SteveStew
SteveStew
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Joined: Nov 4, 2012
November 5th, 2012 at 8:21:15 PM permalink
Hi Everyone,

I have a Craps math question.

I need some help calculating the odds of 5 consecutive hands where a 7 is rolled before either a 6 or 8 after the point is established. We only care about the throws after the point is established. Disregard the Pass Line bets.

Ah, but there is a kicker, let's assume the following data:

6 or 8 comes up 30.28% instead of the random 27.78%
7s come up 14.83% instead of the random 16.67%

The calculation is for analyzing a 6/8 Regression-Progression betting strategy.
The following is inconsequential to figuring out the odds, but I thought I would mention it in case some one was wondering.

Every consecutive hand after a point is established that has a 7 thrown before a 6 or 8 will cause the next hand to increase the Place bet on the 6 and 8. Every hand after a point is established that throws a 6 or 8 before a 7 will cause the next hand to start at the original Place bet on the 6 and 8. If 5 consecutive hands roll 7s before either a 6 or 8 is thrown after the point is established, the progression stops and a big loss occurs. The strategy wins money on every session until the 5 point-7s occur.

Like all Progressions, it is impossible to alter bets to overcome the house odds, but given the slight edge mentioned above, what are the odds that 5 point-7 outs in a row will occur after the point is established? Additionally, will that edge be enough to overcome the house edge over the long term?

Thank you in advance for your help!
ChesterDog
ChesterDog
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Joined: Jul 26, 2010
November 6th, 2012 at 5:35:35 AM permalink
Quote: SteveStew

...let's assume the following data:

6 or 8 comes up 30.28% instead of the random 27.78%
7s come up 14.83% instead of the random 16.67%

...given the slight edge mentioned above, what are the odds that 5 point-7 outs in a row will occur after the point is established? Additionally, will that edge be enough to overcome the house edge over the long term?...



I get 0.38% {=[0.1483/(0.3028+0.1483)]^5} vs the random-case result of 0.74% {=[0.1667/(0.2778+0.1667)]^5}.

And I get a huge (player's) edge of 6.28% for your data, so I would flat-bet the place bets on the 6 and 8 instead of doing a progression.
SteveStew
SteveStew
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Joined: Nov 4, 2012
November 6th, 2012 at 7:27:25 AM permalink
Hi ChesterDog,

Thank you so much for the data! ( I really like the the way you showed the math as well.)

I agree with you, a Progression-Regression would not be the best way to bet that advantage.

I really appreciate your time!

Cheers,

SteveStew
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