Cdtrjyty

I'm at the casino, standing next to a roulette table with a $10 minimum bet. I want to stay here as long as possible, so I'm going to repeatedly make the minimum bet until I run out of money.

I'm playing European roulette, and I'm putting my money on 28 every time. This means that with every spin, I have a 1 in 37 chance of winning $350, and a 36 in 37 chance of losing $10.

I only have $20 in my pocket, so I'm almost certainly not going to be here for very long! (This is distinctly not awesome.) But, on the other hand, there is a small chance that I'll get 35 extra spins, so that's got to count for a little bit.

So, how long, on average, is my money going to last me? Three spins? Four?

Suppose $t(n)$ is the average number of spins you get if you start with $$10n$. We want $t(2)$. If you start with $n$ ten-dollar bills, put $n-1$ in your pocket and play until you are broke, then take the next $10 out and play until you are broke, and so on: this is exactly equivalent to just starting with $$10n$ and playing until you're broke, so $t(n)=kn$ for some $k$. On the other hand, obviously $t(0)=0$ and for $n>0$ we have $t(n)=1+frac{36}{37}t(n-1)+frac{1}{37}t(n+35)$. Substituting $t(n)=kn$ into the latter equation and solving for $k$ we find

$k=37$ -- i.e., the number of spins you get, on average, is 37 times your initial multiple of the minimum stake. So if you arrive with $$20$ and the minimum stake is $$10$ then on average you get to spin the wheel 74 times.

[EDITED to add:] JonMark Perry's answer suggests another way to proceed after establishing that $t(n)=kn$: once you have that you can

go from "you lose $$frac{10}{37}$ per spin on average" to "it takes 37 spins to lose $$10$ on average". But, for me at least, this takes a little more thought to see it's valid than the more straightforward calculation above.

[Meta: to me this seems just "fun" enough to be a puzzle rather than a mere mathematics problem, but I won't be upset if others disagree and this gets closed for being too mathematics-textbook-problem-y.]

Imagine you start with $$370$. You play for $37$ turns and come back with $$360$. You borrow $$10$, and go again for another $37$ turns, and again come back with $$360$, and borrow another $$10$.

You repeat for a total of $37$ big turns, and now you have borrowed as much as you came with, and the bank won't lend you any more money.

So, you survive $37$ big turns with $$370$. $37$ big turns is $1369$ turns, but we only want $frac2{37}$ of this, which is:

74 turns.

Let's say that the value in spins of each $10 is x.

x is equal to 1 (the spin you get for the initial money) plus 35x/37 (350 bucks, 1/37 of the time). From there, it's simple math. Subtract 35x/37 from both sides. 2x/37=1, so 2x=37

thus

on average, your $20 (2x) will net you 37 spins.