Re: Weekend Puzzle

Sukrut wrote:

the expected payout isn't 1/2!

the probability of n tosses isnt 1/2^n, but that of getting a head(or
tail) is.
the probab depends on the number of tosses(n) and is 1/2^(n-1)
if a player wants to play the game, he is assured of winning 1 buck, as
2^(1-1)=1
then, if a head appears, the minimum payout becomes 2, then 4 and so on

thus, the expected payout is
= 1 + 1/2 * 2 + 1/2^2 * 2^2 + ...
= inf
the mean value that can be won is itself infinity

You get the right sum for the wrong reason. There a lot of ways to use
an ellipsis ("...") to reach infinity.

however there is a 50% chance that a tail comes up in the first toss
itself and hence, if the entry fee is greater than 1(=a, say), there is
still a nonzero chance of making a profit
the mean profit = (a-1)*0.5 + (a-2)*(1/2)^2 + (a-4)*(1/2)^3 + ......

again a should be infinity, but practically, if we limit the number of
tosses to a fixed value e.g. 50, we can hope to limit a.

i would pay '1' to be assured of no loss



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.

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