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Andy plays european roulette and bets the lowest amount of 10 dollar at red and he's going to double the amount until he wins. Which is the probability he wins before he has spent all his 1 000 dollar he has brought to the casino?

I think this may be solved using complement event. He may double 5 times (10+20+40+80+160+320=630) and I get P(he succeed) = 1 - (18/37)^5 ≈ 0,97. But the answer should be 0,98.

Anyone can help?

Thanks in advance.

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  • $\begingroup$ I wonder, how can he spend all his money by repeatedly doubling his stakes, if his total purse does not contain a power-of-two multiple of the initial stakes? Do you mean “until he cannot double again”, or will there be a case where he spends as much as he has left even if that doesn't double the stakes? $\endgroup$
    – MvG
    Commented Jul 8, 2013 at 16:19

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The player can make 6 games... Moreover the probability that he don't get red is $19/37$ so the answer is $1-(19/37)^6$.

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    $\begingroup$ if the player looses 6 times then he has no more money to play again... $\endgroup$
    – Emanuele Paolini
    Commented Jul 7, 2013 at 9:44

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