Here is a specific example:
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;
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.
Quote: hwccdealerQuote: wudgedQuote: geoff
If you mean like this point->seven->point->seven then the odds are
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
how can that be correct?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?
There is no being system that will defeat the house edge other than the Martingale with an unlimited bankroll and a no limit table.
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.