110 inside 1st hit down 2nd hit down and off or down on other 2 place bets. Your investing 110 dollars to win 35, (the total win is 70 yes but you still have to it hit a # twice) so does anyone know the math,( not that it matters) lets see you have 18 ways of winning, 6 ways of losing your set goal is win 300 dollars or lose 300 how would you figure the percentage or odds against you

I like to stick my toe in before I do a cannonball. i.e. start low and work my bets up.

Alot depends on the amount of your buy-in…..is 300 dollars 10% of your buy-in or 100% of your buy-in. It is easier to win 10% than 100%.

777

yacraps -

There are advocates of that exact method on this and other boards -

” Two hits and regress/down with your bets ” -

Their premise is the majority of shooters do not get past 4 rolls of the dice -

If you ” need ” action on every shooter -  you get ” action ” with this style of betting - but , early 7 outs will definitely bite you -

How many of those can your bankroll handle ? -

I wait for a shooter to throw a ” repeating box number ” before I wager on him/her -

Nothing is gauranteed that the shooter will not 7-out right away - but at least the shooter has demonstrated either a little control or luck over the dice at this time !

bankroll is 300 hundred you either win 300,or lose 300 so in essence your trying to get 2 #s a hand a total win of 70 dollars a hand what would be the house percent advantage

The house advantage takes effect over time. It is pretty meaningless over a few hands.

Probably the more meaningful analysis is ‘risk of ruin’. I would think the chance for you to lose your 300 are considerably higher than making 300. One of the math people here might be able to give you your chances.

777

basicstrategy777   thanks thats what I am looking for are the math guys I am a little embarrassed I know some math on the game but was hoping for a little more precise figures maybe the answer is to long I know the beginning is 18 ways to win 6 ways to lose (3 to 1 favorite) then you have to figure in the factor betting 110 to win 35 (a 3 to 1 underdog) only until you win twice, then out till next hand for approximately 5 hands (in a perfect world) a early 7 down 110 now need 2 winning hands (70 dollars a hand) to get even and a little ahead if no one answers I understand

God help me, I have to agree with BS. In the long run you will PSO so often that you will usually be fighting an uphill battle just to get even. In the real long run your expectation is total bet times edge. That edge usually belongs to the casino.

D….even a blind pig finds an acorn every now and than.

777

then

basicstrategy777 - 17 January 2012 07:41 AM

D….even a blind pig finds an acorn every now and than.

777

Try ” squirrel ” - they’re the ones that tend to go after acorns a little bit more frequently than pigs do !

Then again - it all boils down to ” reading comprehension ” not even getting into spelling or context !

$nakeEye$ - 17 January 2012 09:24 PM

basicstrategy777 - 17 January 2012 07:41 AM

D….even a blind pig finds an acorn every now and than.

777

Try ” squirrel ” - they’re the ones that tend to go after acorns a little bit more frequently than pigs do !

Then again - it all boils down to ” reading comprehension ” not even getting into spelling or context !

Actually, its hog.  Even a blind hof finds an acorn now and again.
Of course, being cityfolk I wouldn’t expect you guys to know hogs like acorns too.
“http://www.urbandictionary.com/define.php?term=blind hog”
You may have to copy/paste..  clicking it doesnt work.  And I am not responsible for anything if you read the second definition (I never heard the term in that manner and hope I never will again, now please excuse my while I go wash my eyes out with lye soap)

Just play 44 inside,  then after one score,  u got $14 in your kick.  So,  take down the 5 and 9,  and make the 6 and 8 look like $6.  You got $12 at risk and at the worst .........u only make $2 on that shooter.  If u get another hit on the six or eight,  u are up $21 and come down on everything or go to 22 inside.  Then u are only risking $1 to win $28.  Conservative way to stay in action using regressions after the $14 is locked in. 
Anyway…......

yacraps - 14 January 2012 11:57 AM

110 inside 1st hit down 2nd hit down and off or down on other 2 place bets. Your investing 110 dollars to win 35, (the total win is 70 yes but you still have to it hit a # twice) so does anyone know the math,( not that it matters) lets see you have 18 ways of winning, 6 ways of losing your set goal is win 300 dollars or lose 300 how would you figure the percentage or odds against you

I like to think outside the box and give myself a four or five to one advantage over the house on any given roll of the dice, therefore why not Place $100 on the inside and with the $4 used on the odds on the 6 & 8, hop the easy 4 & 10 for $2 each.  Here is what you have now done:  covered all the Place numbers with about 40% less funds at risk; win on any Place number thrown except for the hard 10 and hard 4; give yourself twenty two ways to win instead of eighteen; lose only $4 on any horn number thrown; and lose only $108 on a 7 out.

Now the results:  Easy 4/10 nets $28; 6 & 8 nets $25; and the 5 & 9 nets $31.

Think about it.  One only needs about 4 hits to break even with the totality of the bet of $108.  Of course, if pressing is your game, one can figure out a reasonable pressing gambit to increase the winning if only those rascally dice will cooperate.

falcon/tuttigym/DB+W

You do realize that the odds of throwing a seven are more than 50% in four rolls?

D….....Makes sense to me, and must be the reason crap hands are such short affairs.

If it was 16 % they would last alot longer.

777

What?    The odds for throwing a 7 are the same every roll.!    6 ways to make a 7,  30 ways not-to.  SAME EVERY ROLL! 
just like is laying 6-3 on a No 10/4,  or 2-1 .......the same every roll.