When gambling, do I get my money's worth? (Or: Does the amount I lose per bet determine the number of bets until I lose all my money?) 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?
 A: 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.    
A: 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 $3.5$?
