# Understanding Craps House Edge | What is the math behind craps?

Deriving the player's edge for all the major bets in craps is not difficult if you understand some complex math. Outside of this page the house edge, which is just the product of the player's edge and -1. To avoid multiplying by -1 everything will be referred to in term's of the player's edge, which you can expect to be negative since the house ultimately has the edge on all bets except the free odds.

If you can, work through some of the bets yourself. Not only will this give you a deeper understanding of the odds but hopefully motivate you to refresh or improve your math skills.

Before going on you must have an understanding of the probability of throwing each total in one roll.

The Dice Combinations table. Learn it. Memorize it, know it.

**Number**

2

3

4

5

6

7

8

9

10

11

12

**Combination**

1-1

1-2, 2-1

1-3, 3-1, 2-2

1-4, 4-1, 2-3, 3-2

1-5, 5-1, 2-4, 4-2, 3-3

1-6, 6-1, 2-5, 5-2, 3-4, 4-3

2-6, 6-2, 3-5, 5-3, 4-4

3-6, 6-3, 4-5, 5-4

4-6, 6-4, 5-5

5-6, 6-5

6-6

**Ways To Roll**

One

Two

Three

Four

Five

Six

Five

Four

Three

Two

One

The general formula for the expected return of a bet is:

∑ (probability of event i) × (return of event i) over all possible outcomes.

The player's edge is the expected return divided by the initial bet. For example when betting against the line on a sporting event you have to bet $11 to win $10. Assuming a 50% chance of winning the expected return would be 0.5×(10) + 0.5×(-11) = -0.5 . The player's edge would be -0.5/11 = -1/22 ≈ -4.545%.

An exception to the house edge rule is when a tie is possible. In general ties are ignored in house edge calculations. To adjust for this, when a tie is possible, divide the expected return by the average bet resolved. The "average bet resolved" is the product of the initial wager and the probability that the bet was resolved. In craps the only bets with a tie are the don't pass and the don't come.

Many of the bets in craps win if one particular event happens before another. These bets can take several rolls or more to resolve. If a wager wins with probability p, loses with probability q, and stays active with probability 1-p-q then the probability of winning eventually is:

∑ p×(1-p-q)^{i} (for i=0 to infinity) =

p × (1/(1-(1-p-q))) = p × (1/(p+q)) = p/(p+q).

Throughout this page you will see a lot of expressions of the form p/(p+q). To save space I do not derive the expression each time since it is worked out above.

## Pass & Come

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%.

## Don't Pass & Don't Come

The probability of winning on the come out roll is pr(2)+pr(3) = 1/36 + 2/36 = 3/36.

The probability of pushing on the come out roll is pr(12) = 1/36.

The probability of establishing a point and then winning is pr(4)×pr(7 before 4) + pr(5)×pr(7 before 5) + pr(6)×pr(7 before 6) + pr(8)×pr(7 before 8) + pr(9)×pr(7 before 9) + pr(10)×pr(7 before 10) =

(3/36)×(6/9) + (4/36)×(6/10) + (5/36)×(6/11) + (5/36)×(6/11) + (4/36)×(6/10) + (3/36)×(6/9) =

(2/36) × (18/9 + 24/10 + 30/11) =

(2/36) × (1980/990 + 2376/990 + 2700/990) =

(2/36) × (7056/990) = 14112/35640

The total probability of winning is 3/36 + 14112/35640 = 17082/35640 = 2847/5940

The probability of losing is 1-(2847/5940 + 1/36) = 1-(3012/5940) = 2928/5940

The expected return is 2847/5940×(+1) + 2928/5940×(-1) = -81/5940 = -3/220 ≈ 1.364%

Most other sources on craps will claim that the house edge on the don't pass bet is 1.403%. The source of the discrepancy lies is whether or not to count ties. To be accurate always count ties as money bet. If you do not feel that counting ties as money bet needs to be in your formula then divide by figure above by the probability that the bet will be resolved in a win or loss (35/36). So 1.364%/(35/36) ≈ -1.403%. This is the house edge assuming that the player never rolls a 12 on the come out roll.

## Place Bets to Win

Place bet on 6 or 8: [(5/11)×7 + (6/11)×(-6)]/6 = (-1/11)/6 = -1/66 ≈ -1.515%

Place bet on 5 or 9: [(4/10)×7 + (6/10)×(-5)]/5 = (-2/10)/5 = -1/25 = -4.000%

Place bet on 4 or 10: [(3/9)×9 + (6/9)×(-5)]/5 = (-3/9)/5 = -1/15 ≈ -6.667%

## Place Bets to Lose

Place bet to lose on 6 or 8: [(6/11)×4 + (5/11)×(-5)]/5 = (-1/11)/5 = -1/55 ≈ -1.818%

Place bet to lose on 5 or 9: [(6/10)×5 + (4/10)×(-8)]/8 = (-2/10)/8 = -1/40 = -2.500%

Place bet to lose on 4 or 10: [(6/9)×5 + (3/9)×(-11)]/11 = (-3/9)/11 = -1/33 ≈ -3.030%

## Buy

Buy bet on 6 or 8: [(5/11)×23 + (6/11)×(-21)]/21 = (-11/11)/21 = -1/21 ≈ -4.762%

Buy bet on 5 or 9: [(4/10)×29 + (6/10)×(-21)]/21 = (-10/10)/21 = -1/21 = -4.762%

Buy bet on 4 or 10: [(3/9)×39 + (6/9)×(-21)]/21 = (-9/9)/21 = -1/21 ≈ -4.762%

## Lay

Lay bet to lose on 6 or 8: [(6/11)×19 + (5/11)×(-25)]/25 = (-11/11)/25 = -1/25 ≈ -4.000%

Lay bet to lose on 5 or 9: [(6/10)×19 + (4/10)×(-31)]/31 = (-10/10)/31 = -1/31 = -3.226%

Lay bet to lose on 4 or 10: [(6/9)×19 + (3/9)×(-41)]/41 = (-9/9)/41 = -1/41 ≈ -2.439%

## Big 6 & Big 8

[(5/11)×1 + (6/11)×(-1)]/1 = -1/11 ≈ 9.091%

## Hard 4 & Hard 10

**Note:** The hard 4 and hard 10 pay 7 to 1, or 8for 1. In craps the odds on the cloth are listed on a for 1 basis, including the graphic above.

The probability of a hard 4 on any given roll is 1/36.

The probability of a 7 on any given roll is 6/36.

The probability of a soft 4 on any given roll is 2/36 (1+3 and 3+1).

The probability of winning on any given roll is 1/36.

The probability of losing on any given roll is 6/36 + 2/36 = 8/36.

The probability of winning the bet is p/(p+q) (see above) = (1/36)/(9/36) = 1/9

The expected return is (1/9)×7 + (8/9)×(-1) = -1/9 ≈ 11.111%.

The player's edge is also -1/9 since the bet is 1 unit.

The odds are the same for a hard 10.

## Hard 6 & Hard 8

**Note:** The hard 4 and hard 10 pay 9 to 1, or 10for 1. In craps the odds on the cloth are listed on a for 1 basis, including the graphic above.

The probability of a hard 6 on any given roll is 1/36.

The probability of a 7 on any given roll is 6/36.

The probability of a soft 6 on any given roll is 4/36 (1+5, 2+3, 3+2, and 5+1).

The probability of winning on any given roll is 1/36.

The probability of losing on any given roll is 6/36 + 4/36 = 10/36.

The probability of winning the bet is p/(p+q) (see above) = (1/36)/(11/36) = 1/11

The expected return is (1/11)×9 + (10/11)×(-1) = -1/11 ≈ 9.091%.

The player's edge is also -1/11 since the bet is 1 unit.

The odds are the same for a hard 8.

## Craps 2 & Craps 12

[(1/36)×30 + (35/36)×(-1)]/1 = -5/36 ≈ -13.889%

## Craps 3 & Craps 11

[(2/36)×15 + (34/36)×(-1)]/1 = -4/36 ≈ -11.111%

## Any Craps

[(4/36)×7 + (32/36)×(-1)]/1 = -4/36 ≈ -11.111%

## Any 7

[(6/36)×4 + (30/36)×(-1)]/1 = -6/36 ≈ -16.667%

## Horn

The probability of rolling either a 2 or 12 is 1/36 + 1/36 = 2/36.

The probability of rolling either a 3 or 11 is 2/36 + 2/36 = 4/36.

The probability of roling anything else is 1-2/36-4/36 = 30/36.

Remember that the horn bet is like all four craps bets in one. Even if one wins the other three still lose. The house edge is:

[(2/36)×27 + (4/36)×12 + (30/36)×(-4)]/4 = (-18/36)/4 = 12.500%

## Field

When the 12 pays 2:1 the expected return is:

2×(pr(2)+pr(12)) + 1×(pr(3)+pr(4)+pr(5)+pr(10)+pr(11)) + -1×(pr(6)+pr(7)+pr(8)+pr(9)) =

2×(1/36 + 1/36) + 1×(2/36 + 3/36+ 4/36 + 3/36 + 2/36) + -1×(5/36 + 6/36 + 5/36+ 4/36) =

2×(2/36) + 1×(14/36) + -1×(20/36) = -2/36 = -1/18 ≈ 5.556%.

When the 12 pays 3:1 the expected return is:

3×pr(2) + 2×pr(12)) + 1×(pr(3)+pr(4)+pr(5)+pr(10)+pr(11)) + -1×(pr(6)+pr(7)+pr(8)+pr(9)) =

3×(1/36) + 2×(1/36) + 1×(2/36 + 3/36+ 4/36 + 3/36 + 2/36) + -1×(5/36 + 6/36 + 5/36+ 4/36) =

3×(1/36) + 2×(1/36) + 1×(14/36) + -1×(20/36) = -1/36 ≈ 2.778%.

## Buying Odds

4 and 10: [(3/9)×2 + (6/9)×(-1)]/1 = 0.000%

5 and 9: [(4/10)×3 + (6/10)×(-2)]/2 = 0.000%

6 and 8: [(5/11)×6 + (6/11)×(-5)]/5 = 0.000%

## Laying Odds

4 and 10: [(6/9)×1 + (3/9)×(-2)]/1 = 0.000%

5 and 9: [(6/10)×2 + (4/10)×(-3)]/2 = 0.000%

6 and 8: [(6/11)×5 + (5/11)×(-6)]/5 = 0.000%

## Combined Pass and Buying Odds

The player edge on the combined pass and buying odds is the average player gain divided by the average player bet. The gain on the pass line is always -7/495 and the gain on the odds is always 0. The expected bet depends on what multiple of odds you are allowed. Lets assume full double odds, or that the pass line bet is $2, the odds bet on a 4, 5, 9, and 10 is $4, and the odds on a 6 or 8 is $5.

The average gain is -2×(7/495) = -14/495.

The average bet is 2 + (3/36)×4 + (4/36)×4 + (5/36)×5 + (5/36)×5 + (4/36)×4 + (3/36)×4] =

2 + 106/36 = 178/36

The player edge is (-14/495)/(178/36) = -0.572%.

The general formula if you can take x times odds on the 6 and 8, y times on the 5 and 9, and z times on the 4 and 10 is (-7 / 495) / [ 1 + ((5x + 4y + 3z) / 18) ]

## Combined Don't Pass and Laying Odds

The player edge on the combined don't pass and laying odds is the average player gain divided by the average player bet. The gain on the don't pass is always -3/220 and the gain on the odds is always 0. The expected bet depends on what multiple of odds you are allowed. Lets assume double odds and a don't pass bet of $10. Then the player can lay odds of $40 for a win of $20 on the 4 and 10, $30 for a win of $20 on the 5 and 9, and $24 on the 6 and 8 for a win of $20. The average gain is -10×(3/220) = -30/220.

The average bet is 10 + 2×[(3/36)×40 + (4/36)×30 + (5/36)×24] = 30.

The player edge is (-30/220)/30 = -0.455%.

The general formula if you can buy x times odds then the house edge on the combined don't pass and laying odds is (3/220)/(1+x).