May 9th, 2017 at 2:37:46 PM
permalink

Quote:mustangsally

counting all rolls

I get an average of 4.26 rolls

counting all rolls EXCEPT 5689 on the come out roll.

I get an average of 3 rolls

this should come close to calculated values

The expected number if you don't count comeout rolls is 3, as shown above.

If you do count them:

First, determine the probability of making a point from a comeout roll that establishes a point.

This is 1/4 (the probability that a point will be 4 or 10) x 1/3 (the probability of making that point) + 1/3 (the probability that a point will be 5 or 9) x 2/5 (the probability of making that point) + 5/12 (the probability that a point will be 6 or 8) x 5/11 (the probability of making that point) = 67/165

The probability of sevening out once you have already established a point is 1 - 67/165 = 98/165

The expected number of comeouts where points are established before sevening out is:

1 x 98/165 + 2 x 67/165 x 98/165 + 3 x (67/165)

= 98/165 x (1 + 2 x 67/165 + 3 x (67/165)

= 98/165 x (1 + 67/165 + (67/165)

= 98/165 x (1 / (1 - 67/165))

= 98/165 x (165/98)

= 98/165

The probability that a comeout that is a point number is an inside number is 3/4 (there are 6 ways to roll a 4 or 10, 8 to roll a 5 or 9, and 10 to roll a 6 or 8, so the probability is 18/24 = 3/4), so the expected number of inside numbers on comeouts = 98/165 x 3/4 = 495/392 = about 1.26275,

Add this to the expected 3 inside numbers not on comeouts, and the expected number of inside numbers when you include comeouts = about 4.26275.

May 9th, 2017 at 5:14:24 PM
permalink

Thank you everyone for all the help.

I like to play Craps and plan to play the don't pass with continuous Don't come bets but I want to lay odds also. If anyone can tell me at what point should I lay the odds. I was thinking if the average player rolls 8.5 should I not lay odds after the 4th or 5th roll? Would that not give me some advantage (or rather help me lose less) After the 5th roll there is a 50% chance of the player seven out on the 6th roll. Not sure if wincraps can provide the optimal time to lay odds on the DP and DC's.

One more thing I will only lay odds on the inside numbers and if I lose my odds twice on DC or DP I will take off my remaining odds and leave my flat bets on

Thanks for the input everyone.

I like to play Craps and plan to play the don't pass with continuous Don't come bets but I want to lay odds also. If anyone can tell me at what point should I lay the odds. I was thinking if the average player rolls 8.5 should I not lay odds after the 4th or 5th roll? Would that not give me some advantage (or rather help me lose less) After the 5th roll there is a 50% chance of the player seven out on the 6th roll. Not sure if wincraps can provide the optimal time to lay odds on the DP and DC's.

One more thing I will only lay odds on the inside numbers and if I lose my odds twice on DC or DP I will take off my remaining odds and leave my flat bets on

Thanks for the input everyone.

Last edited by: RouletteProdigy on May 9, 2017

May 9th, 2017 at 6:07:27 PM
permalink

Quote:RouletteProdigyI like to play Craps and plan to play the don't pass with continuous Don't come bets but I want to lay odds also. If anyone can tell me at what point should I lay the odds. I was thinking if the average player rolls 8.5 should I not lay odds after the 4th or 5th roll? Would that not give me some advantage (or rather help me lose less) After the 5th roll there is a 50% chance of the player seven out on the 6th roll.

Excuse me? "After the fifth roll," the chance of the shooter sevening out on the sixth roll is the same as it was for all of the shooter's previous rolls.

Like I said before - the dice don't remember what they did in the past. Especially if, for whatever reason, the shooter switches to new dice!

May 9th, 2017 at 6:19:42 PM
permalink

Quote:ThatDonGuyExcuse me? "After the fifth roll," the chance of the shooter sevening out on the sixth roll is the same as it was for all of the shooter's previous rolls.

Like I said before - the dice don't remember what they did in the past. Especially if, for whatever reason, the shooter switches to new dice!

Don, I get what you are saying each roll is independent. It still does not change that in a series of rolls before a player seven out these figures are real.

Roll Probability of Seven Out

1 0.00000000

2 0.11111111

3 0.22788066

4 0.33264746

5 0.42387109

6 0.50278913

7 0.57095589

8 0.62980865

9 0.68060930

10 0.72445344

Why would I not lose less if I layed odds after a certain number of rolls regardless if each roll has no memory. If I layed odds on roll number 2 the player is less likely to seven out vs. laying odds after the 5th roll the player is more likely to seven out.

One does not mesh with the other. Perhaps this is where most of us non math gamblers go wrong. Just asking.

Thank U

May 10th, 2017 at 2:34:48 AM
permalink

Ah. I'm sure many of us suspected you were working out a strategy based on your question. If you were asking about a bet with a house edge, the only reason you would lose less is by putting less into action overall - if indeed it was less total action. But for an odds bet, there is no edge and it doesn't matter except for how much Variance you run into.Quote:RouletteProdigyWhy would I not lose less if I layed odds after a certain number of rolls regardless if each roll has no memory.

No. If the dice have no memory, then they don't know what roll it is. The chances of a 7-out are the same each time for the next roll. The reason the odds are changing like you show is *not* from the perspective of "the next roll" but from the perspective of such a sequence altogether. You can't make a bet by saying "I want to make a bet that the 7-out [does/doesn't] take 10 rolls"Quote:If I layed odds on roll number 2 the player is less likely to seven out vs. laying odds after the 5th roll the player is more likely to seven out.

They do mesh, you are struggling with what is a counter-intuitive nature of probability to you at this stage. Keep asking questions and keep an open mind and you will find it becomes intuitive.Quote:One does not mesh with the other. Perhaps this is where most of us non math gamblers go wrong. Just asking.

Thank U

One thing to realize is that billions of gamblers before you have studied Craps, examining each angle, every nook and cranny, and bled all over the table with every conceivable concept. If you were on to something here, it would be known already: "just wait till the 7-out is overdue and lay the odds then for better chances" would be the mantra. Of course, if this was true, you would have an edge, wouldn't you? Not just a guy losing less money, but making money. Right from the get-go you would just not be allowed to do it.

"We thank with brief thanksgiving
Whatever gods may be
That no man lives forever,
That dead men rise up never"
And no gambler sees the long run
........apologies to Swinburne for that last line

May 10th, 2017 at 8:38:55 AM
permalink

Quote:RouletteProdigyDon, I get what you are saying each roll is independent. It still does not change that in a series of rolls before a player seven out these figures are real.

Roll Probability of Seven Out

1 0.00000000

2 0.11111111

3 0.22788066

4 0.33264746

5 0.42387109

6 0.50278913

7 0.57095589

8 0.62980865

9 0.68060930

10 0.72445344

Why would I not lose less if I layed odds after a certain number of rolls regardless if each roll has no memory.

I think the probabilities you list are the probabilities of sevening out in that many rolls or fewer. This is certainly true for the third roll.

Obviously, you can't seven out on the first roll, which is a comeout.

The probability of sevening out on the second roll = (the probability of rolling a point number on the first roll) x (the probability of a seven on the second roll) + (the probability of rolling a natural/craps on the first roll) x (zero, since the second roll is now a comeout) = 2/3 x 1/6 + 1/3 x 0 = 1/9, which is what you show.

The probability of sevening out on the third roll = (the probability that the third roll is not a comeout roll) x 1/6.

The probability of the third roll not being a comeout roll is determined as follows:

First two rolls | Probability | Prob of sevening out |
---|---|---|

Neither is a point | 1/9 | 0 |

First one not a point, second one is | 2/9 | 1/6 |

First one is a point, second one sevens out | 1/9 | 0 |

First one is 4/10, second one makes the point | 1/72 | 0 |

First one is 5/9, second one makes the point | 2/81 | 0 |

First one is 6/8, second one makes the point | 25/648 | 0 |

First one is 4/10, second one is not the point or 7 | 1/8 | 1/6 |

First one is 5/9, second one is not the point or 7 | 13/81 | 1/6 |

First one is 6/8, second one is not the point or 7 | 125/648 | 1/6 |

The probability of the third roll sevening out is (2/9 x 1/6) + (1/8 x 1/6) + (13/81 x 1/6) + (125/648 x 1/6) = 227 / 1944 = 0.11676955

If you add this to the probabilities of sevening out on the first two rolls that you list, the total is 0.22788066.

Note that my number includes the cases where you seven out on the second roll.

Also note that this is the probability of sevening out on the third roll not already knowing what the previous two rolls were (i.e. you make the bet that the third roll will not seven out before the first roll, not after the second roll).

After nine rolls, the probability of sevening out on the 10th roll is 1/6 if a point is established or zero if it is a comeout roll. It doesn't matter what the first eight rolls were.