It was reported in the Spring of 2009 that a gambler rolled the dice at a craps table for four hours and eighteen minutes, with 154 rolls in that span. This event took place at the Borgata in Atlantic City. The previous longest roll (that we know of) happened at the California Club in Las Vegas in 1989. [Links: 1, 2 ]

At worst, a craps shooter holds the dice for only two rolls, which might take less than ten seconds. Experienced players might say ten or twenty minutes with the dice is their personal record. A twenty minute roll would almost certainly yield a tidy profit. Thus, a 4-hour roll is a special event.

I was surprised to see the dice were rolled only 154 times in those 258 minutes. That works out to 100.5 seconds per roll. We can be certain the table filled up with bettors at the first signs of a hot streak, and more bettors mean dealers taking longer to handle their various duties between rolls. However, in my experience an average 100 second delay between rolls seems so far outside the norm that I found myself wondering about the veracity of this story. Wizardofoddz.com states that gaming industry sources estimate about 102 rolls per hour is typical for a busy table, or about one roll every 35 seconds.

Another oddity is that Boyd Gaming owns both the Borgota and the California Club (the aforementioned site of the previous record). What are the odds? Further, the news stories indicated that neither the gambler nor the casino divulged how much the roller or other players won.

With three aspects of the story leaving me wondering, I felt compelled to run a numerical simulation to at least clear up the question of what might be won on such an outing by a shooter.

It is usually a trivial programming task to simulate fair (random) dice rolls. Random number generators (aka "RNGs") typically produce a decimal value between zero and one. A programmer may then set it so if the random number lies between zero and 1/6 (0.1666666...), the die has come up showing one, if between 1/6 and 2/6, the die reads two, and so forth. If two random numbers are called from the RNG, and these rules are applied to each, then two theoretical dice can be imagined to have landed on a theoretical craps table.

Random number generators may be tested for the quality of their randomness. I am no expert in the field, but I did take a graduate-level computational physics class in which the professor maintained that an ancient Fortran code, dran2.f, is to this day the best RNG available. Therefore, I wrote my simulation in Fortran and used dran2.f to "roll" the dice.

The difficult part of simulating craps is the accounting. Anyone who has played the game can understand why, but for those who are not familiar, an explanation of the game itself is needed. (Experienced players might now skip ahead a few paragraphs.)

When a new shooter takes the dice, they begin on a "come out" roll. The board is clean, everything is fresh. Bettors may bet on the shooter to either "pass" or "not pass." To pass, the shooter must do one of three things: 1) roll seven, 2) roll eleven, or 3) establish a point and then make it. To not pass, the shooter must do one of four things: 1) roll two, 2) roll three, 3) roll twelve, or 4) establish a point and then not make it.

Points are rolls of 4, 5, 6, 8, 9, and 10. If any point is rolled on a come out, then the shooter continues rolling until they either pass by rolling that same point again, or don't pass by rolling a seven. Shooters roll until they "seven-out" on a point.

Confused yet? Seven is the most important roll in craps in that it wins on come-out rolls, and it is also the dreaded point killer. Seven is also the most likely number to be rolled, statistically. There is a one in six chance of any roll coming up seven.

Now we see why a shooter might only last for two rolls: If the first roll is any point, and the second roll is a seven, then they have sevened-out and are finished. However, rolling only twice is rare. Shooters regularly make dozens of rolls.

So, let's watch a sample shooter and track their results. Our shooter will place $10 on the pass line for every come-out roll.

This shooter rolled the dice 13 times, put out 7 bets, and made a profit of one bet unit ($10). They had four wins and three losses.

Now, earlier we said bettors could bet for the shooter to either pass or not pass. If another bettor had been putting $10 on the "don't pass" line, they would have had exactly opposite results, winning three bets and losing four, to lose $10 in total.

At this point, some may be wondering where the house advantage is. If players can bet either way and attain opposite results, is the game not a 50-50 proposition? It is not. There is one asymmetry between the pass and don't-pass bets: while a come-out 12 is a loss for a pass bet, it is not conversely a win for the don't-pass bet - it is a push.

A pass bet has a 49.29 percent chance of winning, while a don't pass bet has a 47.93 percent chance of winning, and a 2.78 percent chance of pushing (the bet is not won or lost). The don't pass bet actually loses less than half the time, but since it also wins less than half the time, and wins less often than it loses, there is a house advantage.

Now consider that a craps outing can sometimes become quite tedious. A shooter might establish a point and proceed to roll dozens of times without making the point or sevening-out. There might be ten or fifteen players at the table, and so many rolls before a conclusion would probably make them rather antsy. Thus, craps has another ingenious property: every roll can be used as a new come-out roll. If players elect to add such bets while the shooter is attempting to establish a point, they are called "come bets."

Thus, new bets can be made on every roll. At this point it becomes cumbersome to describe all the possible scenarios of play, but suffice it to say, it is possible to have a bet on all six of the different points, with each one able to seven-out on a single roll of seven. On top of that, there is the "odds bet" that can be added on to any bet actively going for a point, and there are numerous side bets.

The odds bet deserves a little attention before proceeding. For any bet that has entered the point stage, an odds bet may be placed behind it. If the point is made, the odds bet is paid off with the true odds for that particular point to be made. For example, there are three dice combinations that yield a roll of four (1-3, 3-1, and 2-2). There are six dice combinations that produce a seven, and so when the point is four, a seven-out is twice as likely (6 divided by 3 equals 2, hence twice as likely) as making the point. Thus, if the point is made, the odds bet is paid 2-to-1. Most casinos allow "double odds," meaning if $10 is bet on the pass line, then the odds bets may be up to $20 (and some casinos may allow a $25 odds bet when the point is six or eight). Odds bets pay 2:1 for points of four and ten, 3:2 for five and nine, and 6:5 for six and eight.

Every last type of bet need not be understood to continue. I will merely say that a little experience at tables shows the side bets are usually not a significant part of play. Most players bet the pass line, place double odds bets, and if a win streak gets going, many make come bets on every roll during points, and add double odds bets to those when they go to points. If one wanted to maximize profit on a long roll, this would be the way to play, therefore, that is what I modeled with my simulation. Given the complexity of a basic description of the game, hopefully readers now see why I said the bet tracking is the major difficulty in programming a craps simulation.

So, given what we have learned, if a shooter is lucky enough to roll 154 times, there is an incredible variety of betting outcomes that may be seen. On one end of the spectrum, they could come out a big loser. They might roll craps (rolls of 2, 3, and 12 are called "craps") several times initially, then they could establish a point, and roll 148 more times without hitting the point, before finally sevening-out. If they made no come bets in the long span trying to hit the point, this player would have squandered an amazing opportunity, and lost money on the historic roll. This would be a highly unlikely event, but it is possible. At the other end of the spectrum, within 154 rolls one might see over a hundred passes made, and if a player were to press up their bets regularly, a large sum of money could be won.

To find out what would be typical for a very long roll, computer simulations are the most efficient route to an answer.

I wrote my program to simulate one billion shooters. Each shooter bet $10 on the pass line, with $10 come bets whenever possible, and double odds placed on every bet in point stage (for points of six and eight, the odds bet was $25). The average shooter made 8.53 rolls and lost $1.82.

The highest roll count was 144, and its profit was $2,330. Interestingly, the longest roll was not the most profitable. A shooter making 136 rolls won $2,890. The worst shooter lost $280.

The following table lists the dice values for the longest sequence, and the running profit.

The Atlantic City shooter made seven percent more rolls than our simulated long shooter, so I would crudely guess at a profit of $2,490 being likely on 154 rolls.

At this point, I would imagine experienced craps players feeling rather disappointed by the story I am telling. A four hour roll is something that happens only in their dreams, and yet here we have a simulation showing the winnings would be modest. Low-stakes video poker and slots players are regularly seen winning jackpots of similar value. We are talking about the world-record longest craps roll! How can this be?

Well, first of all, the roll in Atlantic City really should not have taken four hours. It should have taken about 90 minutes (based on the 102 rolls per hour estimate noted above). So, we should not be in awe of the temporal duration.

On the other hand, we should be in awe of the number of rolls. In my simulation of a billion shooters, none rolled 154 times. So, the Atlantic City roll was an exceedingly rare event. Further, much more money could have been made. Obviously, bettors wagering more than $10 would have won more, with results proportional to bet sizes (a $100 bettor would have made $23,300, or ten times what a $10 bettor made).

Bettors are also likely to press up their bets when they sense a roller is hot. How much could be made by pressing up bets would depend on the table betting limits, and how rapidly bets were increased. Typical tables have an upper limit around fifty times the minimum. In that case, profit could approach being fifty times greater, or $116,500 in the case of our simulated long roll. However, it would take some time to realize the table was hot, and then more time to build up enough profit to bet the table maximum, so in reality perhaps half that figure could be achieved.

What's more, there are also the numerous side bets to consider. What many players fantasize about is something like this: The roller is hot and they have built up a nice stack. When they randomly feel like it is time to throw $5 on the "Yo" bet, it hits for $75 (Yo is a bet that the next roll will be an 11, paying 15:1 on a hit). Feeling lucky, they then make a $25 Yo bet that hits two or three times in a row. This scenario, or others like it with other side bets, could profit over $1,000 in just a few rolls. I would guess nearly every craps player has fantasized something like this, and knowing that such events are possible may be part of what produces high expectations for what would happen in a 4-hour roll.

I did not make efforts to simulate luck on side bets. However, for those interested, I am providing a list of 100,000 rolls (click here) produced in my simulation. Those who wish to do their own experiments may use the list to simulate games. In general, the side bets have larger house advantages than the pass line or don't-pass line, and so side bets are statistically likely to reduce a player's profit in the long run. Nonetheless, lightning might strike with them. In my list of 100,000 rolls, the longest Yo streak is three. The longest streak of any number is eight consecutive sixes.

After imagining amazingly profitable streaks, it might be wise to remind ourselves of just how cruel craps (and gambling in general) can be. Earlier I said the worst shooter in the simulation lost $280. Here is that roll in detail:

The shooter started with a losing three, established a point of nine, then put up all five other points with come bets, along with nine more craps rolls, and finally sevened-out. Wow!

The bottom line from this research is we can estimate that on a very long roll, a craps player is likely to profit about 1.6 bet units per roll. For a low-stakes player, this amounts to a surprisingly paltry sum for such a rare event. If the bet unit is increased over the course of the roll, then naturally the ultimate profit will increase too.

Unfortunately, pressing up bets is not as easy as it sounds in craps. Imagine a player entered a game with a $100 bankroll to make $10 bets. If the roller immediately entered a point and rolled a lot of other points, our bettor would not even have enough to make all the odds bets needed to cover all the points established. Several opportunities for profit could be missed. Further, even if that early frustration was avoided, and a bankroll increased to $500, the player still could not make bets anywhere near the maximum allowed. To cover all six points with double odds, and to have a few bets in reserve, a player's bet unit cannot exceed about five percent of their bankroll. So, even after gaining $400, our bettor would have to stick to $25 bets. Ten thousand dollars would be needed to sustain $500 betting on everything, and getting to $10,000 from a start of $100 is exceedingly rare (as shown by this essay).

Thus, pressing up bets would not be a recipe for getting rich quick. However, an interesting aspect of craps is that the possible variety of events is infinite, while for other casino games this is usually not the case. For example, there are 2,598,960 distinct five-card poker hands. This is a large number, but it still limits the variety of games that can be seen to a finite number. On the other hand, in just seven rolls of the dice, there are about two billion different possible sequences. For more rolls, the numbers become astronomical quickly, and since a shooter could theoretically roll forever without sevening-out, the possibilities are truly endless. Nothing can be ruled out entirely. Someone might some day "break the bank" playing craps, but this simulation shows that to be a highly unlikely event.

Nonetheless, on average I think poker is a more interesting game, and people have better chances of life-changing winnings through poker. Several amateur poker players have now won seats in the World Series of Poker playing low-stakes satellites, and then ended up winning millions in the World Series. Meanwhile, a one-in-a-billion craps event will leave you a far cry from millionaire status.

All Rights Reserved
Posted June 14, 2009
Jon Dokter
Tucson, Arizona

Notes:

I later ran a simulation of ten billion shooters. Two rolls were longer than 154, so we can roughly estimate that one in every five billion shooters will equal or better the Atlantic City roll in duration. A shooter lasting 137 rolls made a profit of $3,425 without pressing bets.

After this writing, Time ran a story on the Atlantic City roll (link here). The article, by Claire Suddath, starts, "It sounds like a homework problem out of a high school math book: What is the probability of rolling a pair of dice 154 times continuously at a craps table, without throwing a seven?" We can be nearly certain Patricia Demauro (the shooter) threw some sevens - as winners between points. Suddath reports a probability that is correct as stated (1 in 1.56 trillion, which is calculated by raising 5/6 to the 154th power), but that does not apply. The question should be, "What is the probability of rolling dice 154 times without sevening out in craps?" That is an entirely different question, and one with no straight-forward answer from a combinatorial equation. There is not a regularly compounding chance of sevening out on each roll because not every roll is eligible to seven-out, and further complicating the matter, whether any particular roll in a series will be eligible to seven-out is effectively random - it cannot be predicted. Simulations are needed to make an estimate.

Suddath refers to Dr. Thomas Cover of Stanford, implying that he was her source on statistics. I have exchanged emails with Dr. Cover, and he would like it to be known that Ms. Suddath only asked him the chances of rolling 154 times without throwing a seven.

home page