I have received is question on an interview.
"If I offered you a game where you would flip a coin until you get a tails and you receive 2^n payout where n is the number of flips, how much would you pay me for this game?"
Meaning if I got 4 heads in a row, My payout would 2^4 or $16.
My Solution: the expected value at all states is equal to (1/2)^n * 2^n = 1, summed across all states would be infinity. Infinity sounds like a stupid answer because why would you pay an infinite amount to receive possible a few dollars. Do you guys have other insights into this question?
This is the St. Peter's paradox, and it's a paradox because, as you correctly state, the expected value of this game is infinite, but it would be stupid to pay a very large amount to play it.
There are a lot of reasons for this. One is risk aversion, and people don't want to risk $100 for a very small chance to make a profit (will happen about 1/64 of the time).
That doesn't fully answer it, because not everyone is risk averse, so why shouldn't a rational person who enjoys gambling play for a massive amount of money? I believe the best response to this is diminishing marginal return of the payout...by the time you get to, say, the 200th 1 that is being factored in, the odds are miniscule (1 in 10^15) and they only continue to add 1 because they are being multiplied by an increasing number.
Since in real life a $1 quadrillion payout has about the same meaning as a $1 quinitillion payout, we shouldn't take the "1"s that we factor into the E(V) seriously after a certain point. I believe different people will pay a different amount for this game based on where they start to make that cut off, in conjunction with their appetite for risk.
Read more about it here: http://plato.stanford.edu/entries/paradox-stpetersburg/
Hope this helps.
With respect to your question, it really is just asking for the mean of geometric distribution (google this if you do not know what it means; alternatively, refer to http://en.wikipedia.org/wiki/Geometric_distribution).
In short, your expected return should simply be 2. You should not pay more than 2 to play this game - it doesn't matter that this game could potentially stretch to infinity.
As has been mentioned, P(X=n) = [(1-p)^(n-1)]p = [(1-0.5)^(n-1)]0.5 = 0.5^(n). This would thus generate E(x)= 1/p = 1/0.5 = 2.
If the price to play this game were 2 simply because you're already at a disadvantage at the first toss (remember the probability of you getting it right on the nth count diminishes as n increases).
everythingsucks, T3H: Thanks for pointing that out - I had erroneously considered the 2^0 case which is impossible. Corrected the reasoning.
Dude, you are so wrong. No matter what, you will get at least 2$.
seconded, totally wrong.
Thanks for your answers. Effectively its a trick question. Even if you consider diminishing marginal returns you're still effectively summing over n states even if at some n sub t - you have an expected value that converges to 0 but never reaches it.
So there is no answer.
Find the smallest n, such that you are (practically) indifferent between 2^n and 2^(n+1). Then you should be willing to bet n.
Everythingsucks: Could you elaborate on that answer?
Suppose I am indifferent between winning 2^23 = 8, 388, 608 and 2^24 = 16, 777, 216, because having around 8 million dollars is more money than I'll ever need, so having anything more than that adds no practical value.
Then for me, personally, this game is equivalent to a game that is as follows: Keep flipping coins until you get tails. Then for n in {1,2,......,22,23}, if the number if flips is n, the payoff is 2^n. If you don't get tails in 23 flips, you still get 2^n. Then the expected payoff of this game is (very very close to) 23$. Which, should then be the fair price of the game.
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