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        riddles >> hard >> Infinite expected value coin-flip game
        
(Message started by: River Phoenix on Feb 21^st, 2006, 5:17pm)

Title: Infinite expected value coin-flip game
Post by River Phoenix on Feb 21^st, 2006, 5:17pm

I play a game with you: You bet X. I flip a coin, and if it is heads I pay you 0. If it is tails then I flip again and if it is heads I pay you X, otherwise I flip again. Now if it is heads I pay you 2X, or flip again, then 4x, 8x, and so forth. Having seen this problem before, you realize that your expected value is infinite. But even if you were risk-neutral, would you really be willing to put your house on the line? I am very curious about this question because it reminds me of the symmetrical random walk in 1 or 2 dimensions, in which you are always guaranteed to return to the origin, but you expect it to take infinitely long. So, suppose that you are willing to lose 20 dollars. Would you rather play the game once and bet X=20, or play the game 20 times and bet X=1? What happens if you let your bet go to 0 and number of times played go to infinity? What if you start with 1 dollar and always bet 1 dollar, and continue to play the game with your winnings forever? What is the probability that you go broke?

I am not sure whether or not the answer to these questions is trivial, or if there is another thread already, but I am very curious to better understand how this game would work.

Title: Re: Infinite expected value coin-flip game
Post by towr on Feb 22^nd, 2006, 12:57am

I'd rather play 20 times with $1 dollar. It decreases the risk, yet has the same expected payoff.

Title: Re: Infinite expected value coin-flip game
Post by River Phoenix on Feb 22^nd, 2006, 2:51am

Right; so all of these questions are easy, except for this: "What if you start with 1 dollar and always bet 1 dollar, and continue to play the game with your winnings forever? What is the probability that you go broke?" I think I could figure this out by myself, but anyway I guess it's not a bad question.

Title: Re: Infinite expected value coin-flip game
Post by towr on Feb 22^nd, 2006, 3:15am

I think that B = 0.5 sum(i=0:inf (B/2)^i)
So B=1.
(B being the chance of going broke, if you hadn't guessed)

Title: Re: Infinite expected value coin-flip game
Post by River Phoenix on Mar 6^th, 2006, 2:40pm

Doesn't that say that it's not so great to repeat the game, after all? ;D

Title: Re: Infinite expected value coin-flip game
Post by towr on Mar 7^th, 2006, 12:17am

Repeating is in that respect no worse than playing at all.

However once 20 would still be worse than 20 times 1 (in regards to risk averseness), I think. Because you're also dealing with variance, skew etc.
The expected pay off may not chance, nor that in the long run you should lose everything. But by decreasing the variance, you do have a better idea of what to expect.

Title: Re: Infinite expected value coin-flip game
Post by River Phoenix on Mar 8^th, 2006, 12:36pm

This is what I thought was so confusing about the game to begin with. It's like the random walk where you always will return to the origin.. but you shouldn't stand around at the origin waiting because you expect it to take infinitely long. Or should you? Since the probability that it takes infinitely long may be very very small. This is what makes it all confusing to me.

The expected value of the game is infinite,but because of variance, it makes sense that more you repeat the game the better off you are, so if I have 20 dollars instead of betting 20 once, I want to bet x 20/x times and let x->0 (is this not true since the game in some sense perhaps has infinite variance?). Since the game has infinite expected value, once all this is done you should do it again with your winnings, and again after that. But if you do this forever you will definitely go broke.
Given that we have all day to play the game, but not an "infinite" amount of time, then what is the best strategy?

Sounds like an addicting casino game if it has infinitely positive expected value yet you'll certainly go broke.

As may be obvious, I have a very limited understanding of infinities, so I'm hoping that a few people will chime in. >:(
By the way, does anybody know a simple random walk with the property I am referring to? For the normal symmetric random walk in 1 dimension, I'm pretty sure the expected time of return is something like 3, but there is such thing as a null-recurrent point which is recurrent but has infinite expected value of time to return.

Currently I have some vague notion of behavior after some arbitrarily large finite number being very different from behavior at infinity, but I'm not sure if that's possible..
My current interpretation is that expected values containing infinities are not a good thing to live your life by. I guess that the case of this game, anybody with a non-zero risk adverseness should not play.

Title: Re: Infinite expected value coin-flip game
Post by cwolves on Apr 7^th, 2006, 7:51pm

Wait a second...let me make sure I have the rules straight here:

I bet x
you flip coin
- if heads: I lose
- if tails: I win x*(2^(number of tails runs-1))

so 50% of the time I win at least what I bet.

how does this possibly have bad odds?

Title: Re: Infinite expected value coin-flip game
Post by towr on Apr 8^th, 2006, 6:28am

on 04/07/06 at 19:51:08, cwolves wrote:

so 50% of the time I win at least what I bet.

25% of the time, you only get as much as you paid. You still have to pay the $X up front after all.
So 50% of the time, you loose $X. And 25% of the time you get it back and break even. Only the other 25% of the time, you'll make a profit.

Winning at least what you bet, doesn't guarantee a positive result. In this case you still have to win 3X for the last 25% to break even on the game as a whole.
Of course with the expected value being infinite that's not a problem in this particular case, but it might be for another game. (For example if the game consists of just two coin tosses, and you get less then 3X for two tails)