# Craps - Other Bets

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.

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.

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.

### 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.

### 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 |

### 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% |

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%.

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.

Average winning pass line bets = 1.244898

Average total pass line bets = 2.52510

Expected points made = 0.683673

Expected rolls = 6.841837

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.

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.

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).

### 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 |

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?

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.