Imagine you go to the casino with &dollar;100 in one-dollar chips and decide to play the roulette wheel. You’ve nothing better to do; you can stay there as long as you have chips left. How long can you expect to carry on playing?

Let’s pick on the 50-50 bets (such as red or black, odd or even), because they’re easiest to work with, and least volatile. Obviously, the worst-case scenario is that you lose your first 100 bets and go home broke. That’s going to happen only once in every $2^{100}$, or about $10^{30}$ attempts. It’s not at all likely.

Even in this simplified version, it’s not an easy problem to solve analytically, so I turned to computer simulations to see how long you’d last. You can do some analysis - for instance, every bet on the roulette wheel has a negative expectation of about a thirty-seventh of a chip - that is to say, if you make 37 one-chip bets, whatever they are, you would expect to win 36 chips back. Naturally, that means you can expect to remain at the table for about 3700 games, right? Wrong.

The median length of time you can expect to remain at the table is only about 3,200, far short of the 3,700 I said. Typical results ((the quartiles, since you ask)) are between about 2,000 and 4,000 spins, although one lucky simulated gambler hung around for nearly 30,000 turns.

### Why is it that the median is so much lower than expected?

It’s because you can’t bounce back from zero. Some of the ways you could lose 100 chips involve first losing 101 chips and bouncing back - but you’re in a casino, not a bank, so you’re not going to get a loan. Once you get to 0, you’re out of the game - and that knocks off an average of 500 turns.

But that’s ok. Roulette is boring, anyway.