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 '19 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". $\endgroup$ – Tanner Swett Feb 20 '19 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 :) $\endgroup$ – Zereges Feb 21 '19 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.]

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    $\begingroup$ Certainly an unexpected or unintuitive solution. I would have expected the answer to be much less. $\endgroup$ – GentlePurpleRain Feb 20 '19 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. $\endgroup$ – Matthew Barber Feb 21 '19 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.


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

  • $\begingroup$ Your probability bound is wrong: the probability to play exactly 37 rounds is the probability of winning once in the first two rounds and losing every other game, or P(37) = (2*1/37*36/37) * (36/37)**35, which is approximately 0.02016. $\endgroup$ – Roland W Feb 21 '19 at 0:35
  • $\begingroup$ You can't say P(N=2+35*W) < P(Won at least 1 of first two rounds)*P(Won W times in all rounds), because the events "win at least 1 of the first two rounds" and "win W times total aren't independent", and because possibilities like "won the first game and the last game" are invalid. $\endgroup$ – user2357112 supports Monica Feb 21 '19 at 0:38
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    $\begingroup$ You can do that, it's an upper bound that may include additional possibilities. But the formula is wrong: After not losing both first two games, we have won at least once already. So there are at most W-1 wins left for the last N-2 games. So the bound should be (1-(36/37)^2) * (N-2 choose W-2) * (1/37)^(W-2) * (36/37)^(N-W) because P(W-2 wins in N-2) > P(W-1 wins in N-2), for W >= 2. But the bound is too loose: at W=2 it is 0.0078 compared to the exact value of 0.0076, but it decays too slowly. The expectation value diverges. $\endgroup$ – Roland W Feb 21 '19 at 1:10
  • $\begingroup$ @user2357117 When the events are not independent, the correct calculation would be an additional -P(both events). Since I omit a negative addend from the large part of the inequality, it still stands. $\endgroup$ – George Menoutis Feb 21 '19 at 6:55
  • $\begingroup$ @Roland W I stand by my choice. There will sure be better answers out there, but at least 1) it is obvious (though not certain or proved) that my formula converges 2) the value provided is much closer to the truth than 34 or 70! I am planning to develop a precise answer....after work. $\endgroup$ – George Menoutis Feb 21 '19 at 6:55

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. $\endgroup$ – Tanner Swett Feb 20 '19 at 19:28
  • $\begingroup$ @TannerSwett; I forgot you get your stake back! $\endgroup$ – JMP Feb 20 '19 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 '19 at 20:01
  • $\begingroup$ @GarethMcCaughan; 2->1 is obvious, and 1->2 is because of the uniformity of roulette $\endgroup$ – JMP Feb 20 '19 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 '19 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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