Craps mathematics hub · Reviewed August 18, 2026
Two fair six-sided dice create 36 equally likely ordered outcomes. That single fact explains why seven appears most often, why 6 and 8 are easier to make than 4 and 10, and how a casino creates an edge by paying less than the true odds.
Sam’s bottom line
You do not need advanced statistics to play craps intelligently. Memorize the number of ways to roll each total, distinguish true odds from the posted payout, and judge wagers by expected cost—not by how exciting the last shooter looked.
The 36 possible dice outcomes
Each die has six faces, so two dice produce 6 × 6 = 36 ordered combinations. “Ordered” matters: 1–5 and 5–1 both total six, but they are separate outcomes.
| Total | Ordered combinations | Ways | Probability on one roll |
|---|---|---|---|
| 2 | 1–1 | 1 | 1/36 = 2.78% |
| 3 | 1–2, 2–1 | 2 | 2/36 = 5.56% |
| 4 | 1–3, 2–2, 3–1 | 3 | 3/36 = 8.33% |
| 5 | 1–4, 2–3, 3–2, 4–1 | 4 | 4/36 = 11.11% |
| 6 | 1–5, 2–4, 3–3, 4–2, 5–1 | 5 | 5/36 = 13.89% |
| 7 | 1–6, 2–5, 3–4, 4–3, 5–2, 6–1 | 6 | 6/36 = 16.67% |
| 8 | 2–6, 3–5, 4–4, 5–3, 6–2 | 5 | 5/36 = 13.89% |
| 9 | 3–6, 4–5, 5–4, 6–3 | 4 | 4/36 = 11.11% |
| 10 | 4–6, 5–5, 6–4 | 3 | 3/36 = 8.33% |
| 11 | 5–6, 6–5 | 2 | 2/36 = 5.56% |
| 12 | 6–6 | 1 | 1/36 = 2.78% |
The frequency pattern is symmetrical: 1–2–3–4–5–6–5–4–3–2–1. A useful memory device pairs numbers equally distant from seven:
| Pair | Ways to roll either number |
|---|---|
| 7 | 6 ways |
| 6 or 8 | 5 ways each |
| 5 or 9 | 4 ways each |
| 4 or 10 | 3 ways each |
| 3 or 11 | 2 ways each |
| 2 or 12 | 1 way each |
Read the 6 × 6 outcome grid
The grid lists all 36 outcomes and prevents a common beginner mistake: treating only the total as if every total were equally likely.
| Die 1 \ Die 2 | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
Trace a diagonal through the grid: seven occupies six cells, while 2 and 12 occupy one cell each. For a fair physical-dice model, every cell has probability 1/36. Totals inherit different probabilities because they occupy different numbers of cells.
This also explains hardways. Total eight occupies five cells, but only 4–4 is a hard 8; the other four combinations are easy 8s. A Hard 8 standing wager therefore competes with four easy combinations and six combinations of seven.
Why seven controls the game
Seven has six combinations, more than any other total. Once a point is established, most line and number bets become a race between that point and seven. Rolls that are neither result do not decide the wager.
| Point | Ways to make point | Ways to roll 7 | Chance point wins the decision | True odds against point |
|---|---|---|---|---|
| 4 or 10 | 3 | 6 | 3/9 = 33.33% | 2 to 1 |
| 5 or 9 | 4 | 6 | 4/10 = 40% | 3 to 2 |
| 6 or 8 | 5 | 6 | 5/11 = 45.45% | 6 to 5 |
For example, a 10 has three winning combinations and seven has six losing combinations. Ignore the other 27 combinations because the bet remains unresolved. The result is 6-to-3, reduced to true odds of 2-to-1 against the 10.
Probability, true odds and payout odds
- Probability is the chance of an outcome: six combinations of seven divided by 36 equals 1/6.
- True odds compare losing possibilities with winning possibilities: six ways to roll seven versus three ways to roll 10 equals 2 to 1 against the 10.
- Payout odds state what a winning wager actually earns. A standard Place 10 pays 9 to 5 rather than the true 2 to 1.
- House edge converts the payout shortfall into average expected loss per initial wager.
Worked example: Place 10 versus true odds
A $5 wager paid at true 2-to-1 odds would win $10. A standard $5 Place 10 wins $9. Across nine resolving combinations, the bet wins three times and loses six times:
- Three wins × $9 = $27 won.
- Six losses × $5 = $30 lost.
- Net expected result = −$3 across nine $5 decisions.
- Total initial action = 9 × $5 = $45.
- House edge = $3 ÷ $45 = 6.67%.
This does not mean every nine decisions will lose exactly $3. It is an expectation averaged across many independent decisions.
The denominator changes with the question
Craps percentages often look contradictory because writers silently switch denominators:
- Per roll: “What is the chance of 10 now?” Use 3/36.
- Per number-versus-seven decision: “What is the chance 10 appears before seven?” Use 3/(3+6) = 1/3.
- Per completed contract wager: “What is the chance Pass wins?” Combine come-out and point branches to get 244/495.
- Per initial wager: House-edge convention normally divides expected loss by the amount placed at the start of the bet.
Before accepting a percentage, ask what event starts the clock, what ends it, and whether pushes stay in the denominator.
Why a “perfect 36-roll distribution” is only a model
The original lesson imagined 36 rolls containing each total exactly as many times as its combinations. That is a useful accounting model, but real sets of 36 rolls almost never match it. The dice have no obligation to produce one 2, two 3s, three 4s, and so on within any session.
Long-run relative frequencies tend toward the underlying probabilities, but a long run does not repair a short run in a scheduled order. A seven remains 1/6 on the next fair roll whether the previous ten rolls contained zero sevens or five.
Ten rolls without seven do not make seven due
On each fair roll, the chance of avoiding seven is 30/36, or 5/6. The chance that a run of ten specified rolls contains no seven is (5/6) to the tenth power, about 16.15%. That is uncommon enough to notice but not evidence of a tilted die or a scheduled correction.
After those ten rolls have already happened, the chance of seven on roll eleven is still 6/36, or 16.67%. The 16.15% describes the entire ten-roll run before it happens; it is not the probability of the next roll once the run is known. Keep the time window and denominator attached to every number you quote.
The Pass Line result in one calculation
The Pass Line wins immediately on 7 or 11, loses immediately on 2, 3, or 12, and otherwise wins if the established point repeats before seven. Combining those branches gives a win probability of 244/495 = 49.293% and a loss probability of 251/495 = 50.707%. At even money, the difference produces a house edge of 7/495 = 1.414%.
See the complete outcome tree and live-table procedure in the Pass Line guide.
What free Odds change—and what they do not
Once a point exists, a Pass/Come Odds wager pays the true point-versus-seven ratio. Its own expected value is zero before rounding or unusual house rules. It does not erase the expected loss already attached to the flat line bet, and it does not make the point more likely to repeat. It increases the amount at risk while lowering the house edge when the calculation uses total combined action as the denominator.
How to use this math at a real table
- Check the table’s posted payouts, commission timing, Field rules, and maximum Odds.
- Translate the bet into winning and losing combinations.
- Compare the payout with the true odds.
- Estimate expected loss as amount wagered × house edge × expected number of decisions.
- Choose a loss limit you can afford before play; do not increase it because a result feels due.
Questions this foundation lets you answer
Why do 6 and 8 usually pay less than 4 and 10?
They are more likely to arrive before seven: five winning combinations versus three. A fair payout therefore needs less profit per winning dollar.
Why can seven be both helpful and harmful?
The rules assign meaning, not the dice. Come-out seven wins Pass; seven after a point loses Pass; seven helps Don’t after a point; and Any 7 covers it for only one roll at an expensive price.
Does a bet with a 50% chance have zero edge?
Not necessarily. You must also know the payout and any push. Probability alone cannot establish expected value.
Can combining losing bets create a winning system?
Changing the outcome pattern can create more frequent small wins or partial hedges, but the expected values of the component wagers still add together.
Should I memorize all 36 cells?
No. Memorize the frequency pattern and point-versus-seven ratios. Use the full grid when auditing a side bet or checking someone else’s claim.
Continue through the math library
- Calculate house edge and expected loss
- Variance, volatility and short-session results
- Craps odds explained for beginners
- Lowest-house-edge bets ranked
- Advanced calculations and probability exercises
- Gambler’s fallacy and independent rolls
Source and calculation notes
- Massachusetts Gaming Commission: Craps and Mini-Craps rules
- All combination counts, conditional point probabilities, Pass Line fractions, and house-edge examples above were independently recomputed from the 36 ordered outcomes.
The “bag of marbles” example was perfect. And I get it that the same odds are in play on every throw, however the likelihood of missing or choosing the same marbles, while infinitely possible, is extremely improbable. The possibility of throwing sevens each toss is also infinite but highly improbable given the other numbers also exist in their relative ratios; as do the marble colors.
No argument here – just seeking another viewpoint on that logic.
….. My bad – obsolete website. Go figure – by the time anyone sees this I’ll have my answer.
Hi Jesse, thanks for your comments.
We’re unsure of what you’re asking, so if you’d like to post a clarifying comment, we’ll be happy to try again. You wrote, “…However, the likelihood of missing or choosing the same marbles, while infinitely possible, is extremely improbable. The possibility of throwing sevens each toss is also infinite but highly improbable given the other numbers also exist in their relative ratios; as do the marble colors.”
We’re unsure of the point you’re trying to make. The example of the bag of marbles was intended simply to explain to our readers who have difficulty understanding the concept that outcomes associated with previous dice roles have no effect on the probabilities associated with the outcomes of future dice roles in a legal game of craps. We agree with your logic that throwing sevens each toss over infinity is highly improbable, but we fail to understand what that has to do with the basic subject about the law of probabilities that future outcomes in a legal craps game are not influenced by previous outcomes; therefore, when you state, “…just seeking another viewpoint on that logic…,” we’re unable to provide any viewpoint without first getting clarification of the point you’re trying to make.