (+10 + 16/5 + 50/11 )/36
= 244 /495
The odds of losing are 100% - odds of winning
1.0 - (244 /295) = 251 /495
The house edge is odds of winning - odds of losing
= 251/495 - 244/495
= 7 /495 which is slightly more than 7/500 = 1.4%
If you no longer want to do fractions in your head there is a fraction formatting code in Excel which is good for numerator and denominator up to three digits.
(A) There are 6 ways to roll a seven and 2 ways roll an eleven on come out roll. So you will your come out bet 8 times out of 36 or (6+2)/36.
(B) There are 3 ways to roll a four or 3/36. At this point the only rolls that you care about are a seven or four or (6+3) rolls. On three of them you will complete your point so 3/(6+3) is the probability that you will complete the point. The odds for landing on a 4 and then completing the point require you to multiply the two probabilities to get 3/36 * 3/(6+3)
(C) To use a similar argument for five you get 4/36 * 4/(6+4)
(D) To use a similar argument for six you get 5/36 * 5/(6+5)
(E) You must double the point solutions to include eight, nine and ten,
The odds of winning
= (A) + ( (B) + (C) + (D) ) * (E)
= +(6+2)/36 + ( 3/36*3/(6+3) +4/36*4/(6+4) + 5/36*5/(6+5) ) * 2 -> you can put directly in Excel
= (+8 + 18/9 + 32/10 + 50/11 )/36
= (+10 + 16/5 + 50/11 )/36
= 244 /495
The odds of losing are 100% - odds of winning
1.0 - (244 /295) = 251 /495
The house edge is odds of winning - odds of losing
= 251/495 - 244/495
= 7 /495 which is slightly more than 7/500 = 1.4%
If you no longer want to do fractions in your head there is a fraction formatting code in Excel which is good for numerator and denominator up to three digits.
Last edited by: pacomartin on Feb 10, 2010
