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Assumed I have infinite money and bet 10 on red or black and every time I lose I will double my bet until I win and then start again with 10 would I make profit?

I did a bit code for that: http://jsfiddle.net/ua84gkaL/5/

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  • $\begingroup$ This may be a duplicate of math.stackexchange.com/questions/1016267/… (I am not sure). $\endgroup$ – quid Jan 25 '15 at 22:34
  • $\begingroup$ Would you please modify your code such that (a) it terminates when the gambler wins (b) or when the gambler runs out of his money. $\endgroup$ – zoli Jan 25 '15 at 22:41
  • $\begingroup$ (a) I don't wanna stop after 1 win, (b) I added a highest Bet field so you see how much money you needed for this run. I don't wanna stop it cause then it's not infinite money $\endgroup$ – nbar Jan 25 '15 at 22:44
  • $\begingroup$ @zoli if you want to use the code for something else I can edit it for you. ? $\endgroup$ – nbar Jan 25 '15 at 22:57
  • $\begingroup$ Yes, I would like to use it in the classroom, if you give me permission. $\endgroup$ – zoli Jan 26 '15 at 0:57
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Yes you would make a profit. Assume you always bet red and to make things minimally simpler let us say you always start with bet $1$ (not $10$) and double when you lost. Let us consider the run until the first red you hit. So there are $n$ black and then red.

First you loose $1 + 2 + 2^2 + \dots + 2^{n-1}$ then you win $2^n$. The former sum is $2^n- 1$. So not matter what happens in the end you win $1$, every time that red comes up.

(This answer assumes a $1/2$ red, $1/2$ black no zeros as rules there are not uniform but it does not change the ultimate outcome for the variants I know of.)

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  • $\begingroup$ ty for the good explanation. To your edit: As far as I can see the odds does not matter $\endgroup$ – nbar Jan 25 '15 at 22:47
  • $\begingroup$ You are welcome. Yes, you are right and we could consider $0$, or the two $0$, just as a loss and assume a smaller then $1/2$ probability of winning. This is however not quite in line with how some roulette variants work (IIRC) so that I preferred the disclaimer. $\endgroup$ – quid Jan 25 '15 at 22:51
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Yes. Your betting system is called the Martingale strategy. It usually fails gamblers, because they do not have infinite money available. However, you do.

As the other user quid writes in his answer, you are essentially going to gain $1$ unit of currency on every trial of the strategy. Consequently, you can make an arbitrarily large gain.

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  • $\begingroup$ ty for the hint with the martingale strategy. infinite money and no table limit is needed to have success. $\endgroup$ – nbar Jan 25 '15 at 22:39
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This is a very interesting topic. If you play until you win, you will have made positive profit by then, because of your strategy.

However, for any given point in time, your (a priori) average wins will be

Earnings = - (total amount of money bet so far)$\cdot\frac{1}{37}$

The thing is that the assumption of infite money makes this situation so paradox. Assuming there is any limit for your money, the casino would always win. The same is true if you assume that there is a bet limit.

All in all, i think the correct answer is: yes, you will make profit. You will win in 18/37 games so if you are going to play until you made a profit of 100, this will require you $10\cdot37/18 \approx 20$ games on average.

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If you have infinite money, what's the point of gambling? If you use this system repeatedly, you will eventually lose. When you do lose, you will lose big.

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  • $\begingroup$ You should have stopped after the first sentence! In particular, your last sentence is diametrically at odds with your first sentence: how can you lose big if you have infinite money? $\endgroup$ – TonyK Jan 25 '15 at 23:48

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