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A fair coin is tossed until heads ($H$) appears but is tossed only up to a maximum of $10^6$ times. If $H$ appears on the $n$-th toss, the player gets $2^{n-1}$ dollars. If all tosses are tails $(T)$ then you lose all of your money. The entrance fee for this game is $10^5$ dollars. Should you play this game?

I have tried to look at similar problems like this but I am not sure how to make the connection from: #of tosses until first heads to the chance that I will get at least one heads in the $10^6$ tosses and then subsequently turning that into an expected value. How can I solve this problem?

Nullspace
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  • What makes you think that $P(\text{only tails})=10^{-6}$? – drhab May 20 '21 at 09:16
  • $P(\text{only tails})=2^{-10^{6}} \approx 10^{-301030}$ – Henry May 20 '21 at 09:18
  • @drhab I just realized we don't have $10^6$ outcomes but we have. $2^{(10^6 )}$ outcomes. My bad. – Nullspace May 20 '21 at 09:18
  • Well, as @Henry made clear that there are much more outcomes. For each of the $10^6$ throws there are two possibilities. – drhab May 20 '21 at 09:23
  • It is a bit of a nonsense. Your calculations will lead to you assuming you might win many many orders of magnitude more than all the money and wealth in the universe – Henry May 20 '21 at 09:23

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Partial answer: Assume the first head occurs on the $N$'th toss (so $N$ is a random variable). Let $X$ be the amount of dollars gained. Then $$ X = \begin{cases} 2^{N-1} & N\le 10^6 \\ 0 & N > 10^6 \end{cases} , $$ and we want to find $E(X)$. I claim that $$ P(N=n) = 2^{-n}. $$ Can you see why? Think about what the individual tosses must be. And now evaluate $$ E(X) = \sum_{n=1}^\infty P(N=n)X(n). $$

Milten
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