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# Average bet resolved per throw in craps

## Introduction

Suppose you want to run a lot of money through an electronic craps machine at minimal cost and variance. Such machines generate points on bets resolved only. This page examines the average bet resolved per throw playing line bets only according to the side, odds taken, and maximum points covered.

The following multiples of odds bets are considered. Recall that when laying odds, you may bet according to the what would result in a win of the multiple allowed.

- 0X: No odds bets taken. This might be appropriate if the odds bet don't earn points.
- 1X: Odds bet equal to pass or come bets taken if a point is established. After a don't pass or don't come the player lays 2x on a 4 or 10, 1.5x on a 5 or 9, and 1.2x on a 5 or 6.
- 2X: Odds bet equal to 2x pass or come bets taken if a point is established. After a don't pass or don't come the player lays 4x on a 4 or 10, 3x on a 5 or 9, and 2.4x on a 5 or 6.
- 3X-4X-5X: On a pass or come bet, odds bet equal to 3x after a 4 or 10, 4x after a 5 or 9, and 5x after a 6 or 8. After a don't pass or don't come the player lays 6x after any point.

The game considers these strategies:

- Player establishes one point only. In other words the player never makes don't pass or don't come bets.
- Player establishes two points only. This would be one pass/don't pass bet plus one come/don't come.
- Player establishes two points only. This would be one pass/don't pass bet plus two come/don't come bets. This is commonly known as the "Three Point Molly."
- Player establishes the maximum points he can. This is accomplished by making a line bet on every roll.

The following table shows the average bets resolved per roll if the player does not make odds bets, according to the maximum number of points bet on.

### 0X Odds

Maximum Points | Do | Don't |
---|---|---|

1 | 0.296230 | 0.296230 |

2 | 0.533214 | 0.533214 |

3 | 0.715104 | 0.715104 |

Maximum | 1.000000 | 1.000000 |

The following table shows the average bets resolved per roll if the player bets 1X the line bet on the odds*, according to the maximum number of points bet on.

### 1X Odds

Maximum Points | Do | Don't |
---|---|---|

1 | 0.493716 | 0.592460 |

2 | 0.888689 | 1.066427 |

3 | 1.191840 | 1.430208 |

Max | 1.666667 | 2.000000 |

The following table shows the average bets resolved per roll if the player bets 2X the line bet on the odds*, according to the maximum number of points bet on.

### 2X Odds

Maximum Points | Do | Don't |
---|---|---|

1 | 0.691203 | 0.888689 |

2 | 1.244165 | 1.599641 |

3 | 1.668576 | 2.145311 |

Max | 2.333333 | 3.000000 |

The following table shows the average bets resolved per roll if the player bets 3-4-5X the line bet on the odds*, according to the maximum number of points bet on.

### 3X-4X-5X Odds

Maximum Points | Do | Don't |
---|---|---|

1 | 1.036804 | 1.481149 |

2 | 1.866248 | 2.666068 |

3 | 2.502863 | 3.575519 |

Max | 3.500000 | 5.000000 |

Footnote:

*: In craps, the maximum the player may lay odds after a don't pass or don't come bet is relative to the win. For example, if 2X odds are allowed, the player may lay 4x on a 4 or 10, 3x on a 5 or 9, and 2.4x on a 6 or 8. Note the win is works out to 2x the don't pass or don't come bet.

## Example

Suppose the player wishes to run through $100,000 on an electronic craps game. The machine counts all bets (including the odds), but only after a bet resolved. The player makes $25 line bets, backs them up with 2x odds, and plays the Three Point Molly (maximum of three line bets) on the don't side.

The 2X table shows us the average units resolved per roll is 1.599641. So, the player would need to make 100,000/25*1.599641 = 2,501 bets on average to complete his goal.

## Methodology

What is the average number of rolls to resolve a line bet? There is always going to be a come out roll.

There is a 6/36=1/6 chance the come out roll is a 4 or 10. After a 4 or 10, 9 rolls will resolve it, for a probability of 9/36 = 1/4. The expected rolls for that to happen is 4 rolls.

There is an 8/36=2/9 chance the come out roll is a 5 or 9. After a 5 or 9, 10 rolls will resolve it, for a probability of 10/36 = 5/18. The expected rolls for that to happen is 18/5 = 3.6 rolls.

There is a 10/36=5/18 chance the come out roll is a 6 or 8. After a 6 or 8, 11 rolls will resolve it, for a probability of 11/36. The expected rolls for that to happen is 36/11 = 3.272727... rolls.

Adding it all up, the expected rolls to resolve a line bet is 1 + (1/6)*4 + (2/9)*3.6 + (5/18)*(36/11) = 557/165 =~ 3.375758 rolls.

Let's look at the case of 0X odds first.

For one point, we have shown it takes 557/165 rolls on average to resolve a pass bet. Thus, the expected bets resolved per throw is the inverse of that, 165/557 =~ 0.296230.

For two points, consider how often the player will have two bets on the table. He will except for come-out rolls after a 7. On average it take 3/2 rolls to establish a point and then 6 rolls to roll a 7, for a total of 7.5 rolls between seven-outs. Of those, 1.5/7.5=80% are come-out rolls. So, wagers increase 80% compared to the one point player. Thus, the average wager is 1.8 * 165/557 = 0.533214.

For three points, consider how often the player will have three bets on the table. He will except after a come out roll when two different points are rolled. How often does that happen? Let's say the point is a 4 or 10. The probability of establishing another point on the next roll is (24-3)/(36-6-3) 21/27. If the point is a 5 or 9, that probability is 20/26. If the point is a 6 or 8, that probability is 19/25. A weighed average of the probability of rolling another point is (6/24)*(21/27) + (8/24)*(20/26) + (10/24)*(19/25) =~ 0.767521. We already established 80% of the time the player will have only one bet on the table. The ratio of time the player will make a bet for a third point is 0.8 * 0.767521 = 0.614017. So, the expected total wager per throw is 0.8 + 0.614017 more than the player making one point only. This is 0.296230 * (1+0.8+0.614017) = 0.715104.

For the maximum points, or a bet every rolls, the answer is simply an average of one unit bet per roll.

For the 1x odds case on the do side, note that 2/3 of the time the player will make a point and thus make an odds bet. The average final wager per come out roll is thus 1 + (2/3) = 5/3. Over the 3.375758 average rolls per come out roll, the average bet per roll is (5/3)/3.375758 = 0.493716. For making two points, multiply that figure by 1.8, much like the case for 0x odds: 0.493716*1.8 = 0.888689. For making three points, multiply the figure for one bet by 1+0.8+0.614017 = 2.414017, for the same reason as 0x odds, which is 2.414017*0.493716 = 1.191840. For maximum points, the average bet per roll will be 1+(2/3)*1 = 5/3.

For the 2x odds case on the do side, note the average wager per pass line bet is 1+(2/3)*2 = 7/3. Thus, the average wager per roll is (7/3)/3.375758 = 0.691203. For 2 and 3 points, multiply that by the same 1.8 and 2.414017 factors as shown for 0x and 1x odds. For maximum points, the average bet per roll will be 1+(2/3)*2 = 7/3.

The logic is the same laying odds. The average bet on the don't pass per roll is the same as the pass at 0.296230. For 1x odds, the average bet is 1+(6/36)*2 + (8/36)*1.5 + (10/36)*1.2 = 2. The average bet per roll is thus 2/3.375758 = 0.592460. For 2x odds, the average bet is 1+(6/36)*4 + (8/36)*3 + (10/36)*2.4 = 3. The average bet per roll is thus 3/3.375758 = 0.888689. For 3x-4x-5x odds, the average bet is 1+(2/3)*6=5. The average bet per roll is thus 5/3.375758 = 1.481149. For more than two and three points bet, multiply by the same 1.8 and 2.414017 factors.

For the maximum points laying the odds, the average bet per roll is 1 with 0x odds. For 1x odds it is 1+(6/36)*2 + (8/36)*1.5 + (10/36)*1.2 = 2. For 2x odds it is 1+(6/36)*4 + (8/36)*3 + (10/36)*2.4 = 3. For 3x-4x-5x odds it is 1+(2/3)*6 = 5.

## Acknowledgement

I would like to thank Ace2 for his help with the math on this page. He discusses it in my forum at Wizard of Vegas.