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ThatDonGuy
ThatDonGuy
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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)2 x 98/165 + 4 x (67/165)3 x 98/165 + ...
= 98/165 x (1 + 2 x 67/165 + 3 x (67/165)2 + 4 x (67/165)3 + ...)
= 98/165 x (1 + 67/165 + (67/165)2 + ...)2
= 98/165 x (1 / (1 - 67/165))2
= 98/165 x (165/98)2
= 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.

RouletteProdigy
RouletteProdigy
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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.
Last edited by: RouletteProdigy on May 9, 2017
ThatDonGuy
ThatDonGuy
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May 9th, 2017 at 6:07:27 PM permalink
Quote: RouletteProdigy

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.


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!
RouletteProdigy
RouletteProdigy
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May 9th, 2017 at 6:19:42 PM permalink
Quote: ThatDonGuy

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!



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

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

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:

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.

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:

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

Thank U

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.

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.
"Baccarat is a game whereby the croupier gathers in money with a flexible sculling oar, then rakes it home. If I could have borrowed his oar I would have stayed." .......... Mark Twain
ThatDonGuy
ThatDonGuy
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May 10th, 2017 at 8:38:55 AM permalink
Quote: RouletteProdigy

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.


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 rollsProbabilityProb of sevening out
Neither is a point1/90
First one not a point, second one is2/91/6
First one is a point, second one sevens out1/90
First one is 4/10, second one makes the point1/720
First one is 5/9, second one makes the point2/810
First one is 6/8, second one makes the point25/6480
First one is 4/10, second one is not the point or 71/81/6
First one is 5/9, second one is not the point or 713/811/6
First one is 6/8, second one is not the point or 7125/6481/6

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.
YoEleven
YoEleven
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June 19th, 2017 at 5:32:11 PM permalink
I am new to the forum , and I have a question regarding numbers and come bets. Does anyone have the odds or calculations for a come bet to repeat ? In other words , including all numbers, 4,5,6, 8, 9,10 that hit during a roll that are come bets, what are the collective odds of a come bet hitting or repeating before a 7 out? I assumed it would be true odds to repeat , and somehow I came up with 1.1 come bet repeated per each 7 out. Does anyone have any calculations on this?

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).

[/spoiler]
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.

Wizard
Administrator
Wizard
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June 19th, 2017 at 6:21:59 PM permalink
Quote: RouletteProdigy

Wizard,

When it comes to math I am a bit thick headed.

If I understand you correctly, the average is 1.619839371 inside numbers after the come out?



That is correct.
It's not whether you win or lose; it's whether or not you had a good bet.
ThatDonGuy
ThatDonGuy
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June 20th, 2017 at 5:57:06 AM permalink
Quote: YoEleven

I am new to the forum , and I have a question regarding numbers and come bets. Does anyone have the odds or calculations for a come bet to repeat ? In other words , including all numbers, 4,5,6, 8, 9,10 that hit during a roll that are come bets, what are the collective odds of a come bet hitting or repeating before a 7 out? I assumed it would be true odds to repeat , and somehow I came up with 1.1 come bet repeated per each 7 out. Does anyone have any calculations on this?


Could you give me an example (with numbers, preferably) of what you mean by "a come bet repeating"? I am not quite sure what you are asking.

For that matter, could you show how you get "1.1 come bets repeated per each 7 out"?
YoEleven
YoEleven
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June 20th, 2017 at 7:17:57 AM permalink
Example. in a roll : 5 , 6, 11, 8, 6, 3, 9, 7 out. The 6 was a come bet that was repeated. I just assumed that the odds on a number repeating would be true odds, such as 6, 8 is 6 to 5 , 5, 9 is 3 to 2, and 4, 10 is 2 to 1. I added up all of the odds collectively and divided by number of average rolls per shooter. Not sure if it is correct, but was wondering , if I am playing pass line and 2 come bets strategy, what is the average or odds of 1 come bet hitting or repeating for a payoff, or 2 come bets hitting or repeating for a payoff or the actual average of all come bets hit or repeated during a roll?

If the roll was 8, 4, 5, 4, 6,9, 5, 11, 7 out , that would be 2 come bets ( 4, 5) were repeated or hit during that roll

Was just curious what are the odds of come bets repeating and paying?

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