How does the expected value change in a lottery when one stops playing after winning?

Question

Using as an example for lotteries in general, the expected value of roulette is negative, which means in the limit of playing for eternity, one is expected to lose a certain amount of money for each dollar bet.

The expected value for the sequence is $ \lim_{n \to \infty } ( \frac{(\sum_n (-\text{loss})*p(\text{loss}) + (\text{win})*p(\text{win}))}{n} ) $.

How does this expected value change if the strategy is stopping after winning for the first time? Assume unlimited time and money like we would when talking about the limit above.

The expected value becomes

$\lim_{n \to \infty}(\frac{S_n}{n})$

Where

$ S_1 = (\text{win})*p(\text{win}) + (\text{loss})*p(\text{loss}) $

$ S_n = (\text{win})*p(\text{win}) + p(\text{loss}) * (\text{loss} + S_{n-1}) $

I feel too rusty to actually compute what that limit would be, and haven't been able to find the result that surely exists somewhere.

For european roulette when betting on colors, win=1, loss=-1, p(win) = 18/37 and p(loss) = 19/37, but I am more interested to see how the expected value looks like in terms of win, loss, p(win) and p(lose) rather than some concrete number.

Answer 1

If you bet one dollar until you win for the first time, the result is a simple geometric series. Or you use a simple equation: with roulette and two zeroes you either win the first time or not, so

E = (1/37) * 35 + (36/37)E - 1

that is,

E = -2

Note that if your bets are going up too fast (like doubling after each bet) some formulas for expected values don’t work anymore.

Answer 2

The amount you "win" if you win at the $k$th step: $W_k = (-\text{loss})*(k-1)+(\text{win})$

The probability of winning at $k$th step: $P_k=(p_{\text{loss}})^{k-1}p_{win}$

Expected value:
$E[W]= \sum_{k=1}^{\infty} P_kW_k$
$E[W]= \sum_{k=1}^{\infty} (p_{\text{loss}})^{k-1}p_{win}[(-\text{loss})*(k-1)+(\text{win})]$

This sum can be split in a geometric series and an arithmetico-geometric series. You can use sum of the terms formulae to calculate final expected value.

Answer 3

$\def\eqdef{\stackrel{\text{def}}{=}}$ If you're betting just one dollar each round, and a win pays even money, then with probability $\ p(\text{loss})^{n-1}p(\text{win})\ $ you will lose $\ n-1\ $ rounds before winning on the $\ n^\text{th}\ $, and your nett winnings in that case will be $\ 2-n\ $ dollars. Your expected winnings will therefore be \begin{align} E(W)&=\sum_{n=1}^\infty(2-n)p(\text{loss})^{n-1}p(\text{win})\\ &=(1-\ell)\sum_{n=1}^\infty(2-n)\ell^{\,n-1}\\ &=2-(1-\ell)\sum_{n=1}^\infty n\ell^{\,n-1}\\ &=\frac{1-2\ell}{1-\ell} \end{align} dollars, where $\ \ell\eqdef p(\text{loss})\ $. As expected , $\ E(W)<0\ $ for $\ \ell>\frac{1}{2}\ $, thus amounting to an expected loss in that case. The expected loss is larger by a factor of $\ \frac{1}{1-\ell}\ $ (i.e. more than twice as large) than that of the $\ 2\ell-1\ $ dollars which you expect to lose from a single bet of one dollar.

For betting a colour on a European roulette wheel, the expected winnings are \begin{align} E(W)&=\frac{37-38}{37-19}\\ &=-\frac{1}{18}\ \ \ \text{dollars.} \end{align}

It's worth pointing out that one of the assumptions on which this calculation is based—namely, that the gambler has an unlimited amount of funds available—is certainly false. If the total amount of money available to the gambler is $\ F\ $ dollars, and he or she sustains a total of $\ F\ $ losses in a row, then no further play is possible. The expected winnings therefore become \begin{align} E(W)&=-F\ell^F+(1-\ell)\sum_{n=1}^F(2-n)\ell^{n-1}\\ &=\frac{\big(1-\ell^F\big)(1-2\ell)}{1-\ell}\ . \end{align} Thus, when $\ F\ $ is very large, the difference between this and the previous result becomes negligible.

There's an extensive literature on this subject. One of the classics is Dubins and Savage's How to Gamble if you Must; Inequalities for Stochastic Processes. A preview on p.2 describes the general class of problems tackled in the book, and the main conclusion:

Suppose the only kind of gamble available to you in your effort to convert a sum of money into a larger target sum is to bet, at fixed, subfair odds, on independent repetitions of some fixed kind of event, such as drawing a red card or a spade. You are free to choose for yourself the amount to stake on each bet, subject only to the restriction that you can never stake what you do not possess. Under these circumstances, one—not usually the only—optimal strategy for you is to play boldly; that is, always to stake on each bet either all the money in your possession or just enough to arrive immediately at the target sum in case you win the bet, whichever of these two stakes is the smaller

"Optimal strategy" here means any betting procedure that maximises the probability of ultimately winning (i.e. reaching the targeted amount).