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Let's assume I play roulette in casino with following strategy:

  1. I bring X\$ amount of money
  2. For each spin I place 1 bet of 1$ on some number
  3. I play until I double up or until I lose everything.

What I my odds of succeeding this strategy? Assume European roulette win single zero.

It's simple to calculate this for amounts up to 18:

If I start with 1\$ my probability is $1/37$

With 2\$ I win if at least one bet succeeds, that is $1-(36/37)^2$

With 18\$ I succeed with $1-(36/37)^{18} \approx 0.389$

Here, I cannot figure out how to extend my calculations to bigger numbers. So, the question is, can simple formula be found for this?

I have done some simulations and if they are correct odds of doubling with this strategy seems to increase to around $0.435$ and slowly decreases after that. This seems little bit unintuitive to me.

And, yes, I do know I should lose money in the long run. What interests me is hoe to do that in "optimal" way

J. Wilkinson
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    google "gamblers ruin" – Peter Oct 14 '17 at 15:58
  • You can get up to about a 49% chance of doubling v 51% chance of busting with a strategy outlined here. https://math.stackexchange.com/questions/1994169/the-bold-strategy-of-a-single-large-bet-is-not-the-best-roulette-strategy-to-d?rq=1 – doug Nov 01 '17 at 22:32

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