# Probability Question- Craps

A local casino (Gold Strike in Tunica) is offering a \$4000 bonus to a craps roller who makes passes on all six numbers (4,5,6,8,9,10) in one “roll” of the dice.

Can anyone figure out what the odds are on accomplishing this feat?

BTW, I sure would like to be on the table if and when this ever happens. I figure if this happens on a crowded table, the casino would lose substantially more than \$4,000

I’m sure I could if I knew the rules of the game. Can you summarize the necessary information?

Some of the probabilities

Establishing 4 or 10 as a point 8.33% for each number.
Establishing 5 or 9 as a point 11.11% for each number.
Establishing 6 or 8 as a point 13.88 for each number.

Making a pass after a 4 or 10 is established 2 against 1
Making a pass after a 5 or 9 is established 3 against 2
Making a pass after a 6 or 8 is established 6 against 5

You can make the passes in any order

Can I ask another question? The probability of rolling a total of 4 with 2 dice is 1/12, equivalent to the percentage you give. However, if I understood your link correctly, this is not the only way of establishing 4 as a point. It seemed to suggest that rolls of 2, 3, 7, 11 and 12 are ignored for this purpose ( for example, you could roll an 11, then a 4, and this would establish 4 as a point). If this is correct, then the probability of establishing 4 as a point is not 1/12, but 1/8. Could you clarify this?

Yes, you are correct, everybody that rolls the dice will eventually establish a point (4,5,6,8,9, or 10). A Come-out roll of 2,3,7,11, or 12) is ignored in this case

So what you are saying is the player establishes his point (let’s say 10 for example). Then the player has to roll each of 4, 5, 6, 8, 9 at least once before he rolls his 10?

and before he/she rolls a 7.

And before he rolls a 2, 3, 11, or 12?

I think that 2, 3, 11 and 12 are only relevant at the beginning. By my reading of the link, only a 7 ends the run.

notfrommensa: Thank you for your reply. Can I ask a few more questions? I think I can solve your problem once I have the answers to these. Suppose I establish 4 as a point. Can I complete another number? For example, if I roll 4, 6, 6 have I won with the number 6? My understanding is that I have not, and that I must keep rolling until I get a 4. Is this correct?

Secondly, I take it that when I have won on the number 4 (say), I then roll again to establish a new point and carry on from there?

Finally, suppose I establish 4 as a point and subsequently roll another 4, completing that number. On my next come out roll, can I establish 4 as apoint again, or do I have to roll until I get a number not yet established?

I’d be willing to bet that only the shooter would get the bonus, not every who had a bet on the table.

No, because presumably he has rolled a point. He would only lose the roll if he then made his point or rolled a 7. I would really doubt it’s a
one time proposition. Any shooter that passes on all numbers throughout the bonus period would probably win the \$4000 regardless of how many times he was shooter. The odds and payoff are certainly low enough to certainly warrant this promotion as I describe it.

I am going to assume once you establish a number and then roll it before a 7, the number is ignored for this special contest.

So each of those six numbers, in some order, must be established and then rolled before a 7.

Given the number of ways a number and 7 can be rolled, I think the chance is as follows:

.333 x .455 x .455 x .4 x .4 x .333=.003673

If my assumption you don’t have to ‘make’ a number more than once is incorrect, the chances of completing all six numbers by in one turn are much less.

Yes, but the casino stands to be forced to pay a large number of place bets that would be made on these point numbers (as well as come and pass bets), and not lost, because the 7 isn’t hit.

Jabba, Sorry for not replying sooner, my isp has been down. To your 1st question. After 4 has been established as your point, then you must roll another 4 before you can establish a another number as your point.

2nd question: You are correct

3rd question: If on a come out roll, you establish a point that has already been “won” you must make a pass on that number again. That is where the calculations get very complicated. Theoretically you can make an infinite number of passes but still lack one or more points to get the bonus.

Ok, I was at that casino last year when they were running the promotion and I’ll be there again in a month or so. When you’re rolling the dice, and you get a point, they mark it. If you get all the points before you crap out, then you win some cash. It’s not as uncommon as you would think. Didn’t happen while I was there, but I was talking to the people and they said it happens about once every couple days.

While I was playing, I managed to get 3 of the numbers before I crapped out. Most people get about that. Closest I saw was 4 though.

So here’s the deal:

When you get the dice you must keep the dice until you have won, at least once, on all the numbers 4,5,6,8,9,10. Then you win the bonus.

The probability of establishing 4 as a point is 1/8 ( i.e. (3/36)/(24/36)). Once you’ve done this, the only numbers of interest are 4 and 7. Thus, once you’ve established 4 as a point, the probability of winning on 4 is 1/3, and the probability of failing is 2/3. Thus the probability of establishing 4 as a point and then winning on 4 is 1/24

Using this method, I get the following probabilities of winning on the various numbers:
4: 1/24
5: 1/15
6: 25/264
8: 25/264
9: 1/15
10: 1/24
This gives a probability of 98/165 that you fail on any particular go.

As you say, the difficulty is now working out the probability of getting all of these before you fail. This has turned out to be trickier than I had anticipated, so I thought I’d throw out my calculations so far to see if they help anyone else, and I’ll get back to you when I’ve made more progress.

Well, I wrote a little Java app that gave me these results. For whatever they’re worth, here you go:

In 5,000,000.0 rolls:
586,412.0 shooters
916.0 would have been winners (got all the points.
2 was rolled: 138,964 times. (2.77928% )
3 was rolled: 277,672 times. (5.55344% )
4 was rolled: 416,003 times. (8.32006% )
5 was rolled: 554,353 times. (11.08706% )
6 was rolled: 694,676 times. (13.89352% )
7 was rolled: 834,423 times. (16.68846% )
8 was rolled: 696,220 times. (13.9244% )
9 was rolled: 554,489 times. (11.089780000000001% )
10 was rolled: 416,436 times. (8.32872% )
11 was rolled: 277,507 times. (5.55014% )
12 was rolled: 139,257 times. (2.7851399999999997% )

percent of shooters getting all the points: 0.15620417044671664%

So if a shooter makes a point, then on the next roll (when he hopes to establish a new point), rolls a 1-1, 1-3, or 6-6, is his turn over? I know he (and everyone else) loses that pass line bet, but does he get to shoot again or do the dice pass to the next shooter?