Assume 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 wrote a bit of code for that.
Assume 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 wrote a bit of code for that.
On
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.
On
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.
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.)