Thread Rating:

sodawater
sodawater
Joined: May 14, 2012
  • Threads: 60
  • Posts: 3142
March 6th, 2014 at 5:46:25 PM permalink
Quote: vegasrvp

Here is a specific example:
Roll 6
6 is the point
no matter how many rolls I am looking for 7 before the point.

Once 7 is rolled a new point is established;

Roll 8
Again no matter how many rolls a 7 is rolled before the point.

That is two 7 outs before a point is won.

The question is what are the odds that no winner will be rolled prior to two 7 outs.


If we call a winner point or loser 7out no point a session then I am looking to find out how many times is average between back to back loser 7 rolls?

Clearer yet. I HOPE.



I believe the question you are asking is what are the chances the next two shooters each 7-out before making a point.


5/12 of the time, your point will be a 6 or an 8. Your chances of losing the 6 or 8 are 6/11. 5/12*6/11 = 5/22

4/12 of the time, your point will be a 5 or a 9. Your chances of losing the 5 or 9 are 3/5. 4/12*3/5 = 1/5

3/12 of the time, your point will be a 4 or a 10. Your chances of losing the 4 or 10 are 2/3. 3/12*2/3= 1/6

5/22 + 1/6 +1/5 = 98/165 = the chance of sevening out once you establish an unknown point.

losing an unknown point two shooters in a row (not counting rolls where you win or lose on the come out)= (98/165)^2 = 9604/27225 = 35.28%

Edited my post due to error pointed out by MustangSally
michael99000
michael99000
Joined: Jul 10, 2010
  • Threads: 9
  • Posts: 645
March 6th, 2014 at 5:52:58 PM permalink
Ok so odds of...a point rolled on first CO roll + a 7 is rolled before that point is hit + another point is established on first CO roll + a 7 is again rolled before that point is hit
wudged
wudged
Joined: Aug 7, 2013
  • Threads: 2
  • Posts: 997
March 6th, 2014 at 5:57:00 PM permalink
Quote: hwccdealer

Quote: wudged

Quote: geoff

If you mean like this point->seven->point->seven then the odds are

(2/3)*(1/6)*(2/3)*(1/6)=0.01234567901



Interesting number. Too bad .01234567890123456789... isn't something so "nice" looking as 1/81.


13717421 / 1111111111



Spoiler is a little off in terms of equaling 81. The actual number would be:


13,717,421 / 1,111,111,101



Yea, I was giving the fraction for 0.0123456789... which is different from 1/81 (which is all digits in order except for the 8)
vegasrvp
vegasrvp
Joined: Jun 15, 2010
  • Threads: 11
  • Posts: 53
March 6th, 2014 at 6:01:11 PM permalink
A friend of mine has come to me with an idea about a system base solely on the don't pass wager with no odds. I tried to stop him right there but he insisted this was viable.

Now I am looking for math to back up my argument.

He wants to use a version of martingale system.

The betting system he has is 1, 1, 1, 2, 2, 4, 4, 8, 8, ect......

The loss moves you up the system and a win triggers a double up on he next roll.

He thinks the 2 / 3 win and 12 push or 7 / 11 loss on the come out roll just move you through the system the same way.

Here is a sample roll

Bet 1 unit come out roll = 6 point
rolling 4, 8, 8 ,9 6 winner lose 1 unit

Bet 1 unit come out roll = 4 point
rolling 5, 6, 5, 8, 7 out winner 1 unit (so next wager is 2 units)

Bet 2 units come out roll = 9 point
rolling 4,12,4,8,6,5,6,6,8,7 out winner 4 units back to beginning

If 2 or 3 hit on the come out roll it is a win, and you either move to double or back to beginning if it finishes a session.

12 on come out is a push not blood

7 or 11 on the come out roll is a loss and move to next level of the system.

His argument is he can win by hitting 2 "7's" or even a 7 out followed by a 3 before hitting 10-15 points winners.

I know this is pretty off the wall, and I know in my gut it is a failure, but I was hoping to have some math to back my argument and I am not even sure this will deter him.
mustangsally
mustangsally
Joined: Mar 29, 2011
  • Threads: 20
  • Posts: 2050
March 6th, 2014 at 6:03:19 PM permalink
Quote: sodawater

losing an unknown point two shooters in a row (not counting rolls where you win or lose on the come out)= (49/78) ^2 =2401/6084 = 39.46%

how can that be correct?

the Wizard says the probability of winning any point = 201/495 = 0.406061
http://wizardofodds.com/ask-the-wizard/craps/probability/

that makes 294/495 the probability of not winning any point

so my earlier post is correct for the next 2 shooters NOT hitting a point
Both 7out
if that is what the OP wants

But the OP wants to know something else about how long will it take?
"out how many times is average between back to back loser 7 rolls"
is this the same as back to back point losers?
I think it might be

but he started another thread where it looks like he wants to know how many pass line winners before 2 point losers in a row
http://wizardofvegas.com/forum/gambling/craps/17326-prove-me-wrong/

I guess this leads down a long and winding road?

I see a new post about a betting system!
yea!
Sally
I Heart Vi Hart
vegasrvp
vegasrvp
Joined: Jun 15, 2010
  • Threads: 11
  • Posts: 53
March 6th, 2014 at 6:03:22 PM permalink
A friend of mine has come to me with an idea about a system base solely on the don't pass wager with no odds. I tried to stop him right there but he insisted this was viable.

Now I am looking for math to back up my argument.

He wants to use a version of martingale system.

The betting system he has is 1, 1, 1, 2, 2, 4, 4, 8, 8, ect......

The loss moves you up the system and a win triggers a double up on he next roll.

He thinks the 2 / 3 win and 12 push or 7 / 11 loss on the come out roll just move you through the system the same way.

Here is a sample roll

Bet 1 unit come out roll = 6 point
rolling 4, 8, 8 ,9 6 winner lose 1 unit

Bet 1 unit come out roll = 4 point
rolling 5, 6, 5, 8, 7 out winner 1 unit (so next wager is 2 units)

Bet 2 units come out roll = 9 point
rolling 4,12,4,8,6,5,6,6,8,7 out winner 4 units back to beginning

If 2 or 3 hit on the come out roll it is a win, and you either move to double or back to beginning if it finishes a session.

12 on come out is a push not blood

7 or 11 on the come out roll is a loss and move to next level of the system.

His argument is he can win by hitting 2 "7's" or even a 7 out followed by a 3 before hitting 10-15 points winners.

I know this is pretty off the wall, and I know in my gut it is a failure, but I was hoping to have some math to back my argument and I am not even sure this will deter him.
sodawater
sodawater
Joined: May 14, 2012
  • Threads: 60
  • Posts: 3142
March 6th, 2014 at 6:14:10 PM permalink
Quote: mustangsally

how can that be correct?



You are right, I messed up adding the fractions at the very end. I've edited my post and my math now agrees with your post.
TerribleTom
TerribleTom
Joined: Feb 18, 2014
  • Threads: 8
  • Posts: 319
March 6th, 2014 at 9:44:44 PM permalink
There is no being system that will defeat the house edge other than the Martingale with an unlimited bankroll and a no limit table.
sodawater
sodawater
Joined: May 14, 2012
  • Threads: 60
  • Posts: 3142
March 6th, 2014 at 10:16:02 PM permalink
Quote: TerribleTom

There is no being system that will defeat the house edge other than the Martingale with an unlimited bankroll and a no limit table.



even that wont work long term
Impmon
Impmon
Joined: Jan 30, 2014
  • Threads: 0
  • Posts: 45
March 6th, 2014 at 10:48:29 PM permalink
Quote: vegasrvp

A friend of mine has come to me with an idea about a system base solely on the don't pass wager with no odds. I tried to stop him right there but he insisted this was viable.

Now I am looking for math to back up my argument.

He wants to use a version of martingale system.

The betting system he has is 1, 1, 1, 2, 2, 4, 4, 8, 8, ect......

The loss moves you up the system and a win triggers a double up on he next roll.



I thought of the same thing, except the progression was: 1 22 333 4444 -- you get the idea. I wrote a program to simm this, and it looked like it was the real deal.

That was until I took a good look at the results and saw how odd the data looked. Turned out that Python's randint function doesn't work right until the system's been on long enough to develop enough entropy. Simms done later in the day showed the expected results: lots of small wins and a few horrendous losses. Actual results with good data were right about a fraction of a percentage point within known expectation for the Don't Pass.

Any Martingale, even a "soft" one, is headed for disaster. This is especially true for Craps, as long hands, or even a series of shooters with a few passes between them, are not all that uncommon. Best way to play is bet the Don't, lay odds, take what you can and get out if you lose two in a row.

Depending on how frequently he plays, he might get away with it for a good long while, but the Day of Reckoning will come.

  • Jump to: