Help with Odds calculation for craps...

[canada_dry]

If I were to make two place bets (6 and 8), then double it each time I lost, for a maximum of six doubles, what are the chances I would lose all six?

$12 on 6, $12 on 8 = $24
$24 on 6, $24 on 8 = $48
$48 ...
$96 ...
$192 ..
$384 ..

Therefore risking $1512 to win $14. Each roll i have 6/36 chance of losing and 10/36 chance of winning. Could someone help with the math.

I realize the inherent risks with a martindale system so i was trying to incorporate it with a betting pattern that has a greater chance of winning than losing (10 ways to hit a six or eight versus 6 ways to hit a seven).

Thanks..

[punkass]

I'm not going to go in the math, but keep in mind, you have to hit twice teach time to be ahead.

[canada_dry]

[ QUOTE ]
I'm not going to go in the math, but keep in mind, you have to hit twice teach time to be ahead.

[/ QUOTE ]

The question precisely asked you to go into the math.

And no you don't have to hit twice each time to be ahead. Just once. Place bets can be taken down at any time.

[Wake up CALL]

$12 on 6, $12 on 8 = $24 win=$14 loss=$24
$24 on 6, $24 on 8 = $48 win=$28 minus prior $24 win=$4
$48 ...win=$56 minus prior $72 loss=$16
$96 ...win=$112 minus prior $196 loss=$84
$192 ..win=224 minus prior $388 loss=$164
$384 ..need I go on?

I love your system, it is like printing your own money except you don't go to jail for this. [img]/images/graemlins/smile.gif[/img]

[Cosimo]

NOTE: You'll need to bet a bit more to break even.

First bet: 12 & 12. If you win, +14. If you lose, -24.

Second bet: 24 & 24. If you win, you only net +4 -- 28 for the win, minus 24 from the previous loss. In order to profit +14 here, you'd have to win $38, ie bet $32.57 on both 6 and 8.

Third bet: 48 & 48. If you win, you're still out $16. You've lost $72 on the two previous bets. At this point, the progression as you posted is just gonna keep you from being as deep in the hole -- if you win. If you lose, you're getting progressively more screwed. You'll lose three bets in a row over 5% of the time.

[ QUOTE ]
If I were to make two place bets (6 and 8), then double it each time I lost, for a maximum of six doubles, what are the chances I would lose all six?

Each roll i have 6/36 chance of losing and 10/36 chance of winning.

[/ QUOTE ]

Chance of winning one roll: 10/16 (62.5%)

Either you win or you lose. The other outcomes don't matter. So just add up the ways of winning (10) and the ways of losing (6) and you get 16. You'll win the first roll about 62% of the time.

Chance of losing one roll: 6/16

The process here is the same.

Chance of losing the second roll: 6/16

Each event is independent; the chance of losing any given roll is the same regardless of which roll it is. Dice have no memory. There are no discards as in Blackjack.

Chance of losing the second roll given that you lost the first roll: 6/16

Trick question; since the first loss is a 'given' then there's a 100% chance that it happened. Dice have no memory, so the results of the first roll are irrelevant.

Chance of losing two rolls in a row: 6/16 * 6/16

This is what you're looking for. The chance of two independent events occuring is the mathematical product of the chance of each event separately.

Chance of losing six rolls in a row: 6^6 / 16^6

That's "six raised to the sixth power, divided by sixteen raised to the sixth power." This is about 359:1, or .278%. Not bad. You could do this maybe a dozen times and walk away with $100, but I wouldn't want to press my luck further than that. If you're looking for a $14-ish profit, you'll need to wager 12, 30, 84, 228, 618, and 1680, for a total bankroll of $5,304. Then you'll have a 99.7% chance of winning $14.

balance; bets. amount if you win, net win (if you win).
-0; 12&12. w14, net +14
-24; 30&30. w35, net +11
-84; 84&84. w98, net +14
-252; 228&228. w266, net +14
-708; 618&618. w721, net +13
-1944; 1680&1680. w1960, net +16
-5304.

[ QUOTE ]
martindale system

[/ QUOTE ]

That's "Martingale".

PS: Did you ask in the Probability forum? There's prolly someone more willing to answer your questions there. And I could have made a mistake.