Hello,The usually proposed solution of the paradox is using some kind of utility function...But, if we just take a look at the expected number of coin tossings which is just 1/p,with p = 1/2 for a fair coin. It means that the expected duration of the game isjust 2 tosses! If on average the game last only 2 tosses, the payout won't be huge!why do we need some sort of utility function to solve it? or maybe I'm missingsomething, is it a wrong solution?Thanks

Computing the "average number of tosses" contains the implicit assumption that all toss counts have equal weight. Because the payout doubles with each toss, the high-toss counts have much higher weight. If the contest is taken literally and the player has constant marginal utility, then the properly averaged outcome becomes infinite.If the player has declining marginal utility or is skeptical that the game provider can or will make the payout (counterparty risks), then the average payoff will be bounded.

Yes, but the doubling payouot won't increase the length of the game, it doesn't depend on the weight you put on each tossing.The game still will last only 2 tosses on average...

Don't confuse the average with the individual. The average game might be two rounds, but some individual games will last much longer than average and produce a payoff much more than does a two-round game.

In practice this will hardly ever happen...

Very true. But when you multiply the "hardly ever happens" times the jackpot payoff if a longer game does happens, the result is a high expected value.How much would you pay for a 1-in-a-1,048,576 chance to win a $1,048,576? A risk-neutral person would pay $1. That the game outcome is determined by flipping a coin 20 times in a row, only awarding the payoff if all 20 flips are heads, and that most times the player loses on the second flip is irrelevant.The point is that all that matters is the sums of the products of all the probabilities and all the outcomes.

This thread is hilarious. Are you an aspiring quant, Andre?

You need a utility function just not to "overbet". The paradox is very closely related to the optimal f(kelly criterion), I wrote a long article here in my old profile, about the kelly criterion whereas I included computer code and detailed formulas - you can take a look here. I can write much more now about the St. Paradox...but Im just busy as hell now...http://www.wilmott.com/messageview.cfm? ... SGDBTABLE=

QuoteOriginally posted by: MatthewMThis thread is hilarious. Are you an aspiring quant, Andre?Yes, Matthew, I am

QuoteOriginally posted by: 4ndre1QuoteOriginally posted by: MatthewMThis thread is hilarious. Are you an aspiring quant, Andre?Yes, Matthew, I am Ask all the question you have here as most people are happy to help. If you don't ask you will end up with misconceptions that I have seen even some PhDs have because they were to shy to ask even the trivial.Cheers,Alk

Yeah, yeah. Just having a bit of fun with the new guy. But in all honesty the mistake of saying that f(E(X)) = E(f(X)) is a bit much.

QuoteOriginally posted by: MatthewMYeah, yeah. Just having a bit of fun with the new guy. But in all honesty the mistake of saying that f(E(X)) = E(f(X)) is a bit much.ha-ha, now I understand your sarcastic mood, Matthew Well, this is not what I wanted to say, even if mathematically it can be interpreted like this.When one is asked how much he/she would pay for the game, does one starts calculating the E(f(X)) (maybe you do, Matthew?), with f his/her utility function?Or one just reasons that anyway the game can't last too long, why should it cost much then? And if you see 20 heads in a row, probably you should start tohave doubts on the fairness of the coin....

QuoteOriginally posted by: giordanobrunoYou need a utility function just not to "overbet".Although a utility function (of the proper type) would bound one's betting, it's not the only framework for estimating a bounds on "overbetting".If one is concerned about betting infinity on this game and losing all their money, shouldn't they also be concerned with expecting the counterparty to pay 2^Infinity?This game has the Gambler's Ruin problem written all over it except that the pay-off structure means that unless the bank/casino has more than 2^(2*(Player_Wealth - 1)), it will be the bank/casino that is ruined first. This factor bounds the bet at quite low levels. If the casino has $1 billion, then the expected value of the game is just under $16. Even if you are betting on the entire GDP of the world, the expected value is less than $24.

Traden4Alpha, thanks for those useful insights.

4ndre1, seems to me, it still falls into the utility function category if you want to use the expected stopping time to explain away the paradox. one common choice is log, and if you starts with zero, the utility function is proportional to the number of rounds the game lasts.