Good afternoon,

I know that in the long run you can never win at roulette. The casino will always win, even if you are using the Martingale technique (that doubles your bet with every loss).

But I am curious about how big the probability is that:

* when you play 60 times on black

* the roulette ball will fall on red or zero 11 times in a row

My quick calculation says is 2,44% ( = ( 2^(60 - 11) * (1 + (60 - 11)) - ( 60 - 1) - (((60 - 2) - (11 + 1))/2 * ((60 - 2) + (11 + 1))) )/ 2 ^ 60), but I am pretty sure it is not totally correct.

Who can correct me please?

...or any other game deliberately designed to make money for the house.

The problem with this silly strategy is the exponentially increasing size of the wager upon loss. Too many consecutive losses and anyone runs out of cash.

If you manage only ONE 11-string of losses - rough estimate ...

p(11 losses in a row): (19/37)^11= 0.000654743

Places for your 11-string to start: 50 => 3.7% -- And we haven't even mentioned getting more than one (1) 11-string.

First impression? Your 2.44% is very much too low.

Shall we quit after the #12 win after the first 11-string? That' a different question.

The problems with the whole thing are:

1) The odd premise that one will eventually score a win. It's just not 100%. It's clearly not 100% with...

2) The finite money supply, since "eventually" is now finite.

3) The house odds.

P.S. Full Disclosure: Keep in mind that I am totally biased against gambling so I might make rough estimates of loss probability that are too high and then feel perfectly fine with my error.