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 $1 quinitillion payout, we shouldn't take the "1"s that we factor into the E(V) seriously after a certain point. I believe different people will pay a different amount for this game based on where they start to make that cut off, in conjunction with their appetite for risk.
Read more about it here: http://plato.stanford.edu/entries/paradox-stpetersburg/
Hope this helps.
With respect to your question, it really is just asking for the mean of geometric distribution (google this if you do not know what it means; alternatively, refer to http://en.wikipedia.org/wiki/Geometric_distribution).
In short, your expected return should simply be 2. You should not pay more than 2 to play this game - it doesn't matter that this game could potentially stretch to infinity.
As has been mentioned, P(X=n) = [(1-p)^(n-1)]p = [(1-0.5)^(n-1)]0.5 = 0.5^(n). This would thus generate E(x)= 1/p = 1/0.5 = 2.
If the price to play this game were 2, you'd play it. Conversely, you wouldn't play if the game cost > 2 simply because you're already at a disadvantage at the first toss (remember the probability of you getting it right on the nth count diminishes as n increases).
everythingsucks, T3H: Thanks for pointing that out - I had erroneously considered the 2^0 case which is impossible. Corrected the reasoning.
Dude, you are so wrong. No matter what, you will get at least 2$.
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.
This wouldn't make sense as, by your argument, you could theoretically have any value for an answer and that would be in no way rational.
As a matter of fact, if you reckon $23 to be fair value, you'd "rationally" want to play this game each time I offered you a chance to play at $23 even though the probability of you making a profit is 1/32 (ie. 2^5 where you need to land your first tail on the fifth flip).
Obviously, the probability diminishes as n tends to infinity and I still stand by the crux of the argument that the fair value should revolve around the probability of you getting it right the first time - clearly, this is not a game where you ideally hope to win on the nth try where n is exponentially large (that would be akin to playing the lottery).
To qualify this further, I was actually asked a similar question at a S&T interview some years back with a follow-up question if, assuming I had a tail after 100 attempts, would I still be keen to carry on with this game if I were allowed to rescind the 100th flip (ie. assume it had actually been a head) in exchange for just 1 more flip (with a payout of 2^101 if the 101st flip is a tail). The caveat, however, is that if the 101st flip yields a head, my payout would only be 2^99.
Clearly, while there's no definitive answer to this question, the paradox presented herein has a practical element and limitation with respect to reality as oftentimes, resources aren't infinite (in the context of S&T) and it would clearly make sense to take the 2^100 on offer than risk losing half your profit by flipping one more time.
everythingsucks: this is essentially the 2^n and 2^(n+1) paradox you mentioned above, but this really has no mathematical connotation involved in the context of an interview question
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