Qestion on Craps

No, my question isn’t “how do you cook them?”

What I’d like to know is whether laying odds on a Don’t Pass bet is a good idea.

For those who’ve never played craps, the shooter throws two dice. If a 7 or 11 comes up, the pass line wins. If a 2, 3, or 12 comes up, the pass line loses. Anything else is his “point”, and the shooter must then roll his “point” again before he rolls a 7. If he does make his point, then the pass line wins and the cycle starts over. (If he rolls a 7, then the pass line loses and the dice are given to the next shooter; that’s called ‘fading’ or ‘sevening-out’.)

If you’re betting the Don’t Pass line, you lose if the first roll is 7 or 11, you win if it’s 2 or 3, and you push if it’s 12. (Sometimes the push is for 2 instead of 12, it will be noted on the table.) And if the shooter rolls anything else, you win your bet if he rolls a 7 before the point comes up again. Since 7 is the most common combination, once the shooter establishes a point you have a positive expectancy on your bet: you get paid even money on something that will happen more than half the time.

You have the option of laying odds once a point is established. If the point is 6 or 8, you lay 6-5 odds. (For the sake of simplicity, let’s say your Don’t Pass bet is $10.) So, on a 6 or 8, you’d put up an additional $12 to win $10; on 5 or 9, you’d have to lay $15 to win $10; on 4 or 10, you’d have to lay $20 to win $10. You would get this $10 in addition to the $10 you put on the Don’t Pass line.

My question is, does laying odds make statistical sense? If your Don’t Pass bet has a positive expectancy at even money, wouldn’t laying true odds make your total bet less of a positive expectancy? Or does the ability to add to your wager and still keep the positive expectancy (even though it has been diminished) help overcome the negative expectancy on the first roll? (Remember that on the first roll, you’d lose on 8 of the 36 possible combinations while winning on only 3.)

I don’t want to get into the math now, but you don’t have a positive expectancy by betting on the Don’t Pass line. The “push” on a 12 turns it into a very slight negative.

I thought I made this clear on my original post: after you get past the first roll, you do have a positive expectancy. I know that overall you have a negative expectancy, I wanted to limit my question to what happens after the first roll.

After the point is established you have a more positive expectancy, so why would you not want to lay odds?

Statistically there is no house advantage when you “lay” or “play” the odds.

Yes it make sense to lay the odds at the maximum allowed. Why? Technically it is not an additional bet, but rather, it is an entirely different bet. You gave the house a slight advantage when you made the Dont Pass Bet. So Why wouldn’t you make a bet that has no statistical house advantage?

Gruven…you are risking more money than you win when you are laying the odds. When the point is 4 or 10 you risk $100 to win $50…Risking $75 to win $50 when the point is 5 or 9…& risking $60 to win $50 when the point is 6 or 8.

Right…thats what I’m saying… Why wouldn’t you lay odds?

Steverino, craps players can debate this endlessly.

I play craps, and I like to get on the Don’t side every now and then. I always lay odds (at least single), unless my bankroll is so low that I’m just interested in killing time. It makes good sense. There is no other bet on the table available to you with less house edge. Free odds bets are, well, “free” in that they pay at the true odds. I think you understand this. Anything else you are thinking of betting on (hardways, a don’t come, anything) would be giving the house an edge. If you come in to gamble, why not lay the bet?

Does laying odds reduce your “advantage” for the Don’t Pass bet (with a point established)? No, it doesn’t. Your flat bet has weathered the come-out roll, and now it’s favored to win. It would be great if you could increase the flat bet at that time, but the house won’t allow it. So lay the free odds. If you had put that total (flat+odds) amount on the Don’t before the come-out, then you’d have risked the greater amount at a disadvantage. It’s a good idea to save that odds money, and lay the odds.

I think that the only hazard in laying odds is that you tend to have a lot more money on the table than you would from the Pass side. A string of bad luck can send you packing pretty quickly.

It’s such a gamble…

Laying odds at craps is the only fair bet in the house. Assuming you’re in the casino to gamble why not risk your money where you don’t have to give the house an edge? Wouldn’t you want to play roulette if there were no zeros on the wheel?

SteverinoAlaReno is absolutely correct. Statistically, in craps, you should only lay odds on pass/come bets, never on don’t pass/don’t come bets.

To illustrate, say you place a $36 Don’t Pass bet (to make the math easy). The shooter rolls a 6 as the point. You now have a choice, lay odds or don’t lay odds?

If you do not lay odds at this point, the expectancy is $37 (+$1). There’s a 6/36 chance you win and get your initial bet back plus an additional $36, a 5/36 chance you will lose your bet, and a 25/36 chance that the bet will stay on the table. So for the Don’t Pass bet without odds after the come-out roll has passed,

Expectancy (no odds) = $72*(6/36) + $36*(25/36) + $0*(5/36) = $37

for a straight Don’t Pass bet without odds, assuming your bet survived the come-out roll. This gives you an edge over the house of 2.778%, again, once your bet survivied the come-out roll.

If you lay odds, say single odds, you put up another $36 in the hopes of winning an additional $30 with this bet (5:6 payout). If you win, you collect $138 off the table, of which $72 was yours to begin with. If you lose, you get nothing. Otherwise, your $72 stays on the table. So, for the Don’t Pass bet with single odds after the come-out roll has passed,

Expectancy (single odds) = $138*(6/36) + $72*(25/36) + $0*(5/36) = $73.

So your expectancy after laying single odds is still only $1. This makes sense, since laying odds is a fair bet and shouldn’t affect your expectancy at all (odds bets have an expectancy of +0%). But now you’ve basically wagered $72 to get the same $1, so your edge over the house has been reduced to 1.388%. Your expectation percentage has been reduced by laying odds on your Don’t Pass bet. Again, this assumes that your initial Don’t Pass bet survives the come-out roll, which is the case we’re dealing with here.

The calculations for all other Don’t Pass Bets after the come-out follow the same principle. The initial bets themselves have a positive expectation and the odds bets have a +0% expectation. Laying odds on a bet with positive expectation reduces the magnitude of the expectation percentage. This is because your expectation in dollars is not affected by odds bets, but by laying down odds, you’re wagering more money.

It’s true that, overall, Don’t Pass bets have a negative expectancy since a 12 roll is considered a push, not a win, and so the casino stays in business. You lay odds when you are facing a negative expectation to minimize the magnitude of that negative expectation. Combined, people conclude that you should lay odds on Don’t Pass bets. And this would be sound advice if you were required to declare whether or not you intended to lay odds on your Don’t Pass bet BEFORE the come-out roll occurred (with the stipulation that you wouldn’t lay down odds if you lost on the come-out, of course). But that’s not the way the game works; you make that decision AFTER the come-out roll occurs. This makes all the difference, as at that point, it’s a totally new decision, and the player is facing different odds and expectations than before the come-out roll.

For similar reasons, you should always lay odds on Pass/Come bets since these bets have a negative expectation once the come-out roll has passed.

…to post past my bedtime.

I knew I would mess up the math somehow and reach the wrong conclusion, and I figured out my error as soon as I hit ‘Submit’ (naturally sigh…).

The value of a push after the come out roll isn’t just the value of the money on the table, it’s more since the shooter rolls again and there’s still the same positive expectation value on the following roll.

BUZZ! Wrong. The value of the (25/36) outcome is more than $36. It’s true value is summation of an infinite series of values.

BUZZ, BUZZ! Wrong again, Caldazar; you fail math. The value of the (25/36) outcome is more than $72.

Sorry about that all :frowning:

However, one thing I still can’t figure out:

After taking into account the infinite series, I still come up with the expectancy of the $36 Don’t Pass no-odds bet as $39.27, or +9.09% and the expectancy of the single-odds bet as $75.27, or +4.55%.

So this still seems to indicate that the laying odds is not a good idea on Don’t Pass/Don’t Come bets…??

This isn’t a question about statistics or probability, it’s a question about money management.

As others have said, laying odds pays off at true odds. One way of looking at this is that you can increase your original bet and not have worse odds. Another way is that you improve your odds by averaging your original bet with the true odds of the odds laid. Another way is to treat is a seperate bet.

The way I see it, it’s akin to someone asking to flip coins with you, heads you win a dollar, tails they win a dollar. Assuming a fair coin and all that, it probably comes down to how conservative you are with your money.

If you have the bankroll, it’s a fine bet, best one in the house. (Unless you count cards of course) Since you’re in a casino and presumedly a gambler, you would probably go for it.

On the other hand, if you were conservative with your money (in which case you should be in an index fund not a casino!) or low on cash, you might not be willing to take the risk of losing a few of these in a row and being wiped out.

I hope my point is clear, since the odds are true, it has nothing to do with strategy, it’s purely how you feel about your money stack. And you probably see that there is virtually no difference between asking the exact same question about getting odds on the “Pass” line – it also comes down to psychological and bankroll factors there.

As far as money management goes, I gamble purely for the fun of it, and only with money specially earmarked for entertainment. But while I’m not looking at craps as an investment vehicle, I’d still like to play the game correctly :slight_smile:

Muttrox, I agree with much of what you said, but I take exception to your statement about taking odds on Pass/Come bets.

If you don’t take odds on Pass/Come, the house edge is 1.4% on your bet. Taking single odds reduces the house edge to 0.8% and taking double odds reduces it to 0.6%. (I have seen some casinos offer 10x odds, but I’ve never calculated the house edge on that…I imagine it’s very low.)

Because of the push on 12, the house edge on a Don’t Pass bet is 1.4%. My question should have been: does laying odds on Don’t Pass reduce the house edge?

When I play craps, I tend to bet Pass/Come and take odds. If the table is cold, I’ll either walk away or play Don’t Pass without laying odds for a while. Playing “wrong” isn’t as much fun, especially when you have a table full of people rooting for the shooter and you’re the only guy on Don’t Pass.


I think you fall under those I mentioned as “Another way is that you improve your odds by averaging your original bet with the true odds of the odds laid.” I think you get those number by just averaging the 50% off the odds with the odds of the original bet, weighted for however many “x”'s time your original bet you are taking and the odds of a given first roll coming up. Harder to say than to actually calculate.

I believe the mathematics is the same for don’t pass odds – that is, take the normal odds for don’t pass, and average them with the 50% for the odds.

(Note: I am aware I’m being loose with terminology, it’s really expected payoff or something I’m speaking of, not odds, which are dependent on the value of the first roll.)

On the other hand, it is not quite the same from a money management side because you have to bet 2x or 3x your normal bet. If (like me) you go in with 20 or so betting units, this is a big factor, because now you have only 7 or so betting units. It’s like moving from a $5 table to a $15 table when I only have $100. My one time in Vegas, I said what the hell, and took 10x behind the line on my first bet. Of course I lost, and that was half my bankroll for the session.

So yes, laying odds on don’t pass reduces the house edge. But it reduces it in exact proportion to increasing your minimum allowable bet.

There’s a number of ways to look at this, reasonable people can disagree about the meaning and implication of the figures. My feeling is, if you have the money to cover it, then yes, taking odds on pass, don’t pass, come, or don’t come, are all good bets and reduce the houses edge. If you don’t have a big bankroll, then it’s arguable.

K? :slight_smile:

Yeah, that’s the way I pretty much look at it, although I’m not sure that this is necessarily an accurate way to look at things.

The thing about Don’t Pass bets that has me a bit puzzled is that while they give the house an edge overall, once your bet survives the come-out roll (which it will fail to do so more often than it succeeds, granted), the edge swings to the player. So why would you want to lay odds and reduce your expectation percentage? The odds bet, in the long run, won’t win you or lose you any money since it pays at true odds. So why play it when you have already have a bet sitting on the Don’t Pass that stands a chance to win you some money (due to it’s positive expectation after having survived the come-out roll)? If the house let you lay some kind of odds bet for Don’t Pass on the come-out roll, it might be a different story, but that’s not the case.

Caldazar, you and I are of a mind on the Don’t Pass bets. The only reason I can think of for laying odds on an established point is that the extra money you’d win as a result would mitigate the losses you’d suffer when the come-out roll is 7 or 11. The question then is whether the mitigating factor outweighs the dilution of the postive expectancy you have on your even-money bet enough to cut into the house edge. Muttrox says it does, and he’s probably right; I was hoping someone here had done the math on it.

I don’t see how an odds bet on Don’t Pass could mitigate losses on the come-out roll. From a probabilistic perspective, the odds bet wins you no money at all since it pays out at odds. It’s the inital Don’t Pass bet, made on the come-out roll and surviving the come-out, that make you money since that’s the case in Don’t Pass that has a positive expectation. And as I said, you can’t lay odds on the come-out to reduce the house edge on the Don’t Pass bet on the come-out roll.

Craps odds calculations can be found at:


However, the author calculates the odds for Don’t Pass/Don’t Come with odds by averaging the overall expectation of the Don’t Pass/Don’t Come with the odds bets (expectation zero). So he always gets than the odds bet is a good idea, since the overall expectation of Don’t Pass/Don’t Come is negative. I still say this isn’t an accurate way to calculate the odds, since you can’t lay odds on the come-out roll.

Intuitively, I think the end result should be that you should always lay max odds on Don’t Pass/Don’t Come, assuming you have the money. But I’ll be damned if I can prove why.

Let me try to explain myself, and see whether you agree with me. It’s somewhat tortured logic, so bear with me.

If you don’t lay odds on a point, you’ll get paid even money. That’ll be the same payout no matter what’s rolled, and because the first roll is a loser 8/36 times and a winner 3/36, the house has an edge, even though any other roll has a better-than-even chance of winning.

If you lay odds, you’ll win more money than you would lose if the come out roll was a bad one. For instance, if you’ve got $10 on Don’t Pass and the point is 5, you lay $15 as your odds bet. If the shooter fades, you get back $20. So, you stand to win more on a point than you stand to lose on the come-out roll, provided you lay odds. That’s why I said that laying odds will mitigate first-roll losses.

But laying odds dilutes the expectancy of your initial bet. With 5 as the point, your bet will win 60% of the time. If you don’t lay odds, that mean you’ll win $10 60% of the time and lose $10 40% of the time. If you lay odds, you’ll win $20 60% of the time and you’ll lose $25 40% of the time. That’s still a positive expectancy, but not as great as it was without laying odds.

So, is the lessening of your come-out roll losses worth the dilution of your expectancy when a point is rolled? (Side question: would it be worth it to lay odds only on 6 & 8, since the dilution on those numbers isn’t as great?)

Anyone want to come to Reno and test it out? :slight_smile:

When people say that “Taking single odds reduces the house edge to 0.8% and taking double odds reduces it to 0.6%,” I think that’s somewhat of a red herring.Yes, it is technically true that odds reduce the house edge when expressed as a percent.

But when you just want the average amount of money you can expect to come away with, the odds bet has zero effect. It does, however, increase the variance of your expected win/loss amount.

I guess if you want to have bet $15 and get the best odds, you would bet $5 on the come-out and take the other ten in odds. However, you have the exact same expectation, on average, as if you had only made the $5 come-out bet. So taking the odds this way is better than betting all $15 on the come-out, but why not just bet the original $5? If you get some entertainment value out of the $10 in odds, then by all means do it, but don’t try to justify it based on the house edge.

Well, since I don’t have that kind of cash on hand to test this, I wrote a program to simulate a craps game instead. The C source can be found at:

Sorry for the crappy code; I’m an engineer, not a programmer.

The game was set up as follows:

A $10 Don’t Pass bet was made before the come out roll. For the odds program, a $12 odds bet was played if the point was a 6 or 8, a $15 odds bet was played if the point was a 5 or 9, and a $20 odds bet was played if the point was a 4 or 10. All rules were standard craps rules.

After 30 million trials for each strategy, the results came out virtually identical:

No odds played:

8.33% of trails won on come-out roll (2 or 3 rolled)
22.23% of trials lost on come-out roll (7 or 11 rolled)
2.78% of trials pushed on come-out roll (12 rolled)
39.62% of trials won on a point roll (7 rolled)
27.04% of trials lost on a point roll (point rolled)

Net loss (no odds): -$3,972,510

Odds played:

8.34% of trails won on come-out roll (2 or 3 rolled)
22.23% of trials lost on come-out roll (7 or 11 rolled)
2.78% of trials pushed on come-out roll (12 rolled)
39.60% of trials won on a point roll (7 rolled)
27.06% of trials lost on a point roll (point rolled)

Net loss (odds): -$3,928,489

Note: Percentages do not add to 100% due to rounding.

This would seem to support the view that the Don’t Pass and Odds bets should be considered totally separate bets, with the Don’t Pass bet having an overall negative expectation and the Odds bet having a zero expectation.

Basically, what CurtC said.

Caldazar, Curt,
Realize that you are considering the betting session on a “per hand” or “per come-out” basis, and that’s why the bottom line is not changed due to odds or not. But consider that you’re gambling with significantly more money with the single odds strategy. If you bet that same amount on the flat, you’d come out farther behind.

I don’t think most folks go into the casino thinking, “I’m going to play 50 hands, then quit”. Usually, you’re thinking (at least, I think you should) that you have X amount of money to gamble with, and maybe some vague time limit. Then laying the odds is the better way.