We play a game as follows:
I place one dollar on the table. I repeatedly flip a coin. Each time the coin comes up heads, I double the money on the table. The first time the coin comes up tails, you take the money and the game is over.
What's a fair admission price for the game?
Would you play the game with me for $100?
(In reply to
re: something missing by Cheradenine)
wild speculation:
define the expectancy expectancy E2 as the number of
trials the games takes to reach a certain expectancy
value. in other words E2 determines how many trials
are necessary to expect on average an expectancy E
(for those trials).
E2 is really a function of cost and target expectancy:
E2(expectancy, cost). my attempt at obtaining this for
a given cost is:
find the biggest term in E(cost) which included in
E(cost) >= expectancy. E2 is then given by this biggest
term. in the case of the above problem, it turns out
that E2(0, cost) = 4^cost.
the answer to the first question is that the fair price
is
log4(x)
where you have x dollars.
the answer to the second question is, if i have
4^100 (1,606e+60) dollars or more, i would play.
re(2): something missing
