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PoBoy321
06-16-2005, 06:43 PM
With a large enough bankroll, could the Martingale System have a ROR of less that 0.1%? My thinking is just that with a bankroll of say, 2048 units (enough to last a run of 10 losses, which as a 47% dog will only happen 0.06% of the time), the likelihood of hitting a long enough streak to bust you is so small that it could be a feasible option.

Of course, this is all theoretical, and any -EV bet is always losing money in the long term, it's just something I've been thinking about and I'm curious.

AaronBrown
06-16-2005, 08:05 PM
With an infinite bankroll, a martingale system is guaranteed to win. That's the one exception to the rule that a series negative expectation always results in a negative expection.

In the finite world, with limited bankrolls and bet limits, the martaingale system buys you nothing. With $2,047 and a casino that will let you bet from $1 to $1,024 on red or black in double-zero roulette; the margingale will give you a profit of $1 1,164 times out of 1,165 and -$2,047 1 time in 1,165. In order to try to win $1, you've taken a negative expectation of $0.76. Much better just to bet once and walk away, then your negative expectation is only $0.06.

riverdance
06-16-2005, 09:00 PM
is your math correct on the 10 losses in a row being .06% ?

it seems i always see a roulette scoreboard with a big run on one color. at least 10 in a row, sometimes more.

AaronBrown
06-16-2005, 09:13 PM
The chance of 10 wins in a row is (18/38)^10, or 0.06%. The chance of 10 losses in a row is higher, (20/38)^10 = 0.16%.

PoBoy321
06-16-2005, 09:49 PM
[ QUOTE ]
The chance of 10 wins in a row is (18/38)^10, or 0.06%. The chance of 10 losses in a row is higher, (20/38)^10 = 0.16%.

[/ QUOTE ]

Whoops, that was my mistake. Thanks for picking up on that.