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?

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    $\begingroup$ This seems a math problem not a puzzling problem $\endgroup$
    – Yout Ried
    Feb 20, 2019 at 17:34
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    $\begingroup$ @YoutRied Perhaps. Looking at the test points at this meta answer, I think that this question has a "clever or elegant solution" and an "unexpected or counterintuitive result". I'm not sure if this can be said to have an "unexpected problem statement". $\endgroup$ Feb 20, 2019 at 17:40
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    $\begingroup$ If you want to stay there longer, bet \$10 on black and other \$10 on red and suppose there is no zero :) $\endgroup$
    – user52799
    Feb 21, 2019 at 8:02

4 Answers 4


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.]

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    $\begingroup$ Certainly an unexpected or unintuitive solution. I would have expected the answer to be much less. $\endgroup$ Feb 20, 2019 at 18:21
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    $\begingroup$ You'd perhaps expect it to be much less because nearly 95% of the time you're walking away after two spins. However, the other tail of the distribution contains some wild and crazy games where you'll play upwards of a thousand and that adds a lot to the expectation value even though it's unlikely. $\endgroup$ Feb 21, 2019 at 1:04

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.


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.

  • $\begingroup$ "Imagine you start with \$370. You play for 37 turns and come back with \$350." If you play 37 spins, the "expectation" is that you lose \$10 36 times and win \$350 once, making a net loss of \$10, so you'll have \$360. $\endgroup$ Feb 20, 2019 at 19:28
  • $\begingroup$ @TannerSwett; I forgot you get your stake back! $\endgroup$
    – JMP
    Feb 20, 2019 at 19:32
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    $\begingroup$ It looks to me as if you're assuming that "it takes an average of N turns to lose \$10" and "on average you lose \$10/N per turn" are equivalent -- the first is what's obvious and the second is what you're using -- but that seems like a thing that needs proving. Or am I missing the point somehow? $\endgroup$
    – Gareth McCaughan
    Feb 20, 2019 at 20:01
  • $\begingroup$ @GarethMcCaughan; 2->1 is obvious, and 1->2 is because of the uniformity of roulette $\endgroup$
    – JMP
    Feb 20, 2019 at 20:11
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    $\begingroup$ Incidentally, if this argument works then it seems to me you don't need the business about "big turns" at all: you lose an average of 1/37 of a unit per spin, "therefore" on average it takes 37 sounds to lose each unit. $\endgroup$
    – Gareth McCaughan
    Feb 20, 2019 at 20:20

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


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


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