Inside Numbers vs 7

Hi, I'm making a theory on Inside Numbers (5,6,8,9 combined) vs 7 and I need great mathematicians here to validate this.

Given the following assumptions:
1. The combined probability of 5 (4 out of 36), 6 ( out of 36), 8 (4 out of 36), 9 (4 out of 36) are 18 out of 36 which is 50%
2. The probability of 7 is (6 out of 36) which is 16.67%

What is the probability of hitting the inside numbers AT LEAST TWICE before hitting a 7? (After a point has been established)
Examples:
1. point 6 established, rolled 8, 9 then 7-out. (2 inside #s made before 7 out)
2. point 8 established, rolled 5, 6. point made (2 inside #s made before 7 out)
3. point 5 established, rolled 6, 8. 8. 9. then 7 out (4 inside #s made before 7 out)
4. point 9 established, rolled 6, 5. 5. 8. 6, 8 then 7 out (6 inside #s made before 7 out)

Appreciate your help. Thank you!

What kind of theory ?

If only Inside Numbers and Red are considered because all other numbers are irrelevant, then the universe has 18 plus 6 relevant outcomes.

Probability of exactly zero Inside Numbers before Red is ( 6 / 24 )

Probability of one or more Inside Numbers before Red is ( 18 / 24 )

Probability of two or more Inside Numbers before Red is ( 18 / 24 ) ( 18 / 24 )

Probability of ( N ) or more Inside Numbers before Red is ( 18 / 24 ) ^ ( N )

Quote: poli2k01

... 1. The combined probability of 5 (4 out of 36), 6 ( out of 36), 8 (4 out of 36), 9 (4 out of 36) are 18 out of 36 which is 50%

Well, before you work on too many calculations for your theory, you might check your figures in parentheses (above) for the 6 and 8. One is missing and the other is wrong. Nevertheless, I think your total probability for a single roll is correct.

@pwcrabb - Thank you for the computation. I think this makes sense.

So given the probability of two or more Inside Numbers before Red is ( 18 / 24 ) ( 18 / 24 ) = 56.25%, what will be the house edge given that I pull down ALL my inside bets ALWAYS after 2 hits and wait it out until either 1. A point has been made OR 2. 7-out?

I must have hit the send button too soon. The correct one is below, but it total is still at 50%

The combined probability of 5 (4 out of 36), 6 (5 out of 36), 8 (5 out of 36), 9 (4 out of 36) are 18 out of 36 which is 50%

Anticipated House Edge for a Place Bet may be differentially calculated based upon duration of anticipated Exposure, after which Takedown is to be executed immediately and permanently. Wizard elsewhere has published the results of the first two of the following. You may find his analyzes and extend them.

1. Exposure until Resolution. This is the textbook result. Few players play this way.
2. Exposure for exactly one roll.
3. Exposure for exactly two rolls.
4. Exposure for exactly ( N ) rolls.
5. Exposure until loss of initial investment. Most players play this way.

Quote: pwcrabb

1. Exposure until Resolution. This is the textbook result. Few players play this way.
…
5. Exposure until loss of initial investment. Most players play this way.
link to original post

I may be totally off base here, but if players press their bets after a win, then I believe neither of the statements that ‘few players play this way’ or ‘most players play this way’ are correct.

Of course, it then gets very complicated…

Thank you. Where did wizard publish his results for the first 2 scenarios?

WizardOfOdds.com

Hello Are you referring to this -> /games/craps/appendix/2/ ?

If yes, given the house edge PER roll are the ff.:
Place 6 and 8 0.46%
Place 5 and 9 1.11%
Place 4 and 10 1.67%

If I put in place bet of 6 once before pulling down bets permanently (until point made or 7 out), house edge is 0.46%
If I put in place bet of 6 twice before pulling down bets permanently (until point made or 7 out), house edge is 0.46% x 2 = 0.96%, correct?
If I put in place bets of 6, 8, 5, 9, total house edge per roll is 0.46% + 0.46% + 1.11% + 1.11% = 3.18%, correct?

Please confirm if my understanding is correct. Once again. Thank you for shedding light on this subject.

Quote: poli2k01

Hello Are you referring to this -> /games/craps/appendix/2/ ?
link to original post

Link for the convenience of our esteemed forum members:
https://wizardofodds.com/games/craps/appendix/2/

Correct on Wizard source but incorrect on your application and extension. Mere arithmetic will not cut mustard. Moreover please know that each possible assortment of negative numbers will also itself be negative. There are no numerical solutions which will force wins at any of the games of craps. Your strategy of Inside Only for Two Hits Only is not exceptional.

Place 8 House Edge for various Exposures before Walkaway
Only one roll 0.4630 percent
Only two rolls 0.8488
Only three rolls 1.1703
Only until decision 1.5152
Until inevitable loss 2.7778