What are the odds rolling every number once before rolling a 7. Thanks for any help I'm way too dumb to figure this out my own. I was curious bc I was lucky enough to do this last night while playing the craps feature bet at the Wynn. They offered 176:1 on it so I know its far greater then that just wondering how big of a dog I was.

5 was the last number I needed to hit the prop so I hedged a little and layed the 5 for $300, but ended up making it a few rolls later. Sick brag I know.

189.1970864:1

Well, you came to the right place for this question. The last time someone asked something similar, I ended up finding a fairly simple solution to a generalized coupon collector problem. In that case, the person wanted to know the average number of rolls needed to roll every number. This required computing the probabilities of certain numbers occurring before other numbers. An R program was written to compute these exact probabilities using the inclusion-exclusion principle because they became too complicated to do by hand. The most complicated probability is the one that you asked - the probability that a 7 doesn't occur before any other number. It involves 1023 terms. Fortunately, it is the last term that the program computes, so all I had to do to get your answer was to have the program output this final probability.

Here is a link to the earlier problem solution. For your question I simply modified this to output the final value of (1-p) which is the probability that a 7 does not occur before any other number. I'm assuming that you mean that every other number has to occur AT LEAST once, not exactly once which would obviously be highly unlikely. Here is a link to the Wizard of Odds' explanation of my solution which you might find helpful in addition to my own explanation. It's the last problem on that page.

Wow great answer, thx for taking the time to lay it all out for me.

So in theory the casino doesn't have a huge edge on this prop, and in certain scenarios you can hedge and lock up a profit.

6.94% which is worse than US Roulette, and worse than most bets on the table including many bets that are generally considered sucker bets.

Except that is not really the best way to look at it. The better way is to calculate your chances of turning one dollar into 175. Using that criteria this prop is less of a sucker bet than most table bets.

Would you say that betting a number in American roulette is less of a sucker bet than betting the pass line in craps without odds because you are more likely to turn $1 into $36 in roulette than on the pass line? Would a hard hop prop bet that pays 33:1 be less of a sucker bet than the pass line bet because it turns $1 into $34 with higher probability?

If the criterion is the probability of turning $1 into $177 (because the prop bet pays 176:1), then this prop bet would also be better than the pass line bet.

I wrote the R program below to compute the probability of reaching a target bankroll from a starting bankroll given the payout odds and winning probability. These can be changed via the 4 numbers at the top. They are currently set for American roulette. It computes the probability of reaching the goal if we bet the whole bankroll until we have enough to reach it in 1 (possibly smaller) bet, then bet enough to reach the goal until we either reach it or can no longer reach it in 1 bet, and then repeat the whole process until we reach the goal, have less than $1, or have repeated this process 1000 times (so it doesn't run forever with negligible increase in probability). The program assumes that we can only bet in integer multiples of $1.

BruceZ gave an ingenious answer to the question you meant to ask, what are the chances of rolling each number AT LEAST once before you get a 7. The question you asked, the chances of rolling each number once before you get a 7 is much easier (and the feat is much harder).

Say you have a random number generator that gives the numbers 1 to 36 with equal probability. We map the random number 1 to the dice roll 2, the random numbers 2 and 3 to the dice roll 3, the random numbers 4, 5 and 6 to the roll 4, and so on. This simulates a dice roll.

If you roll the simulated dice 10 times, there are 36^10 = 3,656,158,440,062,980 possible outcomes. There are 10! = 3,628,800 orders to arrange the 10 numbers from 2 to 12 excluding 7. There is 1 way to get a 2 or 12, 2 ways to get a 3 or 11 and so on. That means there are 5!^2 = 14,400 ways to map each of the 10 dice numbers other than seven to the 36 random numbers. 36^10 / (10!*5!^2) = 69,968 so the odds are 69,967 to 1.

Looking at it another way, only one time in 370 when someone wins the Wynn bet, will she do it without repeating a number.

The question of how to evaluate "sucker" bets is not one mathematics can answer. BruceZ, like most mathematicians, tends to divide the expected loss by the maximum possible loss to rank bets. That's one way to do it, but I consider it arbitrary. I think it makes more sense to start by asking why someone is making a bet. If the goal is to make money in the long run then all negative expected value bets are equally bad, because all will fail at the goal.

If the goal is to have fun, then the appropriate denominator is the amount of fun the bet provides. That's related to the standard deviation of the bet, as well as how long it takes to play (too quick means you don't get much for your expected loss, too slow gets boring) and other subjective features of play. The Wynn feature bet offers a better ratio of expected loss to standard deviation than any bet in roulette.

For example, suppose a player and his wife split up. The player makes this bet all night at the craps table, his wife bets red on roulette all night. At the end of the night, the craps feature bet player has probably won once and lost a little money; but he has a reasonable chance of having won twice and being up a lot, or even won more than twice. He also has a reasonable chance of having lost every bet. His wife has a fairly predictable small loss.

The wife like rebet winnings many times over, the craps bettor might have done no rebetting, and in any case did much less than his wife. The expected loss fraction does not apply to the stack of chips you buy, but to the chips you bet. Games with a small house edge in which players rebet winnings a lot mean the casino takes a larger fraction of the chips purchased than games with larger house edges in which players do not rebet.

Every dice roll mattered a lot to final outcome for the craps bettor, each roulette bet was insignificant to the wife. He was hanging out with a bunch of guys yelling slang and jargon; and drinking; with lots of complicated table markers moving around. She was probably hanging out with a quieter mixed sex crowd watching a pretty piece of machinery generate random numbers.

Who had more fun for their money? De gustibus non est disputandum. But I think it's fair to say that the craps player did more gambling for his money.

If you have x dollars and you need to turn it into x+n dollars (and will go broke trying) and you can bet fractional amounts (to precisely hit your goal) what x and n makes pass line bets lesser negative EV than the Wynn bet?

Great post!

My BF always says I always make bets with a low *thrill* factor.
Such a boring way to play and gamble.

thrill factor = SD/EV

So last time I played in a casino I added 345X odds at Craps while betting the pass line instead of my usual 1X odds.

Too much thrill for me.

Sally

This appears to have a very interesting answer that I did not expect. My program is saying that the following are all of the integer values of n/x less than 1000 for which the pass line is less negative EV than Wynn:

1-27
30-32
496-524

That is, it is saying that if the ratio of your stop win to your stop loss is equal to one of these values, then you are better off betting the pass line without odds than betting the Wynn bet. I also checked values of 0.1, 0.2, up to 0.9, and in all of those cases the pass line is better.

I've been trying to understand why the better bet reverses 4 times. I suspect that it's due to the fact that certain stop wins will allow a bet to hit or fail with little or no money left over, which would save bets by eliminating additional passes or decrease the size of bets, which would improve the EV. The other bet could have lots of money left over to make more bets or larger bets. If I eliminate all passes but the first, then the island of values from 496-524 goes away. Let me know if the results are in line with what you were expecting.

I also did the analysis for betting a number in American roulette. There are values of n/x for which roulette will be better than both Wynn and the pass line. Yet people consider roulette to be a sucker's game. Aaron Brown says that the Wynn bet has a "better ratio of standard deviation to expected loss than any bet in roulette". Well the ratio is certainly higher, but that doesn't always make it better for the player in this scenario depending on his stop win and stop loss.

I also did the analysis for the 33:1 hard hop prop bet in craps. There are values of n/x for which that bet is better than both the pass line and Wynn. Yet about these prop bets one noted expert opines, "Players with any sense at all will avoid them completely".

We are assuming that bets can be subdivided indefinitely, but in the real world there is always a minimum chip, even if it is only a penny. This can change which bet is better. For example, I looked at betting a single number to playing red in single 0 roulette, and the bet which is better for turning 1000 into 2000 with only integer bets is different from the bet that is better for turning 1000 into 2000 with a bet that can be subdivided indefinitely. In American roulette, the result is same either way; betting a number is better.

What are the odds of rolling a 2, 3, 4, 5, and a 6 at least once with 2 dice before rolling a 7? Would that be half of 189:1?

Thanks, ECinOC

36.945:1.

There are only 31 terms to compute instead of 1023 as before. The first 5 are 6/7 + 6/8 + 6/9 + 6/10 + 6/11 which are the 5 probabilities that a 7 occurs before each of the 5 numbers 2-6. Then we subtract the probabilities for 10 combinations of 2 numbers, add the probabilities for 10 combinations of 3, subtract the probabilities for 5 combinations of 4, and add the probability for all 5. That's the probability that a 7 occurs before we get all the others, so we subtract that from 1.

To compute this exactly with the original R script I linked, it was only necessary to change in_36 to c(1,2,3,4,5,6) and have the script output the final value of 1-p. I created the new script below which would answer the general question: "What are the odds of rolling <list of numbers> at least once with 2 dice before rolling <number>?". Simply put list of numbers in the array nums on the first line. The order you list the numbers doesn't matter as long as the one that has to come last is listed last (7 in this case).



Output:

> Odds.to.1
[1] 36.94503

Thanks Bruce for the informative reply. Just to be clear, in the computation above, if for instance an 8 is rolled, it is a non-event. By that I mean the numbers 2, 3, 4, 5, and 6 need to be rolled before a 7 but those aren't the only numbers that can be rolled. If a 9 is rolled, that is fine for the sake of the odds I requested. It is only a 7 that cannot be rolled. Is that consistant with your computations?

And to recap, the odds of rolling a 2, 3, 4, 5, 6, 8, 9, 10, 11, and 12 (of course not in any specific order) before rolling a 7 is 189:1?

Again, many, many thanks for taking the time and sharing your considerable intellect,
ECinOC

Yes, all correct.

Thinking about this a bit further, I guessing that the odds of rolling a 2, 3, 4, 5, and 6 at least once before rolling a 7 are the same as rolling a 8, 9, 10, 11, and 12 at least once before rolling a 7, correct?

Does this mean that in the same 36.9 rolls you should roll *both* the lower set of numbers and the upper set of numbers or *either* the lower set of numbers or the upper set numbers before rolling a 7?

Thanks so much for any thoughts on the subject...
-ECinOC

Yes.


On average 1 in 37.9 attempts you will roll the lower set before a 7, and 1 in 37.9 attempts you will roll the upper set before a 7, but you only roll BOTH before a 7 about 1 in 190 attempts, and you will roll EITHER lower or upper (or both) 1 time in 21 attempts.

P(lower set) =~ 1/37.9

P(upper set) =~ 1/37.9

P(both sets) =~ 1/190

P(either set) =~ 1/37.9 + 1/37.9 - 1/190 =~ 1/21


That's assuming that you always play until you hit a 7 or both and not stop once you've made one or the other. 1 time out of 10 times that you hit the lower you will have already hit the upper, and vice versa. That's because hitting both is 5 times less likely than hitting lower, so you would hit both 1 in 5 times you hit lower, but only half of those times will the upper occur before the lower.

Bruce, that's puzzling to me. If you should roll either the upper set or the lower set in 38 rolls, you should be able to roll both in 38 rolls. Example: One guy is betting the upper set while at the same time another guy is betting the lower set. Both can expect to hit their bets in 38 rolls. So why would it take 42 rolls for both to hit? Either way, the bet is a loser as it only pays 35 to 1. This give the house a 7.9% advantage, correct?

With the 2 players, the probability someone wins on each attempt will be about 2/38, so the average number of attempts until the FIRST win will be about 38/2 = 19, and in fact it is 21 (the difference of 2 is due to the fact that both can win on the same attempt). Now once the first player wins, it will take an average of an additional 38 attempts for the other player to win. So the average attempts for both players to get a win is the sum of the average attempts for each player 21 + 38 = 59.

Another way to look at it is that it takes an average of 38 for the lower guy to win, and half the time he will win first, so half the time we will have both, but half the time it takes another 38 for the upper guy to win, so on average it takes 38 + 0.5*38 = 57. The difference of 2 between that and the 59 is due to the times that both win on the same attempt. We took that into account in the first calculation by using 21 instead of 19.


No, if lower pays 35-to-1, then the house has a 5% advantage. With probability 1/37.9 you will receive a total payout of 36, and it cost 1 to play, so your EV is

1/37.9 * 36 - 1 = -5%.

Even the all-bet where you have to get all 10 numbers only has a house advantage of 6.94%. It pays 176:1 and has a winning probability of 1/190.2, so

1/190.2 * 177 - 1 = -6.94%

It make sense that the house edge for the 35-to-1 payout is smaller since the payout odds are smaller.

I just played this Bonus Craps bet downtown Las Vegas at both the Golden Nugget and at Binion's
where when I walked up to the table a shooter had just hit the All, Small and Tall (he bet all 3) with $10 on each
(this being the max bet). A very nice win. (175 for 1 and 35 for 1)

The Dealers where I played said it is really a min $3 bet ($1 on all 3 bets) as you would be foolish to just bet $1 on just one or two of them. (like the All)
I tried it a few times and lost every time. I just could not get into the bets.

It got me to thinking about the ev and combined house edge for always betting all 3 bets as most players do when making these bets.

ev for $3 total bet = -$0.05110846
(this is slightly higher than a $3 pass line bet at -21/495)
I come up with a house edge of -0.017036153
(-$0.05110846 / 3)

I see only 3 possible outcomes
win all 3: 0.005257704 net: $242
win either Tall or Small: 0.047450116 net:$32
lose all 3: 0.94729218 net:-$3

this looks right but seems not right.
Sally

Ah
I still see only 3 possible outcomes
win all 3: 0.005257704 net: $242
win either Tall or Small: 0.042192412 net:$32
lose all 3: 0.952549884 net:-$3

ev: -$0.2351281 / 3 = -0.078376033 house edge

the small: 0.02635391 but not the All - 0.005257704 = 0.021096206
the tall: 0.02635391 but not the All - 0.005257704 = 0.021096206

back to having fun
Sally

I agree with your computation. Good luck.

Regarding the All Small bet, rolling the 2,3,4,5,6 each at least once before the 7:

BruceZ, you seem to have the best handle on this. Unfortunately, I don't have the R program and need a different approach to the answer.

I came up with the following: There are 5! = 120 permutations of the numbers 2,3,4,5,6 each beginning with 21 meaningful possibilities which decrease as each number is made. Totalling the probabilities of all 120 permutations gives the correct answer. For example, the prob of the first permutation would be:

p(2,3,4,5,6) = 1/21 * 2/20 * 3/18 * 4/15 * 5/11 = 9.62 x 10-6

After the 2 is made, it no longer matters and is removed from the pool of significant events. So the prob of making the 3 becomes 2/20 rather than 2/21, etc.

Summing all 120 permutations, I get p(win) = 2.63539 x 10-2

This answer agrees with others I've seen here and elsewhere. But now the question I've been struggling with: how do I compute the average number of rolls per decision (which includes both winning and losing decisions)?

With a simple bet like say the Place4, the p(win) = 3/36 and p(lose) = 6/36 and therefore the average number of rolls per decision is 1/(3/36 + 6/36) = 4. However, the All Small bet is far more complicated and the answer has thus far eluded me.

Here's what I have so far:

Since I'm now also considering the p(lose) there are now 720 permutations of 2,3,4,5,6,7 to consider. The avg number of rolls/dec should be an average of the sum of rolls per decision for each permutation mutliplied by the probability of making the numbers in the permutation.

The avg number of rolls until the first significant event is 1/(1/36 + 2/36 + 3/36 + 4/36 + 5/36 + 6/36) = 36/21 = 1.714286

At this point we've either lost p(7) = 6/21 or we remove the rolled number from the pool of significant events and continue to the next event.

So let's say that the 2 rolled first, the average number of rolls until the next significant event would be 1/(2/36 + 3/36 + 4/36 + 5/36 + 6/36) = 36/20 = 1.8

If the second event were a 7 then the rolls/dec would be 1.714286 + 1.8 = 3.514286. If not a 7 then the rolled number is again removed from the pool of significant events and we continue to the next event. This continues until the 5th significant event which is either a win or a loss and there are no further events.

So for each significant event in these 720 permutations the avg number rolls/dec = 1/(p(significant event))

Running through all 720 permutations and taking an average I get 7.913 rolls/dec which is not correct. By simulation I get approx 5.84.

Can you offer a solution without the R program?

Thanks,
Steen

Fyi you can execute R code here without installing R. Just have to change the language in the drop-down menu to R, paste the R code then click Evaluate.