Not too long ago, on a cruise ship in the Caribbean, I was convinced that Craps is *by far* the most exciting game in the casino. Maybe it’s the dice constantly flying through the air. Maybe it’s the loud shouts from players and dealers: “Dice out!”, “Six, winner!”, “Odds on my ten!”. Or maybe it’s the camaraderie formed by winning and losing with (almost) everyone at the table. Whatever the case, I had to learn how to play.

After an hour or so of getting advice from my dad and uncle, and watching my money fly on and off the table, I was able to figure the game out for the most part. But after the cruise, I sought out some more information from the plethora of online resources about craps.

Unfortunately, roughly a year later, I’m finding that much of this conventional wisdom just doesn’t add up. And I’d like to change that. Let’s start with the most foundational bet in Craps: the Pass Line bet.

If you’ve never played craps before, I encourage you to learn the basics with a little Google searching. For those in need of a refresher, the pass line bet takes place in two phases and always pays even money. **Phase 1:** (Point is OFF) bet wins if 7 or 11 is rolled, bet loses if 2, 3, or 12 is rolled. Any other roll moves the bet to the next phase by setting a point to be the number rolled. **Phase 2:** (Point is ON) bet wins if the point number is rolled again, and loses if 7 is rolled.

Given the two-phase nature of the pass line bet, calculating the “house edge” can be a little tricky. Furthermore, it shouldn’t be completely clear what I mean by “house edge” anyways. For now, let’s assume that the house edge is the percent of a bet the house expects to win once that bet is made. It turns out that the pass line bet has a house edge of 1.41%. I.e. on a $10 pass line bet, you will *on average* lose about 14 cents. Of course, you either win $10 or lose $10, which is one reason why thinking in terms of house edge is quite difficult.

The other reason that house edge is tough to think about is that most bets in craps are on the table for multiple dice rolls. A pass line bet could be in play for 1 roll, or 100 rolls (although the latter is unlikely). Should this change how we think about the “house edge”? I think so.

To illustrate my point, let’s consider a more realistic scenario. You walk up to a craps table with $500. There are 4 other players at the table with you making a few bets. You place a pass line bet for $5 down, wait for a win/loss on it, and continue placing $5 pass line bets. What are your winnings after an hour of play? I couldn’t find this answer anywhere, so I made a craps simulator in Python. The results of 100,000 simulations (assuming 144 rolls/hour) are shown below:

So, where in this simulation does that 1.41% house edge show up? Nowhere that I can find. And that’s the problem with measuring craps bets with the way we have defined house edge (and the way it’s defined on just about every craps website you will find). In my search, I only saw one website thinking about this problem, and that was Wizard of Odds. There, they define a house edge in terms of bet made, bet resolved, and per roll, which helps to break down the problem and make it more realistic for a craps session. In theory, if I know there’s a house edge of 0.42% per roll (per Wizard of Odds), and I can estimate about 144 rolls per hour at a table of 5 players (also per Wizard of Odds), then I can expect $5 * 144 * 0.0042 = $3.02 in losses per hour, right?

Well, yes and no. In the above simulation, the actual average loss per hour was $2.32. This would correspond to a house edge of 0.32% per roll. So where’s the discrepancy? One small source of discrepancy is the fact that “only” 100,000 simulations were run. Believe it or not, we really need more like 1,000,000 simulations (that’s 144,000,000 dice rolls) for results up to that accuracy. The other, more significant, source of discrepancy is that in my simulations, most of these sessions end with a bet left on the table. It’s uncommon that on the 144th roll, the pass line bet wins or loses. Further, the pass line bet is more likely to win during the first phase, but more likely to be on the table during the second phase, so it’s no surprise my simulation **understates** the expected losses.

To confirm my suspicions, I ran 1,000,000 simulations of sessions that end at 144 rolls or when all bets are resolved, whichever comes later. After dividing the average loss by the number of rolls and the bet amount, the 0.42% house edge per roll shows up perfectly.

To summarize these results, defining the house edge in craps is hard. I’d say that 1.41% is not the wrong answer, but it is a bad one. For me, the average loss per hour is much better, and much more useful when I’m playing. However, this needs to account for bets still on the table. Further, since most players have the pass line bet as the starting point of their betting strategy, it helps to think of this bet as your “cost to play”, and at $2.32 an hour on a $5 table, that’s not too bad. (Side note, you can easily convert this to whatever table minimum you play at, a $10 table would cost $4.63 an hour). For completeness, let me mention a few more useful metrics for craps betting. I plan to explore more complicated strategies in terms of these metrics.

Average Winnings ($/hour) | Typical Winnings ($/hour) | SD of Winnings ($/hour) | Low 25% of Winnings ($/hour) | High 75% of Winnings ($/hour) |
---|---|---|---|---|

-2.32 | -5.00 | 32.65 | -25.00 | 20.00 |

Typical winnings refer to the median result, or what happened to the 50th percentile person when sorting the winnings. SD is the standard deviation of winnings, which is important for understanding the **risk** or **variance** of a bet. Risk is a double edge sword. Without it, you’re almost guaranteed to lose. With it, you have a chance at coming out ahead, but your potential losses could be magnified.

Low 25% and High 75% refer to the 25th and 75th quantiles of the winnings. A nice interpretation is that you have roughly a 25% chance of leaving after an hour with less than $-25, a 25% chance of leaving with more than $20, and a 50% chance of being somewhere in the middle. For the average gambler, understanding this range for a strategy is (in my opinion) more useful than knowing the house edge for a bet.

All of these metrics should scale up with your bet too. So at a $10 table, simply multiply these numbers by 2, at a $15 table, multiply by 3, etc. That’s just one more perk of thinking in terms of $/hour.

The house edge is hard to understand in craps, and even harder to define. No matter how you split it, it’s hard to figure out exactly how much you expect to lose without running simulations. Personally, I think losses measured in $/hour are most useful. After all, who counts the number of rolls at the table? But, unless you plan on playing craps for 100,000 rolls, the house edge isn’t going to matter much to you anyways. I’d encourage any serious gambler to think in terms of a range of outcomes, especially low 25% and high 75%. Knowing that half the time I’ll be somewhere between $-25 and $20 after an hour, I’m more than willing to make a pass line bet and partake in the excitement at the craps table.

Want to know how good your craps betting system is? Stay tuned for more posts, as I plan to detail popular strategies in terms of these metrics and compare them based on how the dice shake out. If you can’t wait for those posts, reach out on twitter @Sean__Kent or via email spkent@wisc.edu. I’d love to hear what you want to see next and any feedback you have on this post.

Next week, I plan to discuss all of the misconceptions of the **odds bet** in craps. Everyone seems to have very strong opinions on these, and we’ll see how they shake out in the simulations.

**House edge per bet made**

To find the **house edge per bet made**, we need to consider \(\text{Win/Loss %}*P(\text{outcome})\) for all possible outcomes. Note: \(P(\text{outcome})\) denotes the probability of some outcome occuring. Let \(W\) be the event of a win (which pays 100% of the bet) and \(L\) be the event of a loss (which pays -100% of the bet, i.e. a loss of 100%). Since we have two phases, this turns out to be

\[ \begin{aligned} &1*P(W \text{ in Phase I}) + (-1)*P(L \text{ in Phase I})\\ &+ 1*P(W \text{ in Phase II}\mid \text{No W/L in Phase I})*P(\text{No W/L in Phase I}) \\ &+ (-1)*P(L \text{ in Phase II}\mid \text{No W/L in Phase I})*P(\text{No W/L in Phase I}) \end{aligned} \]

Being a little more specific in the possible outcomes for Phase II, we get \[ \begin{aligned} & P(7,11) - P(2,3,12)\\ &+ P(W \text{ in Phase II}\mid \text{Point is 4 or 10})*P(\text{Point is 4 or 10}) \\ &- P(L \text{ in Phase II}\mid \text{Point is 4 or 10})*P(\text{Point is 4 or 10}) \\ &+ P(W \text{ in Phase II}\mid \text{Point is 5 or 9})*P(\text{Point is 5 or 9}) \\ &- P(L \text{ in Phase II}\mid \text{Point is 5 or 9})*P(\text{Point is 5 or 9}) \\ &+ P(W \text{ in Phase II}\mid \text{Point is 6 or 8})*P(\text{Point is 6 or 8}) \\ &- P(L \text{ in Phase II}\mid \text{Point is 6 or 8})*P(\text{Point is 6 or 8}) \end{aligned} \]

We note that a Win or Loss in Phase II depends on what the point is. In general, \(P(\text{Point is 4 or 10}) = \frac{3+3}{36} = \frac{6}{36}\), \(P(\text{Point is 5 or 9}) = \frac{4+4}{36} = \frac{8}{36}\), \(P(\text{Point is 6 or 8}) = \frac{5+5}{36} = \frac{10}{36}\). Further, we can figure out the Win probability in Phase II by considering outcomes that win over outcomes that either win or lose. That is, \(P(W \text{ in Phase II}\mid \text{Point is 4 or 10}) = \frac{P(4)}{P(4)+P(7)} = \frac{P(10)}{P(10)+P(7)} = \frac{3/36}{3/36 + 6/36} = 1/3\). Similarly, \(P(W \text{ in Phase II}\mid \text{Point is 5 or 9}) = \frac{4/36}{4/36 + 6/36} = 2/5\) and \(P(W \text{ in Phase II}\mid \text{Point is 6 or 8}) = \frac{5/36}{5/36 + 6/36} = 5/11\).

All together, the house edge is \[ \begin{aligned} &\frac{6+2}{36} - \frac{1+2+1}{36} \\ &+ \frac{1}{3}*\frac{6}{36} - \frac{2}{3}*\frac{6}{36} \\ &+ \frac{2}{5}*\frac{8}{36} - \frac{3}{5}*\frac{8}{36} \\ &+ \frac{5}{11}*\frac{10}{36} - \frac{6}{11}*\frac{10}{36} \\ =& -0.014141 \end{aligned} \]

And there you have it, a house edge of 1.41%. There’s a reason this math is not trivial–casinos do not want you to think about these things and know their edge. *Even more reason to just run simulations…*

**Average Passline Rolls**

To calculate house edge per roll, we first need to find out **how many rolls a pass line bet is on the table** for (on average). Fair warning, this includes a lot of math and statistics.

The first step is to break the problem down into two phases, as we have done before. Once a pass line bet is placed, that bet wins and loses in Phase I if a 2, 3, 7, 11, or 12 is rolled, which happens \(\frac{1+2+6+2+1}{36} = \frac{12}{36}\) of the time. In that case, we observe 1 roll. Otherwise, a point is set.

Let’s say that the point is a 6. Now, in Phase II, we will either win, lose, or keep rolling. The chance we win or lose is \(P(6)+P(7) = \frac{5+6}{36} = \frac{11}{36}\). Otherwise we keep rolling, adding to the count. In the real world, this phenomenon is quite common. There’s some probability that we end our streak, and a complementary probability that the streak keeps going. For example, think about the *number of lottery tickets you buy until you win*, the *number of apples you purchase until you find one with a worm in it*, or the *number of emails you have to sift through to find something that’s not spam*. To a statistician, these are all instances of the same probabilistic phenomena, called a Geometric Random Variable.

The good news is that this is a well studied phenomena, and statistians even have an expression for the average number of such events until we observe the end of the streak. They found the average to be \(\frac{1}{p}\), where \(p\) is the probability the streak ends. In our case, the streak ends with a probability of \(\frac{11}{36}\), so the expected rolls after the point is 6 is \(\frac{36}{11} \approx 3.273\). All we have to do now is consider the other point numbers and the probabilities that they happen.

Point Number | Probability of Point | Probability of Win/Loss | Expected Number of Rolls |
---|---|---|---|

4 | ^{3}/_{36} |
^{9}/_{36} |
4.000 |

5 | ^{4}/_{36} |
^{10}/_{36} |
3.600 |

6 | ^{5}/_{36} |
^{11}/_{36} |
3.273 |

8 | ^{5}/_{36} |
^{11}/_{36} |
3.273 |

9 | ^{4}/_{36} |
^{10}/_{36} |
3.600 |

10 | ^{3}/_{36} |
^{9}/_{36} |
4.000 |

Last step: put it all together. We multiply the probability of an outcome by the expected number of rolls for that outcome. This involves taking the sum over all values in the above table, where we multiply the second column by the last column + 1. Adding 1 comes from the fact that it took 1 roll to establish the point. In a formula, it looks like this:

\[ \begin{aligned} \frac{12}{36}*1 + \sum_{\text{Point}} P(\text{Point})*(\text{Expected Rolls} + 1) \end{aligned} \]

A little bit of algebra gives us a final answer of 3.3758. Thus, the house edge per roll should be 1.41% / 3.3758 = 0.4189%, as we saw previously.