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.

  • $\begingroup$ @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

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.

  • $\begingroup$ 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

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.