Roulette: double-up probability

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

• 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. math.stackexchange.com/questions/1994169/… – doug Nov 1 '17 at 22:32