# Craps - Other Bets

I made the mistake of making a pass line bet in craps after a point was established. The next roll was a seven and I thought I won but the dealer said I lost. Can the player make a pass line bet after a point is rolled? What should have happened?

Yes, you can make a pass line bet at any time. However, you give up the come out roll, which twice the chance of winning as losing. Making a late pass or come bet is called a "put" bet.

What happens to a place bet to win on a six if the shooter makes his point other than six? What happens to an initial come bet when the shooter makes his point on that roll? What happens to a come bet after it is moved to a number (lets say 5) and the shooter makes his point of lets say four. As per "craps strategies from crappers delight" I have read that the average shooter rolls the dice five (5) times before sevening out. Your comments.

I hope it will answer your question if I say that place, buy, and odds on come bets are temporarily turned off for come out rolls, unless requested otherwise.

The average shooter throws the dice 8.53 times, including the seven-out roll. For more information on that, visit my odds table for the probability of every number of throws from 2 to 200.

I have heard it said that with one come bet a player should call off the odds on a come out roll, but with two or more the player should leave them on. The rational is that with two or more on the odds of rolling one of the come points is greater than rolling a 7, but with one bet the player is more likely to win than lose.

The player should always leave the odds on regardless of how many come bets are active. When considering the options, looking at the probability of winning alone is not enough. Yes, with one come bet the odds of losing the come bet are greater than winning, however the potential win is greater than potential loss. The reason the player should always leave the odds on are because it is a bet with zero house edge. By turning the odds off the player is making the overall game more heavily weighted towards the bets with a house edge, thus increasing the overall ratio of the expected loss to the total amount bet.

I like your site very much. It is very informative. Thanks for putting out your thoughts. I noticed a betting strategy for craps suggested at Crappers Delight called "classic regression". In it he suggests, placing a 6 and 8, after a point is established. Then taking it down after one of them is hit. He said there are 10 combined ways to make the 6 and 8, but only 6 combined ways to make the 7. It sounds logical, but I've seen where you are able to show, that what appears logical on the surface is not so bright once it is analyzed. What are your thoughts on this strategy and what would the true odds be, if you did take the bets down after one hit?

This is similar to a question I got last week. Yes, it is true that there are ten ways to roll a 6 or 8, and six ways to roll a 7. However, one must not look at the probabilities alone, but weight them against the payoffs. The place bet on the 6 and 8 pays 7 to 6 odds when fair odds would pay 6 to 5. By making six unit place bets on the 6 and 8, and taking the other down if one wins, the probability of winning 7 units is 62.5% and the probability of losing 12 units is 37.5%. If the player must cover both the 6 and 8, then the place bet is the way to go. This rate of return isn't bad but could be better. For the player who puts a priority on minimizing the overall house edge, the best strategy is to make combinations of pass, don't pass, come, and don't come bets, and always take the maximum allowable odds.

What is the house advantage on put bets with 20 times odds. Should the house allow put bets at these odds, for example someone could take $1000 with $20,000 on every number. Could you explain this to me? Thanks, Great site.

For other readers, let me explain that a put bet is making a pass or come bet after a point has already been established. The player may choose the point to be established on the put bet. While the player can make an odds bet immediately on top of the put bet, the opportunity to win on the initial roll is lost. The effect is the same as making a place bet or buying odds, but the house edge is different depending on the multiple of odds allowed. In the case of 20 times odds the house edge of the put bet on the 4 and 10 is 1.59%, on the 5 and 9 is 0.95%, and on the 6 and 8 is 0.43%. At this high level of odds allowed, which is much greater than the norm, all put bets are better than the corresponding place or buy bets. This option should never be taken at a casino that offers less than 5 times odds. At 5 times odds exactly, the put bet on the 6 and 8 is slightly better than the place bet. At 10 times odds or greater, all put bets become better than their corresponding place or buy bets. I shall add something to my craps section about the put bets, thanks for the idea.

What is a put bet in craps?

A put bet is like a come bet made after a point has already been established. I don't recommend put bets because the value from come bets comes from the first roll, before any point is established. On a the first roll of a come bet the probability of winning is 22.22%, and the probability of losing is 11.11%.

However, if you must cover other points quickly, the odds are generally equal or better on place bets and buy bets on the 4 and 10 (assuming commission paid only on a win). Only if you can get 10x odds, and take them, do put bets become a better value than the better of place and buy bets. Since so few places offer 10x or better odds, it is safe to say to just avoid put bets completely.

What is the better system, or which gives me the better chance to win on craps? On the come out roll, I bet $10 on the don’t and $10 on the do, and then when a point comes out I lay full odds against the number. Or is it better to just play the don’t pass, and then lay the odds. I think getting passed the come out roll will increase my chances of winning.

The better system is to bet on the don't pass only and take full odds. Yes, betting on both does increase you chances of winning on any one bet. However you are suffering a higher combined house edge by betting on both the pass and don't pass and it will cost you in the long run.

Which casinos pay 3 to 1 odds on a field bet of 12?

Most casinos in Las Vegas do, except those owned by Caesars Entertainment (formerly known as Harrah's). In Reno and Lake Tahoe they pay 2-1 on the 12, but 3-1 on the 2.

I played craps for the first time the other night and went from $70 to $700 with small bets on the pass odds and field bets. I then lost it all down to $6 because my bets were too large (by the dealers suggestion), and gained it back to $1000 after slowing down. For this being the first time it seems like a very easy game to win if you have patience, was it beginners luck?

Yes, it was luck. It helped that you stuck to the low house edge bets. However, next time, make the line bets with odds only, and don't bet the field, especially if it pays 2 to 1 only on both the 2 and 12.

What is the calculation of the house edge for a put bet on the pass line (after the come out roll and the point is established) with full odds (eg. 2x, 10x, 100x) for the 6&8, 7&9, 4&10 and any combination there of? I became very curious after I saw this bet allowed at Binions-LV last time I was in town.

Since you asked I added information about the put bet in my craps section. That should tell you what you want to know.

I have read about a few casinos that pay 3-1 on both the 2 and 12 in a field bet. What Las Vegas casinos offer this bet?

I've never heard of a Las Vegas casino paying 3-1 on both the 2 and 12, but the Santa Ana Star in New Mexico does (at least as of this writing in 2013). This rule results in a house edge of exactly 0.00%.

Concerning the 5% vig on buy bets and lays how would the odds change if $1 was charged for $20-$39, $2 for $40-$59, $3 for $60-$79 and $4 for $80-$99 without round up. Your information you give is outstanding.

Thanks for the compliment. The formula for the house edge in buy and lay bets is the commission divided by the bet plus commission. In this case, the best bet is to bet $39 for the $1 commission. On the buy bet the house edge would be 1/40 = 2.5%. Assuming you can lay $78 to win $39 on the 4 and 10, and still only pay $1, the house edge would be 1/79=1.27%. I'll leave the other situations as an exercise for the reader (I hated it when my math books would say that).

I just finished reading your section on strategy for craps with great interest. I understand that better the pass line and come bets with full odds is a good strategy. My question is "does the house edge change at all when playing a strategy of pass line with full odds and making a maximum of two come bets with maximum odds?" In other words, how does time (more rolls) and having more money at risk affect these odds if at all? Or should a person stick with just the single bet with full odds? This seems to be a favorite strategy for most knowledgeable players I have met at the dice tables.

The house edge is the same regardless of how many come bets you make assuming you always take the maximum allowable odds and leave the odds turned on during a come out roll. How many come bets you make should be a matter of personal preference.

I love to play craps and would like your opinion on a conventional method of play. Pass line and two come bets with full double odds or with one come bet? Does having three different bets working superior to two?

As long as you are backing up your pass and come bets with full odds, it doesn't make any difference how many come bets you make. However, it does reduce the overall house edge to keep the odds on your come bets working on the come out roll.

In craps, does the house edge change if you make a don't pass bet then remove it if the point is 6 or 8? What if you remove it if the point is 6,8,5,or 9?

You should never remove a don't pass bet after a point is made! Once a point is made of 6 or 8 the don't pass has equity of 9.09% of the bet amount, which you would be throwing away by taking the bet down. The equity of a don't pass bet on a point of 5 or 9 is 20%, and on a 4 or 10 is 33.33%.

What is the difference between making a come bet then taking the odds and a put bet? They sound the same to me in your explanation of put bets. Since you recommend against put bets you also recommend against come bets followed by the odds or am I not understanding you correctly?

A put bet is like a come bet on a particular point. In other words the put bet does not get the benefit of winning on a 7 or 11 on the first roll, but the bet immediately gets "put" on the point of the bettors choice. The bettor may also immediately take odds on the put bet. Put bets are generally not a good idea because the player is twice as likely to win as lose on the first roll of a come bet, and the player is voluntarily forfeiting that roll with a put bet. However if the odds allowed are high enough a put bet, backed up with maximum odds, can have a lower house edge than a place or buy bet. My craps section states how high the odds need to get for this to happen. My craps advice in general is to stick to the line bets (pass, don’t pass, come, and don’t come) and backed up the odds.

Our $3 craps game pays $4.50 on a place bet of a 5 or 9, and $5.50 for a place bet on the 4 or 10. Could you tell me what the house advantage is on these bets? (I'm especially curious about the 5 or 9 since we are actually paying true odds for a place bet.)

There are 4 ways to roll a 5 (1+4, 2+3, 3+2, and 4+1) and 6 ways to roll a 7 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1). So the probability of rolling a 5, given that a 5 or 7 was rolled, is 4/(4+6) = 0.4. The expected value of the place bet on 5 is (0.4*$4.50 + 0.6*-$3.00)/3 = 0. So you're right there is no house edge on the 5, or the 9. On the 4 and 10 the expected value is ((1/3)*$5.50 + (2/3)*-3)/3 = -0.0556. In other words the house edge is 5.56%. I take it this is only true of $3 bets, the casino rounding the winnings up from $4.20 to $4.50. As I reported in the last issue the player can gain an advantage in blackjack if the dealer pays $4.00 for a blackjack on a $2.50 bet.

You have a great site. Thank you for spending so much of your time helping people to learn the math behind the games. I think I am missing something on your ’Buy’ bets payouts, however. For example, the ’Buy 4’, which you have listed as 39:21. This implies that a winning player would have 21+39, or 60, if they win. But a winning player only has 59 (because they pay the $1 fee). Should this not be 38:21? Then you lose 21 or have 59. I can’t wait to find out exactly how much of a doofus I am.

Don’t be so hard on yourself. Very few Internet casinos implement the buy bet odds correctly. When I say the odds are 39-21 I mean that if you bet $21 (including the commission) you will get back $39 plus the original $21 if you win. So if you win you get back $60. Consider making a $20 buy bet, plus the $1 commission. The $1 is gone forever, but if you win you get back $40 in winnings plus the original $20, for a total of $60. Either way you risk $21 and get back $60 if you win.

I have two friends that have a bet on which game (craps or baccarat) have the best odds for the player. Could you help me settle this. They are both casino workers and are sure they are right.

It depends on how the games are played. If optimal strategy is compared to optimal strategy then craps is better. By betting only the line bets and taking maximum odds the combined house edge in craps is well under 1%. The best you can do is baccarat is bet on the banker at a house edge of 1.06%. However it wouldn’t surprise me if the actual house edge in craps is higher, due to all the sucker bets players make.

I’ve just starting learning the game of craps. In craps, the Don’t Pass is a better bet for the player than the Pass Line bet. But the few times I’ve played in the casinos, most people seem to be betting the Pass Line and not the Don’t pass. I’m either not correct on the odds b/t the two bets or is there some reason most players are taking the Pass Line bet over the Don’t pass line bet?

That is a good question. It is obviously more fun to go with the crowd than against it. The question is why does the crowd favor the pass line? Perhaps it is just tradition. Maybe when people first started playing craps in private games the don’t pass wasn’t even an option.

I do not understand why you should lay the odds on the don’t pass or don’t come bets. It seems that you have already dodged the 7 and ll bullet, so the bet is now in your favor. Why would you dilute a bet that is already heavily in your favor with a large (relative speaking)bet at true odds? It seems that you are working in the houses favor by reducing the house edge on the entire bet.

I understand that taking the odds on the pass side reduces the overall house edge, however I don’t understand how laying the odds can reduce the house edge on the don’t side. I’m very curious? By the way, I discussed this with several casino bosses and dealers yesterday and they all had opinions, but not reasons for these opinions. Thanks for your time.

Let’s say you have a $10 don’t pass bet and the point is a 4. You have a 2/3 chance of winning the bet, so the expected value is (2/3)*$10 + (1/3)*-$10 =$ 10/3 = $3.33. Now consider adding a $40 odds bet on top of it. Now you have a 2/3 chance of winning $30 and a 1/3 chance of losing $50. The expected value of both bets combined is (2/3)*$30 + (1/3)*-$50 = $10/3 = $3.33. So either way your expected gain is 3 dollars and 33 cents. With the don’t pass alone the player edge is $3.33/$10 = 33.33%. With the don’t pass and odds the player edge is $3.33/$50 = 6.67%. So, yes, the player edge as a percentage drops by making the odds bet. However that player edge is effective over more money. The way I think gamblers should view the house edge is as the price to pay for entertainment. If you want to pay as little as possible then taking or laying the odds is getting entertainment for free.

At the craps table, I’ve observed that 80% of the people bet on the pass line. From what you say, the don’t pass line has a slightly lower house edge. So why don’t most people bet accordingly?

Most player probably don’t know the don’t pass is a slightly better bet. Even those that do probably bet the pass bet because it feels better to be part of a group than to go against it.

I know from your chart that PUT bets w/10x odds beats ALL place/buy bets. Lets talk about PUT vs. COME. A ’put’ on the 6/8 with 10x odds has a house edge of .83% while a come bet w/10x odds is .18%. You’re giving the house an additional 0.65% (.83-.18) edge by using a PUT vs. a COME bet. In return, you have the potential of getting more $ since you win on every 6/8 hit vs. winning every other time the # is made if you did a COME. That’s my theory at least. Now in the long run, who comes out ahead with more $? I say the guy w/the PUT bet comes out with more $ since you are hitting twice as often. With a come bet, you have make the #, then wait till it hits again, cutting your hits in half. If there’s something wrong w/my logic, please let me know!

I get challenged a lot on my support for come bets with odds. Those against come bets are always quick to say a number has to hit twice with a come bet and only once with a place or put bet. That is not a good way to look at it. First of all with come bets you have a 8/36=22.22% chance of winning on the first roll as opposed to only 11.11% of losing. You get no such advantage on the first roll of a place or bet bet. Even if you do roll a point number it can be any number. In other words there are six point numbers that can win on a come bet, and only one on a place or put bet. Ultimately the reason the come bet with odds beats place or put bets is the house edge is less.

I saw a bet in craps called the "Fire Bet" that paid if player made 4 to 6 different points. What is the scoop on that?

The following table shows what each number of points pays, the probability, and contribution to the total return. The probabilities were determined by random simulation. The exact probability of making all six points is 0.000162.

### Fire Bet

Points Made |
Probability |
Pays |
Return |

0 | 0.594522 | -1 | -0.594522 |

1 | 0.260503 | -1 | -0.260503 |

2 | 0.101038 | -1 | -0.101038 |

3 | 0.033364 | -1 | -0.033364 |

4 | 0.008776 | 10 | 0.087764 |

5 | 0.001633 | 200 | 0.326582 |

6 | 0.000164 | 2000 | 0.328063 |

Total | 1 | -0.247017 |

The lower right cell shows an expected loss, or house edge, of 24.70%. It is my understanding the only allowed bet amount is $2.50, so the expected loss per bet would be about 62 cents.

You say the house edge on the pass line bet in craps is 1.414%. Is there any coincidence that this number is the square root of 2?

Just a coincidence I assure you. The exact house edge in craps is 7/495, which by definition must be a rational number. In fact I would argue the house edge in all casino games must be a rational number because there are a limited number of possible outcomes in all games, resulting in a house edge of a perfect fraction. 2 is not a perfect square thus the square root of 2 must be irrational by definition. Therefore the two numbers can not be equal. To be specific the house edge on a $100 pass line bet would be $1.41414141... The square root of 2 is 1.4142135623731...

Mr. Wizard, great site. You are the one true expert on the internet. Because the odds in dice are based on the probability of winning as compared with the payout, I question whether it makes sense to take odds on bets that have a low a probability of winning. For example, the odds on the 4 and 10 are 2 to 1. Is it really a good bet to back up the pass line when the 4 or 10 shows. If someone offered a payout of 10,000 to 1, with the odds of winning 5000 to 1 (and you could only make one bet), it would appear to be a good bet, but the chances of winning are so slim that in reality it is probably a suckers bet. Is there any validity to what I am saying?

Thanks for the compliments. As you know I strictly look at the expected value of a bet. However, if taken to an extreme it doesn’t make sense at some point. If you made the proposition to me in your example I would only bet about $200 on it, although it has a 100% player advantage. The reason is if I won I would win $2,000,000 and I don’t really need more than that. However in craps a 2 to 1 win is not going to change my lifestyle if I win. 2 to 1 in craps is not a big long shot so I say get aggressive on the odds. If you’re uncomfortable with bets that have a low probability of winning you might take up betting the don’t pass and laying odds.

First, great site. During a recent visit to Harrah’s, they gave me an option of either $100 match play or $50 in slot play. In your opinion with which is the best to take. (I took the match play). Also, for the match play would it be better to play all $100 on one hand, or multiple smaller hands (10 x $10 hands). Thanks

Thanks for the compliment. I recommend taking the match play. I’m sure the $100 in slot play was on specially designated machines. From anecdotal evidence I believe these free play slots are extremely stingy, set to pay back about 25%. That match play is worth about 48 cents on the dollar. I recommend betting in on the don’t pass in craps. The reason I favor that over blackjack is that blackjack has a lower probability of winning, thus reducing the value of the match play. For further explanation please see my October 30 2001 column.

Hi Michael. Why is it better to make Come bets with odds than to make Place bets? My math indicates that you make more on a place bet when betting equal units. On 4, to instance, if I were to place $10 I’d win $18. On the come bet I would only win $15 ($5 on the initial bet and $10 on the odds). The other advantage to place bets is that I get to choose which numbers I want to bet and that I win the first time that it is rolled. Am I missing something?

Yes, you are. You are forgetting that a come bet wins on the first roll 22.22% of the time and loses 11.11% of the time. So you are missing the extra value of the first roll of a come bet. However if you had a crystal ball that told you that the first roll would result in a point number then you would be right.

Dear Wizard, Can you please explain to me how the house advantage on craps place bets are calculated. For instance, how does the nine to five payout on a four/ten place bet work out to a 6.67% house advantage when the true odds are two to one? No matter how I do it, I can’t come up with that 6.67% figure. This is driving me nuts. I would greatly appreciate an explanation.

I prefer to calculate the house edge as 1-(pr(win)*payout - pr(lose)). In this case it would be 1-((1/3)*1.8 - (2/3)) = 6.67%. However if you know the fair payout and the actual payout a convenient formula for the house edge is (f-a)/(f+1), where f=fair payout and a=actual payout. In this case (2-1.8)/(2+1) = 0.2/3 = 6.67%.

Thanks for the great site. You recently stated that the average craps shooter lasts approx. 8.5 throws. I normally bet the pass with full odds followed by come bets with full odds. Does it make more sense to quit making come bets after, say four throws given the long term probability that the thrower will 7 out in only three to four more throws?

You’re welcome, thank you. The dice do not have a memory so at four throws you do not get any closer to sevening out. You could roll 1000 non-sevens and still be no closer or further away from a seven than you were the first throw. There is no optimal number of come bets, just make as many as you find the most fun.

Last night I was at a casino in Louisiana and after playing for a while I noticed that the Big 6/8 bet was replaced with an Over/Under 7 for one roll bet. The crew member said it was new and paid even money. He then mentioned that he thought the field was a better bet. Was he right?

No! The probability of winning is 15/36 = 41.67%, for a house edge of 16.67%. Even if the 2 and 12 both pay 2 to 1 on the field the house edge is only 5.56%. The over/under 7 bet ties for the any 7 bet as the worst bets on the table.

Hi, if person A makes 1000 consecutive bets on the pass line without backing up his bet, and person B makes 1000 consecutive bets on the pass line and he takes 100X odds whenever possible, doesn’t each person lose the same amount of money?

Yes. I can just imagine the follow up question to be why I recommend taking the odds if doing so doesn’t help to win more. What I suggest is betting less on the pass so that your need for action is mostly met by a full odds bet. For example if you are comfortable betting about $90 per bet, and the casino allows 5x odds, then I would drop the pass line bet to $15 and bet $75 on the odds. That will lower the overall house edge from 1.414% to 0.326%.

At the showboat in Atlantic City there’s a new bet on the layout where the big 6/8 was. I’m wondering what the odds were on this one roll bet. 6-7-8 pay even money, hard 6/8 pay double. Thanks.

The following table shows the house edge is 5.56%.

### Low Bet

Total | Combinations | Probability | Pays | Return |

Hard 6,8 | 2 | 0.055556 | 2 | 0.111111 |

Soft 6,8 | 8 | 0.222222 | 1 | 0.222222 |

7 | 6 | 0.166667 | 1 | 0.166667 |

All other | 20 | 0.555556 | -1 | -0.555556 |

Total | 36 | 1 | -0.055556 |

If you have reason to believe that the seven is weighted and is coming up more than it should, does that favor the don’t or the pass side of craps?

The fewer the sevens the greater the odds favor the pass line bet. The following table shows the house edge according to the percentage of sevens, assuming the probability of all other numbers is proportional to the fair probability.

### House Edge in Craps According to Seven Probability

Seven Probability | Pass House Edge | Don’t Pass House Edge |

15.000% | -0.666% | 3.499% |

15.333% | -0.202% | 3.024% |

15.667% | 0.237% | 2.574% |

16.000% | 0.652% | 2.148% |

16.333% | 1.044% | 1.744% |

16.667% | 1.414% | 1.364% |

17.000% | 1.762% | 1.005% |

17.333% | 2.089% | 0.667% |

17.667% | 2.395% | 0.349% |

18.000% | 2.682% | 0.051% |

18.333% | 2.949% | -0.227% |

I’m walking through a casino and see a craps table with a shooter winning lots of money. I feel lucky and want to place a bet. What can/should I do? Do I have to first make a pass line bet? Can I make a come bet? Can I make either bet with odds? Or, do I wait until the next round of play?

First, it doesn’t make any difference that the shooter is making lots of money. Your odds are the same on an ice-cold table. The past does not matter. However if you are going to play then wait patiently for a come out roll. Never make a pass bet after a point has been established.

A follow-up to the recent question about the person who walked up to the hot craps table and wanted to play. I understand, of course, why the history of the table means nothing. However, why wouldn’t you recommend he place a come bet right away? My understanding is that it would result in the exact same expectation as waiting for a new come-out roll, but the eager better wouldn’t have to wait.

You’re right. A come bet would have been just as good and require no waiting. I should have added that.

During a 4-hour layover in Vegas, what’s my best strategy to double a $2000 bankroll? what game, large or small bets, etc.?

First I would take a taxi to the Hard Rock, the closest major casino to the airport. I’m not sure how much odds the Hard Rock allows in craps but I would guess 3-4-5. If that is the case then bet 1/7 of your bankroll on the don’t pass bet, or $275 to round down. If a point is established then lay the maximum on the odds, or $1650. If you win you’ll be a lot closer to your goal, the amount will depend on the point. Win or lose bet the lesser of 1/7 of your bankroll and 1/7 of how far you are from $4000. If you get close to either extreme just get it over with and bet everything if you’re low, or whatever you need to close the gap on $4000 if you’re high, and forget about the odds. Four hours should be enough time. However don’t dilly dally. The lines at security can get pretty bad. If your outbound flight is in terminal C be sure to ask an agent about the secret entrance.

Great site keep up the excellent work. I am looking for a game/games that are suitable to get up one unit. i.e. If i bet 10 dollars I would like my original 10 back plus the 10 I bet. I belives the craps pass line would be the best bet to do this. I dont mind going down a little money if I have to, but I would expect to get positive at least the amount of my original bet sometime before the end of the shoe or game. Is craps the answer...or is there a card game that may be better? Thanks for your time and effort.

Thanks. If your goal is to win just one unit I agree craps is the best place to start. The don’t pass is slightly better than the pass. However if you lose your first bet I would switch to blackjack. Only when exactly where you started would I go back to craps. This is because you won’t need to double or split to win just one unit, and a hit/stand only blackjack game has a house edge of about 2.5%.

Re buy bet vig. in craps. Most online casinos on a $10 4/10 buy bet return 19.00, they claim a 5% vig. I always thought this was on the bet amount, not on the win amount which means the return should be 19.50 which is how Bodog does it. Are the other casinos wrong?

Yes. Assuming the commission is paid only on a win then it should be applied to the bet amount, not the win amount.

As a craps player I enjoy playing the more player friendly version of crapless found in Tunica. With their version of buy bets the vig is paid only on a winning bet. With that I am wondering on the ev of a $10 3 or 11 paying $29 after vig, or a $10 2 or 12 paying $59 after the vig.

A $10 buy bet on the 3 or 11 would pay $29.50, and on the 2 or 12 it would pay $59.50. The house edge on the 3 or 11 is 1.25%, and on the 2 or 12 it is 0.71%.

I recently visited Imperial Palace, Las Vegas and played craps, and was very surprised to find out that horn bets there pay 15 FOR 1 and 30 FOR 1 (14 to 1 and 29 to 1 respectively) How dramatically does this change the house edge, and do you know of any other casinos that do not pay the standard of 15 to 1 and 30 to 1?

As if the normal odds were not bad enough. Shame on the IP. Lowering the 2-number horn bets from 15 to 14 increases the house edge from 11.11% to 16.67%. Lowering the 1-number horn bets from 30 to 29 increases the house edge from 13.89% to 16.67%. No, I don’t know of any other casinos that do this, but I don’t look for this sort of thing either.

What is the "buffalo" bet in craps?

I asked the Bone Man of Next Shooter.com. He said it is all the hard ways and either the 7 or 11. It is referred to either "buffalo -- seven" or "buffalo -- eleven."

I understand the put bet is a good bet when the odds are 10x or more. So, if the point is a 6 or 8, bet $5 plus $50 free odds for example. My question is, why wait for the point? Why not bet on the passline, and then if the point is a 6 or 8, bet the full 10x odds? This way you get the advantage of the odds, plus the advantage of the come out 7 / 11 roll. Am I missing something?

I agree. I don’t recommend making put or place bets, because as you said, the odds are better making a line bet and then maximum odds. However, some people absolutely must bet on the points directly. If that is a given, I advocate making the best bet between the place, buy, and put, which I explain in greater depth in my craps section.

I am a crap dealer in a casino that offers the fire bet (pay table A, 20.83% edge). The limits on the fire bet are $1-$5 (for players and dealers), but the dealers are limited to $1000 payout. What does that do to the house edge?

That is very tight to limit the dealers like that. On a $2 bet the house edge goes up to 29.02%, and a $5 bet it is 41.94%.

I was at Foxwoods casino this weekend playing craps. The dealer accepted a $3 commission on a $75 buy bet. What would this make the house edge?

Assuming the commission is always payable, then the house edge would be 4% at a $3 commission (3/75), compared to 5.33% at a $4 commission.

It seems to me 3-4-5x odds are a scam because the player is paid out the same regardless if the point is 4 or 6. If a player bets 2x odds, regardless of the point, at least he is rewarded a higher payout. What is your opinion on this?

With 3-45-x odds, you don’t have to bet as much on a point of four as a six, yet the reward is the same. With all gambling, you shouldn’t just look at the reward, you also have to look at how much you are risking.

What is the house edge of the “anything but seven” strategy in craps? Under this strategy, after a point is rolled, the player bets the field and places the 5, 6, and 8, unless the point is already one of those numbers. Also, what is the house edge of the “anything but the point” strategy, in which the player does the same thing after a don’t pass bet?

According to my calculations, the “anything but the point” strategy will have a loss of 0.11988 units per pass bet resolved. The expected amount bet, counting only resolved bets, is 5.09865 units. So, defining the house edge as the ratio of expected loss to total bets resolved, it is 2.351%. Under the “anything but the point” strategy, the house edge is 9.19394/5.09865 = 2.341%, which includes the push on 12 as a resolved bet.

What would be the house edge of a pass and don’t pass bet after a point is established?

After a pass bet, on a point of 4 or 10, the house edge swings to 33.33%. On a 5 or 9, it is 20%, and on a 6 or 8, it is 9.09%. The player has these same advantages on a don’t pass bet, once a point has been established.

I was playing craps at Harrah’s in St Louis, and noticed they have added place bet positions for the 2, 3, 11, and 12 to the table. I don’t remember what they paid. Do you know the odds for these bets? Thanks.

In Crapless Craps the 3 and 11 pay 11 to 4. Using the same formula, t=3, and a=2.75, so the house edge is 0.25/4 = 6.25%.

Most of our casinos collect the commission on buy bets on a win only. They only charge $1 for wager of $20–$39. Does this actually make buying the 4 or 10 for $33–$39 a better bet than placing the 6 or 8?

Let’s find the breakeven point. The expected value of placing the 6 or 8 is [(5/11)*7 + (6/11)*-6]/6 = -(1/11)/6 = -1.52%.

Let b be the buy bet. The expected value is [(1/3)*(2b-1) + (2/3)*-b] / b = (-1/3)/b

Equating the two bets:

-1/66 = (-1/3)/b

3b = 66

b = 22

So, at a bet of $22 the odds are the same. The odds are better on the buy bet for bets of $23 to $39.

Playing craps at Barona casino I had three come bets working with odds. After the pass line point won, the come out was a seven so I expected to lose my come bets and have my odds returned to me. Instead, the boxman said for a price I could keep my come bets & odds up. Surprised, I said yes and paid up $15 since the table was hot. Opinion? Is this available in Vegas?

I assume the come bets were $5 each. Under this assumption, you were actually making put bets, which you can do at any time, at any casino that allows them. Put bets are generally allowed in Nevada, and generally not allowed in Atlantic City. I normally don’t recommend put bets, because you skip the come out roll on the come bet, which has a 22.2% chance of winning, and only a 11.11% chance of losing. My advice is to start over with new come bets.

What is the average number of winning pass line bets a shooter will make?

Here you go:

Average winning pass line bets = 1.244898

Average total pass line bets = 2.52510

Expected points made = 0.683673

Expected rolls = 6.841837

Recently, I was in a casino in Oklahoma, playing craps. There are a few rule changes to the "normal" craps rules. Instead of dice, the casino uses a deck of 54 cards (Aces through 6). The stickman will ask you for a number between 1-3. He’ll then burn that number of cards and then put the next two face up. That becomes the dice roll. After approximately half to 3/4 of the deck has been used, a new deck is brought in and the old deck is shuffled.

Also, if you want to make a bet on the table, you’ll have to pay a dollar ante to the casino. You pay only $1 per come out roll. Once the point is established you can bet as much/little as you’d like without another payment of ante. The table limits are from 5 dollars to 300 dollars.

If the dealer went through 39 cards (out of 54) before re-shuffling the deck, you can count/see 26 of those cards. Previously, you’ve said that if there are a lot of 5s and 6s left in the deck, you would bet the "yo 11" bet. Can you develop a more effective strategy and way for betting in this casino? I truly feel that this game is beatable. Would a count of high/low, like counting cards in blackjack, work? Thanks.

I still say hop bets, like the yo-11, are the way to go. Using chips, you could keep track of how many cards of each face are left in the decks.

With 26 unseen cards, if any one face had 6 left in the deck, you would have a 43.1% advantage on a hard hop bet (two of the same face), assuming it paid 30 to 1. With only 5 left, the house would have a 4.6% advantage.

The easy hops are even more exploitable. If the two dice sides in majority have at least 10 left combined, both with a minimum of 3 left, out of 26 unseen cards, then make an easy hop bet on those two numbers. If two numbers have 5 left, you will have a 23.1% advantage. If one has 4 and one has 6, you will have an 18.2% advantage. If one has 3 and one has 7, you will have a 3.4% advantage. All this assumes easy hop bets pay 15 to 1. None of the above takes into consideration the $1 fee. As long as you are making large bets, it won’t make much difference.

My question is on craps. I know that the Fire Bet is a lousy bet, but I bet it anyway, when I am rolling. Well, I got lucky and hit my four points, and was on the fifth point. I had won $75, and was on my way to winning $750, if I hit the 5th point. My other bets were $5 on the pass and $20 to win $30 on the 5. Having established the 5th point, which was a 5, I realized that I had a 2/5 chance of hitting it, for a net win of $785. I also realized that I had a 3/5 chance of not hitting it, for a net win of $25. If I wanted to hedge my bets, what is the largest win I could lock in? Also, what are your thoughts on this strategy.

Unless life changing amounts of money are involved, I disapprove of hedging, per my seventh commandment of gambling.

I'm going to ignore the fact that if you hit the 5 you could hedge more to lock in an even larger win, and just look at this as if it ended after a 5 or 7. At this point your net will will be $785 or $50. You should start by taking down the odds bet. That will change the scenario to winning $755 or $70. Then you should lay the odds on the 5. Let b represent your lay bet against the 5. If you lose the bet, you’ll have $755-$b. If you win the bet, you’ll have $70 + (19/31)×$b. So, equate the two sides, and solve for b:

755-b = 70 +(19/31)×b

685 = (50/31)×b

b=424.7

That will lock in a win of $330.30. So, if rounding were not an issue, then lay $424.7 against the 5. However, rounding always is an issue, so I would lay $403 against the 5 ($390, plus $13 commission on possible win of $260).

How does the house percentage change when I buy the 5&9 for $50 if I "pay the vig after I win"? If I place them for $50, then I get $70 in return. If I throw the same $50 out to the dealer and say "buy it," then I’ll win $75 minus the $2 vig, for a $73 win. It seems that buying the 5&9 in this manor would lower this house edge. Paying after you win on buy bets is a very common practice these days. The player could buy the 5&9 for as low as $20 and still have a little edge. Place for $20, win $28. Buy for $20, win $30 minus the $1 vig =$29. Thanks for any insight.

Comparing buy bets to place bets, on points of 6 and 8, the place bet always has the lower house edge. On points of 4 and 10, the buy bet always has the lower house edge. On points of 5 and 9, it depends on whether the commission is always paid or only paid on wins. If the commission is paid up front, as is usually the case, then place bets are better; otherwise, buy bets are. To be more specific, the house edge placing the 5 and 9 is 4.00%. The house edge buying the 5 and 9, when the commission is paid up front, is 4.76%. When it is only paid on wins, the house edge is 2.00%.

If I make a pass or come bet every single roll in craps, then how many active bets will I have at any one time? I ask because I’d like to know what a fair average bet should be for such a player.

The average number of bets such a player will have on the table for any one throw is 3.6. If the player is betting $10 at a time, for example, then a fair average bet would be $36.

At the Boulder Station, there is a side bet in craps called the "Replay" bet. It pays if the shooter makes the same point at least three times. If the shooter achieves a win on two or more different numbers, only the highest win is payable. I’m including the pay table. What are the odds on this bet?

I checked, and indeed they do have that bet there. The return table below shows a house edge of 24.8%.

### Replay

Event | Pays | Probability | Return |

4 or 10 four or more times | 1000 | 0.000037 | 0.036892 |

5 or 9 four or more times | 500 | 0.000207 | 0.103497 |

4 or 10 three times | 120 | 0.000524 | 0.062847 |

6 or 8 four or more times | 100 | 0.000698 | 0.069815 |

5 or 9 three times | 95 | 0.001799 | 0.170927 |

6 or 8 three times | 70 | 0.004294 | 0.300609 |

Loser | -1 | 0.992441 | -0.992441 |

Total | 1.000000 | -0.247853 |

At the City of Dreams casino in Macau they have an "any hardway" bet, which pays 4 to 1. What is the house edge on that?

I get a probability of winning of 2/11, for a house edge of 9.09%.

This question was raised and discussed in the forum of my companion site Wizard of Vegas.

If the Fire Bet were offered in Crapless Craps, what would be the probability of winning?

As a reminder, in Crapless Craps the 2, 3, 11 and 12 do not immediately resolve a pass line bet, but are considered points, much like the 4, 5, 6, 8, 9, and 10.

The first step in my solution requires calculating the probability of any given outcome of the passline bet, as follows.

### Crapless Craps Possible Outcomes

Event | Formula | Probability | Fraction |
---|---|---|---|

Come out roll | 1/6 | 0.166667 | 1/6 |

Point 2 win | (1/36)*(1/7) | 0.003968 | 1/252 |

Point 3 win | (2/36)*(2/8) | 0.013889 | 1/72 |

Point 4 win | (3/36)*(3/9) | 0.027778 | 1/36 |

Point 5 win | (4/36)*(4/10) | 0.044444 | 2/45 |

Point 6 win | (5/36)*(5/11) | 0.063131 | 25/396 |

Point 8 win | (5/36)*(5/11) | 0.063131 | 25/396 |

Point 9 win | (4/36)*(4/10) | 0.044444 | 2/45 |

Point 10 win | (3/36)*(3/9) | 0.027778 | 1/36 |

Point 11 win | (2/36)*(2/8) | 0.013889 | 1/72 |

Point 12 win | (1/36)*(1/7) | 0.003968 | 1/252 |

Point 2 loss | (1/36)*(6/7) | 0.023810 | 1/42 |

Point 3 loss | (2/36)*(6/8) | 0.041667 | 1/24 |

Point 4 loss | (3/36)*(6/9) | 0.055556 | 1/18 |

Point 5 loss | (4/36)*(6/10) | 0.066667 | 1/15 |

Point 6 loss | (5/36)*(6/11) | 0.075758 | 5/66 |

Point 8 loss | (5/36)*(6/11) | 0.075758 | 5/66 |

Point 9 loss | (4/36)*(6/10) | 0.066667 | 1/15 |

Point 10 loss | (3/36)*(6/9) | 0.055556 | 1/18 |

Point 11 loss | (2/36)*(6/8) | 0.041667 | 1/24 |

Point 12 loss | (1/36)*(6/7) | 0.023810 | 1/42 |

If you add all the ways to lose, you get 7303/13860 = apx. 0.526912.

The next step in my solution to this problem uses calculus. It relies on the fact that the answer would be the same if there was a random period of time between pass line bets resolving. Let's call the mean time between bet resolutions 1 and distributed by the exponential distribution, meaning it has a memoryless property.

Let x represent the time since the shooter started his turn.

The probability the shooter did not get a point 2 win is exp(-x/252). Thus, the probability that get got at least one point-2 wins is 1-exp(-x/252).

The probability the shooter did not get a point 3 win is exp(-x/72). Thus, the probability that get got at least one point-3 wins is 1-exp(-x/72).

The probability the shooter did not get a point 4 win is exp(-x/36). Thus, the probability that get got at least one point-4 wins is 1-exp(-x/36).

The probability the shooter did not get a point 5 win is exp(-2x/45). Thus, the probability that get got at least one point-5 wins is 1-exp(-2x/45).

The probability the shooter did not get a point 6 win is exp(-2x/45). Thus, the probability that get got at least one point-6 wins is 1-exp(-x/72).

Note these probabilities are the same for 8 to 12, so we can square them to show they have been achieved twice each.

The probability the shooter did not lose is exp(-7303x/13860).

The probability of losing is 7303/13860.

We can solve for the problem by integrating from t = 0 to infinity of the probability of the product of all winning requirements have been met, the losing outcome has not been met, and the probability of losing given a bet has been resolved.

The function being integrated is exp(-7303x/13860)*(1-exp(-x/252))^2*(1-exp(-x/72))^2*(1-exp(-x/36))^2*(1-exp(-2x/45))^2*(1-exp(-25x/396))^2*(7303/13860).

Put that into an integral calculator like the one at integral-calculator.com. Remember to put in the limits from 0 to infinity. The answer will be what is expressed as the answer above.