This question came up when I asked on Puzzling.SE, How long will my money last at roulette? The basic question is: if I take $\$20$ to a roulette table which has a house edge of $1/37$, and I bet $\$10$ on each spin until I run out of money, what's the average amount of time before I lose all of my money?

The reasoning used in a couple of the answers is:

  • I start with $\$20$ and I bet $\$10$ per spin.
  • The expected value of amount I lose on each spin is $\frac{\$10}{37}$.
  • Therefore, the expected value of the number of spins required in order to lose all my money is $\$20 / \left (\frac{\$10}{37} \right) = 74$.

However, one of the commenters pointed out that this argument needs some additional justification, writing:

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? – Gareth McCaughan♦

I'm pretty sure that the two statements are equivalent. In other words, for any "roulette-like" bet, the following equality holds:

$$E(\text{number of bets I make before losing $\$N$}) = \frac{\$N}{E(\text{amount of money I lose in one bet})}$$

However, I also know that probability often behaves in counterintuitive ways, so I wouldn't be very surprised if this equality turned out to be wrong.

Is the equality correct, and if so, how can it be proved?

  • $\begingroup$ The denominator on the right side should be 10N, not N $\endgroup$ – user581844 Feb 21 '19 at 3:26
  • $\begingroup$ Sorry, I meant numerator $\endgroup$ – user581844 Feb 21 '19 at 4:07
  • $\begingroup$ @user581844 The way I have it (with N in the numerator, not 10N) seems to be correct. $\endgroup$ – Tanner Swett Feb 21 '19 at 5:03
  • $\begingroup$ Since I'm quoted here, I'd like to clarify: I wasn't questioning whether the equality in question is true in this particular case (it certainly is, and the proof isn't difficult), merely suggesting that it needs proving and that the person whose answer I was commenting on didn't seem to have noticed that there was anything to prove. (I think it's entirely possible that there's some perspective that does make it so immediately obvious that only a few words are required.) $\endgroup$ – Gareth McCaughan Feb 21 '19 at 11:00

We may as well say that you start with a bankroll of $2$ and bet $1$ each time.

Let $E_B$ be the expected number of rolls until the bankroll is gone, if the current bankroll is $B$. We have $$ E_B=1+{18\over 37}E_{B+1}+{19\over37}E_{B-1}$$ or $$18_{B+1}-37E_B+19E_{B-1}=-37\tag{1}$$ This is a straightforward second-order linear difference equation with constant coefficients, but we only have one initial value: $E_0 = 0.$ However, we know that $E_B=BE_1$ by linearity of expectation. Suppose I have a bankroll B. I put $B-1$ in my pocket and play with the remaining $1$ until it is gone. Then I take another $1$ out of my pocket and play until it is gone and so on. The expected number of plays until the whole bankroll is gone is clearly $B\cdot E_1.$

Substituting $E_B=tB$ into $(1)$ we find that $t=37$ so $E_B=37B$ and the suggested answer is correct.

| cite | improve this answer | |

If you re-write the formula as:

$$E(\text{amount of money I lose in one bet}) = \frac{\$N}{E(\text{number of bets I make before losing $\$N$})}$$

it might make more sense (note that one of the E's returns a number, the other an amount of money).

We can look at the two sides as asking, on the one hand,

how many goes knowing the average go is $k$ do I need to make $N$ (answer $\frac{N}{k}$)

and on the other hand as,

if I know that the average of $k$ goes is $N$, can I assume that the average of one go is $\frac{N}{k}$.

A very similar concept is that if I throw $100$ dice and find the average is $350$, and I throw $101$ dice and find that the average is $353.5$, can I assume the average of $1$ die is $3.5$?

| cite | improve this answer | |
  • $\begingroup$ "The sum is correct because in this case we have $E(\sum X)=\sum E(X)$." - Which side is the summation in this case? Neither side particularly looks like a sum of expected values to me. $\endgroup$ – Tanner Swett Feb 21 '19 at 2:58

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.