Price of Option in Betting Game

Question

We both put 20 USD into a box. Then, we each generate a number in the interval (0,1) with uniform distribution. The person with the higher number wins and takes 40 USD, whilst the loser is left with 0 USD. I offer to sell you an option that allows you to regenerate your number after you see both of our numbers. What is the price of the option?

This is my approach. After we get our number, there is a 50% chance that yours is higher than mine. If this is the case, I choose to regenerate my number to try and win, which gives me another 50% chance of winning. However, if my number is initially higher than yours (50% chance), then I do not use the option. This gives the expected probability P of winning of:

P = 0.5(1) + 0.5(0.5) = 0.75

Hence price of option would be (0.75*40) - 0.5(40) = $10

However, the answer is 20/3. May someone please explain where I am going wrong?

Answer 1

Let $U_1$, $U_2$ and $U_3$ be the three numbers the probability of you to win $$P(U_1> U_2) + P(U_3 > U_2 > U_1) = \frac12 + \frac16 = \frac23$$

Answer 2

The error is here:

If this is the case, I choose to regenerate my number to try and win, which gives me another 50% chance of winning.

No, it doesn't. It would give you an extra $50\%$ chance of winning if you were allowed to regenerate both numbers, but you only get to regenerate yours. This gives you a $1-X$ chance of winning, where $X$ is my number. And my number was the larger of the two numbers initially generated, so $X$ will typically be higher than $1/2$. (In fact it has a triangular distribution with mean $2/3$ but Kroki's answer gives a more elegant way to see what the chance of winning is.)