Craps House Edge: How Each Bet’s Edge Is Calculated

The craps house edge is the average share of each wager the casino keeps, and it differs sharply from bet to bet. The Pass Line costs 7/495, or about 1.41%; the Don’t Pass costs about 1.36%; the Field costs about 5.56%; and Any 7 costs about 16.67%. Each figure comes from counting the 36 ways two dice can land and comparing the result with what the bet pays.

This guide shows that counting step by step. It derives the Pass Line edge from scratch, works through the Don’t Pass, Field, Any 7 and Place 6 bets, explains why free odds lower the percentage cost without lowering the dollar cost, and ends with a comparison table. All examples assume standard payouts, which vary between casinos, and no bet in the game has a positive expected value.

What House Edge Means in Craps

House edge is expected loss divided by the amount wagered. A 1.41% edge means that, on average, the casino keeps about $1.41 of every $100 wagered on that bet over many bets. It is an average, not a prediction for any one roll, and it does not describe how often a bet wins. The same idea drives the roulette house edge calculation, applied here to dice.

Every craps edge starts with the same table: the 36 equally likely combinations of two dice. Counting the ways to make each total gives each number’s probability, a counting skill covered in probability in gambling.

The 7 is the most common total, with 6 ways, which is why many bets either rely on it or race against it. Standard rules are assumed throughout: the Pass Line pays 1 to 1, a 12 is barred (a push) on the Don’t Pass, the Field pays 2 to 1 on 2 and 12, Any 7 pays 4 to 1 and a Place 6 pays 7 to 6.

How to Calculate the Pass Line House Edge

The Pass Line wins in two stages. On the first roll (the come-out), a 7 or 11 wins, a 2, 3 or 12 loses, and any other total becomes the point. After a point is set, the bet wins if the point repeats before a 7. For a point with w ways to roll it, the chance that it repeats before a 7 is w ÷ (w + 6), because only those two totals end the bet.

where:

P(7 or 11) = 8/36 from the table above

P(point) = ways to roll that point ÷ 36 (points are 4, 5, 6, 8, 9, 10)

w = ways to roll the point; w ÷ (w + 6) = chance it repeats before a 7

Step 1: Come-out win = 8/36 = 440/1,980.

Step 2: Point 4 or 10: (3/36) × (3/9) = 1/36 each, so 2/36 = 110/1,980 for both.

Step 3: Point 5 or 9: (4/36) × (4/10) = 2/45 each, so 4/45 = 176/1,980 for both.

Step 4: Point 6 or 8: (5/36) × (5/11) = 25/396 each, so 25/198 = 250/1,980 for both.

Step 5: P(win) = (440 + 110 + 176 + 250) ÷ 1,980 = 976/1,980 = 244/495 ≈ 0.4929.

Step 6: EV per $1 = P(win) − P(lose) = 2 × (244/495) − 1 = −7/495 ≈ −0.0141.

The Pass Line edge is 7/495, about 1.41%. The bet wins slightly less than half the time (49.29%) and pays even money, so the average result is slightly negative. This is the same logic as the formula in the expected value formula guide, with the probability found by counting dice outcomes.

Worked Examples: Don’t Pass, Field, Any 7 and Place 6

Example 1: Don’t Pass (12 barred). Suppose you bet $10. The bet loses when the Pass Line wins (244/495), pushes on a come-out 12 (1/36), and wins otherwise.

Step 1: P(lose) = 976/1,980; P(push) = 1/36 = 55/1,980.

Step 2: P(win) = 1 − 976/1,980 − 55/1,980 = 949/1,980.

Step 3: EV per $1 = (949 − 976) ÷ 1,980 = −27/1,980 ≈ −0.01364.

Step 4: House edge ≈ 1.36%, so about $0.14 on a $10 bet.

Example 2: the Field bet. The Field wins on 2, 3, 4, 9, 10, 11 and 12, which is 16 of 36 combinations, and loses on 5, 6, 7 and 8 (20 of 36). Suppose the 2 and 12 pay 2 to 1 and the rest pay 1 to 1.

Step 1: EV per $1 = (1/36)(2) + (1/36)(2) + (14/36)(1) − (20/36)(1).

Step 2: = (2 + 2 + 14 − 20) ÷ 36 = −2/36 ≈ −0.0556.

Step 3: House edge ≈ 5.56%. Some casinos pay 3 to 1 on the 12, which changes this figure.

Example 3: Any 7 and Place 6. For Any 7, the chance is 6/36 = 1/6, the payout is 4 to 1, and EV per $1 = (1/6)(4) − (5/6) = −1/6, an edge of 16.67%. For a Place 6 with the 7-to-6 payout, the chance of a 6 before a 7 is 5/11, so EV per $1 = (5/11)(7/6) − (6/11) = −1/66, an edge of about 1.52%.

Free Odds: Lower Percentage, Same Dollar Cost

The free odds bet pays exactly true odds, so its edge is 0%. For a point of 4 or 10, the chance of repeating before a 7 is 3/9 = 1/3, and the odds pay 2 to 1, so EV = (1/3)(2) − (2/3) = 0. For 5 or 9 it pays 3 to 2 and for 6 or 8 it pays 6 to 5, both with EV of exactly 0. The bet can only be placed after a Pass Line bet is made and a point is set, and casinos limit its size.

Adding odds therefore does not remove the Pass Line’s cost. The dollar expected loss on each $1 line bet stays at 7/495, about 1.41 cents. What falls is the percentage of total money wagered, because the odds add action at zero edge. A point is set 2/3 of the time, so with odds equal to m times the line bet, the expected total wagered is 1 + (2/3) × m, and the combined edge is (7/495) ÷ (1 + 2m/3).

where:

7 ÷ 495 = Pass Line edge, about 1.41%

m = odds taken as a multiple of the line bet, on every point

2 ÷ 3 = probability that a point is established

The table shows the percentage falling while the dollar loss per line bet stays fixed. An hour of play still costs the same expected amount per line bet; it simply spreads over a larger total stake. To convert any edge into an hourly cost, use the method in the expected loss per hour guide.

Craps House Edge Chart by Bet

Each row of the chart follows the same pattern: count the dice outcomes, find the win probability, and compare the payout. The Place 4 and 10 edge comes from (1/3)(9/5) − (2/3) = −1/15, and the Place 5 and 9 edge from (2/5)(7/5) − (3/5) = −1/25. Assumptions are the standard payouts listed earlier.

A lower edge does not mean a bet wins often, and a high win probability does not mean a low edge. Any 7 wins 1 time in 6 but costs the most in the table, as explained in true odds and payout odds.

Common Mistakes With Craps House Edge

Treating the edge as the chance of losing is the first error: a 1.41% edge does not mean 1.41% of bets lose. Believing free odds are “free money” is the second; they carry 0% edge but require a Pass Line bet that carries 1.41%.

Ignoring the barred 12 is the third: without that push, the Don’t Pass would win 1,004/1,980 of the time and lose 976/1,980, a player edge of 28/1,980 ≈ 1.41%, so the rule is what keeps the casino ahead. Mixing up per-bet and per-roll edges is the fourth, since a Pass Line bet can take several rolls to resolve. The fifth is assuming a pattern of bets beats the game. Combining bets blends their edges but never creates a positive one. Gambling should be treated as paid entertainment, and every figure here is a long-run average.

Quick Reference and Check Your Answer

Use this checklist to compute any craps edge: list the dice outcomes that win, lose and push, find each probability, multiply by the payouts, and sum the results to get EV per $1.

If a total of probabilities does not reach 100%, a case is missing, usually a push or a come-out outcome that was skipped.

Related Concepts and Next Steps

To understand the payout side of every row, read true odds and payout odds. For turning an edge into a cost per hour, see expected loss per hour, and for the general method, the expected value formula guide. The roulette house edge calculation applies the same approach to a wheel, and probability in gambling covers the counting skills behind the dice table.