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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 $t(n)=1+frac3637t(n-1)+frac137t(n+35)$. 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 $$frac1037$ per spin on average" to "it takes 37 spins to lose $$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.]

To take another approach:

Suppose that you have 37 people standing around the table, each one betting on the number closest to them. Then every round, they lose $360 and win $350, for a net loss of $10. The time it takes for them to lose $740 ($20 per person) is 74 turns.

Less than 2.66

I will skip a lot of the calculations to provide a ceiling(maximum) to the answer.

First, let's calculate the probabilities of playing N rounds P(N).
Obviously, you will play N=2 rounds + 35*W rounds, where W is the amount of rounds you won.
So N=2+35W

P(2)=P(Lost both rounds)=(36/37)^2=0.9466

P(N=2+35*W) < P(Won at least 1 of first two rounds)*P(Won W times in all rounds)=(1-(36/37)^2) * N * (1/37)^W * (36/37)^(N-W)

The P(N) is strictly less than the product of the two probabilities I've mentioned, because the factor (N) that the binomial distribution
provides for the second part considers wins that would happen after money got negative. For example, if we played 72 rounds, the product mentioned
includes the probability of having the first win in the first round, and the second win in the 72nd round, which is not useful to our game as
we'd have ran out of money beforehand.

The average is the sum of probability * value. Here, "value" is the number of turns itself. Thus,

AVG=P(2)*2+ SUM_over_W(P(N)*N)

The contribution of each number of wins goes down by ~100 each time,
so taking the first few addends is correct. I put some results in excel and got:

+------+--------+-------------+---------------------+
| Wins | Rounds | Probability | Contrib. to average |
+------+--------+-------------+---------------------+
| 0 | 2 | 0.946676406 | 1.893352812 |
| 1 | 37 | 0.019885997 | 0.735781895 |
| 2 | 72 | 0.000412005 | 0.029664383 |
| 3 | 107 | 6.51897E-06 | 0.000697529 |
| 4 | 142 | 9.21103E-08 | 1.30797E-05 |
| 5 | 177 | 1.22241E-09 | 2.16367E-07 |
| 6 | 212 | 1.55885E-11 | 3.30477E-09 |
| 7 | 247 | 1.93371E-13 | 4.77626E-11 |
| 8 | 282 | 2.35054E-15 | 6.62852E-13 |
| 9 | 317 | 2.81321E-17 | 8.91788E-15 |
| | | SUM | 2.6595097 |
+------+--------+-------------+---------------------+

Thus, you can be sure the average number of turns you are going to play is less than 2.66

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 $frac237$ 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.