Its not a good play after doing the math. Def thought the odds would be better.

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I think doing this on the craps table on the don't pass bet would be more successful

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It's all relative, as there is no expected value that would ever turn positive without counting cards.

I trjed this on the field in craps. Busted out. Was not pretty

The golf pro at the course I play got heavy into this for like 2 months. He was going to the casino a couple times a week, using the system, and waking out with a couple hundred bucks each time. He kept telling everyone about it and I was like no way this ends well.

Apparently he had gotten up a couple grand, went back one night, and good old 0 and 00 crashed his house of cards. Ended up losing everything that he had made and then some.

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In the end, I played for about 6 hours over a couple days (I would get bored) and ended up winning around $600. There were 2 times where I made it to the 7th spin with $320 on the table.

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Another way to look at it. You were basically two coin flips away from walking away with nothing.

Well using this system to win a few bucks a time or two is a pretty decent play. Highly unlikely you'd lose 7x consecutively in 10 or so spins so you'd make a few bucks there.

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Well using this system to win a few bucks a time or two is a pretty decent play. Highly unlikely you'd lose 7x consecutively in 10 or so spins so you'd make a few bucks there.

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This statement right here is why casinos are a multi-billion dollar industry.

The best case with $10 hands over 10 spins is that you've won every spin and walked with $100. Remember, you're not doubling your bet in response to winning; winning just means your next bet is $10. Winning 10 hands in a row is WAY less likely than losing 7 in a row and your BEST outcome if that happens (with $10 hands playing a Martingale) is only $100.

If you are playing 10 hands and Martingale out to table maximum on a $10 initial bet (assuming $1k table max), that's 7 hands and you've just lost about $1250 (depending on whether you started losing on hand 1,2,3, or 4). If that happens to you just ONE time, you've gotta win over 120 more $10 hands than you lose in the long run just to erase that one bad streak. Do you know what casinos are REALLY good at doing? Making sure their games will not allow players to win 120 hands more than they lose in the long run.

The Martingale system is fatally wounded simply by the difference in bankrolls that you and the casino hold. Once the casino throws in table maximums, the Martingale system is not only fatally wounded, but its organs are pulled out of its body and shown to it one by one as it's dying a quick, brutish, horrible death.

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There is only one way to have the odds at roulette and it involves a team game with someone actiling like a drunk betting the green squares (35/1, so he can afford to lose) and you betting odd/even or red/black. You actually have the odds in your favor.

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Lolwut?

To answer your original question in simpler math terms:

If the maximum loss is 255 dollars/units then you can bet up to 8 times. The probability of losing eight bets in a row is (19/37)8=.004835. So, you have 99.52% of winning one dollar/unit, and 0.48% of losing 255 dollars/units. Those are not odds any sharp would ever take.