How far can you go with the Martingale Roulette strategy

Question

I am currently arguing with my friends that just learned the Martingale roulette technique for the casino that consist of doubling your initial bet as much time that you lose (with a budget of course). So if you have 5000\$ you could initially bet 1\$ on a color (with 48% aprox. chances of winning) and if you lose you bet 2\$, then 4\$, then 8\$... up to the point where you don't have the money to double on and you lose it all.

I am certain that in the end you are losing using this technique. Yes your probability of winning a percentage of your budget are great up until a limit of wins using this technique. I'm almost certain that you hit that limit before 5000 times ( winning more than what you were willing to lose ). So in general even with millions of people who would use this the casino is still winning in total and with any maximum budget (except infinity of course). My friend base a lot of their thoughts behind the fact that this is banned from a lot of casinos along with all sort of techniques or strategies but i think it's more of a collateral damage and in general you still lose with that.

So I'm asking your help to explain why I am or they are wrong about this. Thank you !

Answer 1

You have the right idea. Your expected loss on any bet is $\frac {1}{19}$ of your wager in an American casino and $\frac {1}{37}$ in Monte Carlo. Your betting strategy does not change this.

If you had an infinite endowment (and the casino had no maximum bet size) even if you are down a very large sum of money at some point, eventually your double-or-nothing bet will hit and almost surely you will eventually hit a win and be up one dollar. But you don't have an infinite endowment and the casino will not allow you to make bets of unlimited size.

Supposing you have a budget of $\$5,000$ for example, if you hit a streak of 12 consecutive losses, you can not afford to double your bet. We can calculate your expected loss with a strategy of play to the first win, or walk away after losing 12 in a row.

$(1 - \frac {20}{38}^{12}) - \frac {20}{38}^{12}(2^{12}-1)\approx - 0.85$ in Vegas.
$(1 - \frac {19}{37}^{12}) - \frac {19}{37}^{12}(2^{12}-1)\approx - 0.38$ in Europe.

Spending enough time employing this strategy, almost surely you will bust.

Answer 2

The basic issue with the martingale is that probabilistically, "anything that can happen will happen."
If I have a nonzero probability of exhausting my bankroll via the martingale strategy, then eventually it will happen.
With a larger bankroll, I can make the probability of going broke very small, but as long as my bankroll is finite, I can never make the probability zero entirely.
Sooner or later, it will happen to me.

As regards the expected value, Wikipedia has a thorough analysis. If I'm playing an unfair game (i.e. my win probability is under $0.5$) then my net expectation is always negative, because those infrequent times that I lose the amount I budgeted for the martingale, cancel out all the gains I make.

Not to mention, as Milo and Doug point out, the casino is going to have a maximum bet at some point. So if I don't exhaust my bankroll, at a certain point I'm not going to be able to cover my losses by doubling my bet, because the casino isn't going to let me bet that much.