Making a probabilistic function for equal ballance

Question

This is the game:

• Then, you choose a number from 1.00 to +infinity
• Then, the other player choose a random number from 1.00 to +infinity
• If you choosed a number smaller or equal the other players's, you win a prize of your choosen number
• If you choosed a number bigger than the other player's, you win nothing.

You, as a player, can choose any strategy on this game.

Answer 1

If I understand the game right:

• I choose a number $X$
• The plane chooses a number $Y$ (which I assume is chosen independently of $X$?)
• If $X \leq Y$, my payout is $T = X - 1$ (the winnings less the initial bet)
• If $X > Y$, my payout is $T = -1$ (i.e. I lose my bet)

Then we can first consider what happens to the payout for any individual value of $X$:

$\begin{eqnarray} \mathbb{E}(T | X = x) & = & (x - 1)\mathbb{P}(Y \geq x) - \mathbb{P}(Y < x) \\ & = & (x - 1)(1 - F_Y(x)) - F_Y(x) \end{eqnarray}$

where $F_Y(x)$ is a not-quite-standard version of the cumulative distribution function of $Y$, i.e. $F_Y(y) = P(Y < y)$.

If you want a fair game, then you want this expectation to equal 0 for any valid value of $x$, which corresponds to the strategy of always picking the same value every game. This works out to be $F_Y(y) = 1 - \frac{1}{y}$.

If you're implementing this by taking a random draw $P \sim Unif(0, 1)$ then you can just invert the formula to get $Y = \frac{1}{1 - P}$.