Unfair Wager

Julie proposes the following wager to her sister Louise.
She suggests they play a game of chance to determine who will wash the dishes.
For this game, they shall use a generator of independent random numbers uniformly distributed between $0$ and $1$.
The game starts with $S = 0$.
The first player, Louise, adds to $S$ different random numbers from the generator until $S \gt 1$ and records her last random number ‘$x$’.
The second player, Julie, continues adding to $S$ different random numbers from the generator until $S \gt 2$ and records her last random number ‘$y$’.
The player with the highest number wins and the loser washes the dishes, i.e. if $y \gt x$ the second player wins.
For example, if the first player draws $0.62$ and $0.44$, the first player turn ends since $0.62+0.44 \gt 1$ and $x = 0.44$.
If the second players draws $0.1$, $0.27$ and $0.91$, the second player turn ends since $0.62+0.44+0.1+0.27+0.91 \gt 2$ and $y = 0.91$.
Since $y \gt x$, the second player wins.
Louise thinks about it for a second, and objects: “That’s not fair”.
What is the probability that the second player wins?
Give your answer rounded to $10$ places behind the decimal point in the form 0.abcdefghij.

This is a combination of probability with calculus.

Let’s denote by P(x) the probability that the total of Julie’s draws exceeds 1 before she draws a number less than x. Since the only way this can happen is if all of Julie’s draws are larger than x, and since each draw is uniformly distributed in [0, 1], it follows that P(x) = (1-x)^2 (we square it because there are two draws). We’re assuming that the total without the last draw is less than 1, so the first draw which is less than x must make the total exceed 1; hence, P(x) is also the probability that Julie’s last draw is less than x.

It’s clear that the probability that Julie wins is the integral from 0 to 1 of P(x) dx, since these are all the possible values of Louise’s last draw, and Julie wins if and only if her last draw is less than Louise’s. Thus, we calculate:

∫₀¹ P(x) dx = ∫₀¹ (1-x)^2 dx
= [ -(1-x)^3/3 ] (from 0 to 1)
= 1/3 (rounded to 10 decimal places 0.3333333333)

This means that the second player, Julie, has a winning probability of about 0.3333333333. So in fact, the game is not fair, and Louise would be wise to refuse this wager!

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