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?

  • 5
    $\begingroup$ I guess, for the case that you already lost the first time, your opponent's number is more likely to be large and not uniformly distributed, and you don't have $50\%$ chance of winning the second time? $\endgroup$
    – peterwhy
    Commented Jul 26, 2023 at 3:24
  • 1
    $\begingroup$ @peterwhy is correct. If the other player's number is higher than your (first) number, then you will choose to regenerate; but your probability of winning will be less than $50\%$. In particular, the probability that your second number is the highest of the three drawn so far is just $1/3$. So $P=0.5\cdot 1 + 0.5 \cdot (1/3)=1/2 + [1/6]$. Hence the price of the option would be $[1/6] \cdot 40 = 20/3$. $\endgroup$
    – mjqxxxx
    Commented Jul 26, 2023 at 4:02
  • 6
    $\begingroup$ You pay for the option before, so using it does not cost anything after $\endgroup$
    – Anon
    Commented Jul 26, 2023 at 11:12
  • 1
    $\begingroup$ @Anon The value of a pre-game option depends on the probability you will exercise the option and the probability you will win if you do. That second probability depends on your opponent's random number and your first random number and is clearly not 50%. $\endgroup$
    – matt_black
    Commented Jul 26, 2023 at 14:33
  • 3
    $\begingroup$ @nicola You pay before the game begins, to have the option of rerolling. If you paid after seeing the initial rolls, then you would not be buying an option to reroll, rather you would have the option to buy a reroll. $\endgroup$
    – kaya3
    Commented Jul 26, 2023 at 16:30

2 Answers 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.)


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

  • 1
    $\begingroup$ This answer is implicitly using the fact that it doesn't matter whether the third number is chosen before you exercise the option, or if it's chosen at the beginning $\endgroup$
    – Carmeister
    Commented Jul 26, 2023 at 13:11
  • $\begingroup$ @nicola I don't understand your comment. $\endgroup$
    – Kroki
    Commented Jul 26, 2023 at 14:02
  • $\begingroup$ That was clarified in the comments. My understanding was that you could pay just in the instances you want a reroll; actually, it looks like you buy before the game just to have the option of rerolling and in this case it's obvious that you reroll any time you have lost the first roll. $\endgroup$
    – nicola
    Commented Jul 27, 2023 at 4:42

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .