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A sponsor provides two prizes for a raffle. The first prize winner gets to choose a probability $p$ from $\left[\dfrac{1}{e},1-\dfrac{1}{e}\right]$. A sequence of independent coin flips with probability $p$ for a head are then made. The winner receives $£10$ for each flip up to and including the first head. The same coin is tossed in another independent sequence and the second prize winner receives $£5$ for each flip up to and including the first head in this sequence.

(i) Write down the p.m.f.s of the amounts of the two payments $U$ for the first prize and $V$ for the second prize.

I don't know how to write down the pmfs so that we can differentiate between the first and second prize winner.

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2 Answers 2

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Since the distribution is discrete (only multiples of 10 or 5 are possible), we have: $$f_U(n10)=P(\mbox{n-1 times tail followed by one time head})=p^{(n-1)}(1-p)$$ For the same reason is: $$f_V(n5)=p^n*(1-p)$$ This question is highly related to the geometric distribution. In fact it is almost the same.

Since $E(U)=10\frac{1}{p}$ (use the mean of the geometric distribution to show this), we have that E(U) is maximal when p is minimal (1/e). Since U and V are independent (we toss again for the second winner), we have $E(U/V)=E(U)/E(V)=10\frac{1}{p}\frac{1}{5/p}=2$ and thus constant. This is intuitive because you would guess that the first person gets 2 times as much as the second person.

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  • $\begingroup$ okay this is pretty much what I had actually, but then I need to maximise E(U) and E(U/V) and I'm not sure how to calculate those $\endgroup$
    – Tom
    May 28, 2013 at 14:04
  • $\begingroup$ I edited my answer to answer this question. $\endgroup$ May 28, 2013 at 14:26
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Given $p$, and for $n\in\mathbb{N}^*$, what is the probability that the first player wins £$10n$? (each possible outcome for his reward being of that form). In other terms, what is the probability that in the first sequence of $p$-biased coin tosses, the first $n-1$ tosses are tails, and the $n^\text{th}$ is head?

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