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A game is played as follows: A random number $X$ is chosen uniformly from $[0, 1]$. Then a sequence $Y_1, Y_2,\ldots$ of random numbers is chosen independently and uniformly from $[0, 1]$. The game ends the first time that $Y_i > X$. You are then paid $(i-1)$ dollars. What is a fair entrance fee for this game?

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  • $\begingroup$ (i1) dollars? Do you mean (i-1) dollars? $\endgroup$ – k1next Nov 14 '14 at 21:46
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    $\begingroup$ Every time Y is smaller than x you receive a dollar. $\endgroup$ – Schidu Luca Nov 14 '14 at 21:48
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Let $Z$ represent the payment. Then $$\Bbb{P}(Z = k|X = x) = \Bbb{P}(Y_1 \le x, Y_2 \le x, \ldots , Y_k \le x, Y_{k+1} > x) = x^k(1 − x) .$$ Therefore, $$\Bbb{P}(Z = k) = \int_0^1 x^k(1 − x)\operatorname{d}x=\left[\frac{x^{k+1}}{k+1}-\frac{x^{k+2}}{k+2}\right]_0^1=\frac{1}{k+1}-\frac{1}{k+2}=\frac{1}{(k+1)(k+2)}.$$

Thus, $$\Bbb{E}(Z) =\sum_{k=0}^{\infty} k\Bbb{P}(Z = k)=\sum_{k=0}^{\infty}\frac{k}{(k+1)(k+2)},$$ which diverges.

Thus, you should be willing to pay any amount to play this game.

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Hint: If I told you $X$, can you calculate the value $V(X)$? Then since $X$ is selected uniformly, the overall value is $\int_0^1V(X)\ dX$

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