# How long will my money last at roulette?

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?

• This seems a math problem not a puzzling problem Feb 20, 2019 at 17:34
• @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". Feb 20, 2019 at 17:40
• If you want to stay there longer, bet \$10 on black and other \$10 on red and suppose there is no zero :)
– user52799
Feb 21, 2019 at 8:02

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.] • Certainly an unexpected or unintuitive solution. I would have expected the answer to be much less. Feb 20, 2019 at 18:21 • 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. 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.

• "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. Feb 20, 2019 at 19:28
• @TannerSwett; I forgot you get your stake back!
– JMP
Feb 20, 2019 at 19:32
• 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? Feb 20, 2019 at 20:01
• @GarethMcCaughan; 2->1 is obvious, and 1->2 is because of the uniformity of roulette
– JMP
Feb 20, 2019 at 20:11
• 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. 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 thus on average, your$20 (2x) will net you 37 spins.