Craps: Working Out the House Edge

Question. In craps, a "pass line" bet wins if the first roll (the come-out roll) is 7 or 11, loses if it's 2, 3, or 12, and otherwise the rolled number becomes "the point" — the shooter keeps rolling until either the point comes up again (bet wins) or a 7 comes up first (bet loses). What's the house edge on this bet? This is a genuinely multi-stage conditional-probability problem, and it's a favorite because it forces you to combine several probabilities correctly rather than just compute one.

Stage 1: the come-out roll

With two dice, sums have these counts out of 36: 7 has 6 ways, 11 has 2 ways, so $P(\text{win on come-out}) = 8/36$. Sums 2, 3, 12 have $1+2+1=4$ ways, so $P(\text{lose on come-out}) = 4/36$. The remaining $24/36$ of rolls establish a point of 4, 5, 6, 8, 9, or 10.

Stage 2: given a point is established, what's the win probability?

If the point is $p$, the shooter keeps rolling until $p$ or 7 appears, ignoring all other numbers (they're just re-rolled). Given that only $p$ or 7 matters, the win probability is

$$P(\text{win} \mid \text{point}=p) = \frac{\text{ways to roll } p}{\text{ways to roll } p + \text{ways to roll } 7} .$$

For point 4 (3 ways) vs. 7 (6 ways): $P = 3/9 = 1/3$. Point 5 (4 ways): $P = 4/10 = 2/5$. Point 6 (5 ways): $P = 5/11$. By symmetry, points 8, 9, 10 mirror 6, 5, 4 respectively.

Worked example: assembling the total win probability

Total win probability sums the come-out win, plus for each point value, the probability of establishing that point times the conditional win probability given that point:

$$P(\text{win}) = \frac{8}{36} + \sum_{p} P(\text{point}=p) \cdot P(\text{win}\mid \text{point}=p) .$$

Plugging in each point (4 and 10 each have 3/36 chance to be established, with win prob 1/3; 5 and 9 each 4/36 with win prob 2/5; 6 and 8 each 5/36 with win prob 5/11):

$$P(\text{win}) = \frac{8}{36} + 2\left(\frac{3}{36}\cdot\frac{1}{3}\right) + 2\left(\frac{4}{36}\cdot\frac{2}{5}\right) + 2\left(\frac{5}{36}\cdot\frac{5}{11}\right) .$$

Computing each term: $\frac{8}{36} \approx 0.2222$; $2 \times \frac{3}{36}\times\frac{1}{3} = \frac{2}{36} \approx 0.0556$; $2 \times \frac{4}{36}\times\frac{2}{5} = \frac{3.2}{36} \approx 0.0889$; $2 \times \frac{5}{36}\times\frac{5}{11} = \frac{4.545}{36} \approx 0.1263$. Summing: $P(\text{win}) \approx 0.2222+0.0556+0.0889+0.1263 \approx 0.4929$, or about $49.29\%$.

Since this is an even-money bet, the house edge is

$$\text{edge} = 1 - 2 \times P(\text{win}) \approx 1 - 2(0.4929) = 0.0141 ,$$

about 1.41% — matching the well-known real-world figure for the craps pass line, one of the best bets on a casino floor precisely because the multi-stage structure keeps the edge small compared to, say, roulette's 5.26%.