We have a fair coin. The game ends when we get a head. the payoff is $2^N$ where N is the number of flips until we get the head. What is the expected value of the game? or, how much would you pay to play?
The series goes to infinity. We can get the head at the 1st,2nd,...,Nth flip. The exp. payoff would thus be:
$E(x) = \frac{1}{2}2 + \frac{1}{4}4+...+\frac{1}{2^N}2^N$
$E(x) = \sum_{k=1}^n 1 $
How much would I pay? I don't know honestly
This question as said in the comments is a famous paradox, I would recommend reading some of the tagged literature to get a quantitative understanding, but I will provide a few qualitative things to point about.
Yes the expected value is infinite, does this mean you should mortgage your house, sell your TV and flog all your possessions to play it once? Surely the sum of all that isn’t infinite? No you shouldn’t, obviously not.
The notion of expected value works great when used as an indication for purchasing with “relatively small” values, lowish variance and a few other factors.
The truth is we shouldn’t think of money as linear or symmetric. That is loosing $\$10,000$ and winning $\$10,000$ shouldn’t be equally bad / good. Likewise winning ten billion dollars isn’t a thousand times better than winning ten million dollars. In fact for a lot of normal people they would be sort of the same?
In a nutshell this question deals with an infinite amount extremely unlikely but extremely large values and as such expectation is somewhat misleading. Account for any finiteness to the game, for example consider the all the capital in circulation and the game becomes worth about $50$ bucks. Definitely not worth selling your house for!
You should look until utility functions, specifically logarithmic for easy calculations in this question to better understand it!