# What's the expected value in this jackpot winning experiment.

If there exists a fair National Lottery, that someone bets £1, Jackpot increases by £1, and there is p chance that he wins a Jackpot. If a Jackpot is won, it is reset to 0. repeats.

We can easily model this as a Bernoulli trials and the average Jackpot size is 1/p.

Now, if we observe that there are £M bets staked on the National Lottery, and it is won n times. What's the average Jackpot size in this case?

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The average jackpot is $\frac{M-J}{n}$ where $J$ is the amount left in the pot after the $M$ trials. This is a linear function of $J$, so given $E(J)$ the expected average jackpot is $\frac{M-E(J)}{n}$.

Let $j(x, y)$ be the expected jackpot after $x$ trials of which $y$ are wins. Since the trials are memoryless the probability that the last trial is a win is $\frac{y}{x}$, so we get a recurrence

$$j(x, y) = \begin{cases} \text{undefined} & \text{if } x<\min(y, 0) \\ 0 & \text{if } x=y \\ \frac{x-y}{x} \cdot (1 + j(x-1, y)) & \text{otherwise} \end{cases}$$

We can expand that as $$j(x,y) = \frac{x-y}{x} \left(1+ \frac{x-1-y}{x-1}\left(1+ \frac{x-2-y}{x-2}\left(1+\ldots\left(1+\frac{1}{y+1}\right)\ldots\right)\right)\right)$$

Then by induction from the inside out, $j(x, y) = \frac{x-y}{y+1}$, so $E(J) = \frac{M-n}{n+1}$ and the expected average jackpot paid out to the $n$ winners is $\frac{M-1}{n+1}$

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i doubt that it has nothing to do with p, i had thought we have to apply some posterior on that. (Bayesian inference) Is it valid that each trial has equal chance to win a jackpot regardless of how low the prior p is? –  colinfang Feb 14 '12 at 17:12
@colinfang, each trial has the same $p$ a priori, so by symmetry each is equally likely to be one of the $n$ winners. If that doesn't convince you, we have $\binom{M}{n}$ possible sequences of results; each with probability $p^n(1-p)^{M-n}$ a priori but $1/\binom{M}{n}$ a posteriori, and you can count the number of them in which the last trial is a win. –  Peter Taylor Feb 14 '12 at 17:26