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Suppose there's a lottery. Each ticket sold has probability $p$ of winning, and they are all independent of each other. The size of the jackpot is $j$. If there are $n$ winners, each winner gets a payoff of $j/n$ dollars. The total number of tickets sold is $t$.

What is the expected value of a lottery ticket? Also, given that I win, what is the probability that I have to share the jackpot with at least one other person?

PS - I think I know the answer, but have failed to convince someone else, so I'm looking for a third-party to give an answer.

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  • $\begingroup$ @FlybyNight $n$ is a random variable whose distribution is determined from $p$ and $t$. I believe you do have enough information. $\endgroup$ Nov 29 '12 at 17:46
  • $\begingroup$ For the expected value, I believe you should get $$\frac{j}{t}(1-(1-p)^t) $$ unless my quick chicken-scratch has an error (which has nonnegligible probability). $\endgroup$
    – cardinal
    Nov 29 '12 at 17:58
  • $\begingroup$ @cardinal I hadn't thought of it that way, but it's a good point and looks right to me, thanks. $\endgroup$ Nov 29 '12 at 18:00
  • $\begingroup$ "and they are all independent of each other" may be the source of the disagreement, since it is not the way usual lotteries work. $\endgroup$ Nov 29 '12 at 18:56
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Probability that no one wins: $(1-p)^t$.

Expected value for a particular ticket: $ \dfrac j t (1-(1-p)^t)$. (Pot times probability that the pot is distributed.)

Probability that you win alone: $p(1-p)^{t-1}$

Probability of shared win: $p-p(1-p)^{t-1}$

Probability of sharing given that you win: $\dfrac{\text{prob of shared win}}{\text{prob win}}=\dfrac{p-p(1-p)^{t-1}}p = 1-(1-p)^{t-1}$

This is just one minus the probability of everyone else losing because of independence.

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  • $\begingroup$ Just an aside: While your parenthetical remark on the expected-value result is a good heuristic and provides useful intuition, it is not really a rigorous line of reasoning to arrive at the result. $\endgroup$
    – cardinal
    Nov 29 '12 at 18:33
  • $\begingroup$ @cardinal: Oh yes, it is. (Note that every lottery ticket is created equal in this probability distribution.) $\endgroup$
    – Phira
    Nov 29 '12 at 18:35
  • $\begingroup$ Yes, I agree it can be formalized via an exchangeability argument. $\endgroup$
    – cardinal
    Nov 29 '12 at 18:41

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