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Someone pointed out this puzzle in Kahneman's Thinking, Fast and Slow. I paraphrase.

A game rewards participants based on how long it takes to obtain the first heads in a toss of a fair coin. If the first toss results in heads, then the player gets reward $r(1) = \$2$. If the first heads appears only in the second toss, then the player gets $r(2) = \$4$. And $r(3) = \$8$ if the first heads appears on the third toss, and so on. How highly can one price the ticket before this game stops making sense for a player?

It seems to me that the solution depends on what I choose for the random variable. If I choose the payoff as the random variable $X_P$, then the expected reward is the expected payoff, i.e. $E[\mathsf{reward}] = E[X_P] = \Sigma_{i=1}^{\infty} (\frac{1}{2})^i \times r(i) = \frac{1}{2}\times\$2 + \frac{1}{4}\times\$4 + \ldots = \infty$. This is the approach discussed in the book.

However, if I choose as random variable $X_N$ the number of tosses it takes to obtain the first heads, then my expected reward is the reward associated with the expectation $E[X_N]$ of the number of tosses leading up to and including the first heads, i.e. $E[\mathsf{reward}] = r(E[X_N]) = r(2) = \$4$ [cf. this].

While the first analysis suggests that a player ought to play the game no matter the cost of ticket, the second analysis pegs the ticket price strictly below $\$4$ for the game to be attractive for a player.

Is my alternative analysis correct? If yes, then how do I decide which is the correct approach to such puzzles?

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Principle: $\mathrm{E}(f(X)) \ne f(\mathrm{E}(X))$ given the function $f$ is not linear.

In your case, $r(\mathrm{E}(X_N)) =2^{\mathrm{E}(X_N)} \ne \mathrm{E}(r(X_N)) = \mathrm{E}(2^{X_N}) = \infty$, because $2^x$ is not linear function of $x$.

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  • $\begingroup$ True. If the reward function was linear in the number of tosses until the first heads, then the two expectations would turn out to be equal. The question, however, is which of the solutions would you favour to argue about the correct ticket price? $\endgroup$ – N. CHATURV3DI Jun 23 '17 at 9:43
  • $\begingroup$ When you are gambling, you want to get money. So it is very clear you need expectation of the money you will win, i.e., $\mathrm{E}(2^{X_n})$. $\endgroup$ – user158565 Jun 23 '17 at 14:55

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