Problem 436

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 > 1 and records her last random number ‘x’.
The second player, Julie, continues adding to S different random numbers from the generator until S > 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 > 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 > 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 > 2 and y = 0.91.
Since y > 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

不公平的打赌

朱莉和她的妹妹露易丝打了个赌。
她建议来玩一个几率游戏，来决定谁去洗盘子。
在这个游戏中，她们将使用一个独立随机数生成器，这些随机数等概率地分布在0到1之间。
游戏开始时S = 0。
先手玩家露易丝向S中不断加入生成器生成的随机数，直到S > 1，并记下她最后得到的随机数’x’。
后手玩家朱莉继续向S中不断加入生成器生成的随机数，知道S > 2，并也记下她最后得到的随机数’y’。
得到较大数的玩家获胜，而输掉的一方要去洗盘子，例如，如果y > x，则后手玩家获胜。

例如，如果先手玩家得到的随机数是0.62和0.44，先手玩家结束操作，因为0.62+0.44 > 1，而她得到的随机数是x = 0.44。
如果后手玩家随后得到0.1，0.27和0.91，后手玩家结束操作，因为0.62+0.44+0.1+0.27+0.91 > 2，而她得到的随机数是y = 0.91。
因为y > x，后手玩家获胜。

露易丝考虑了一会儿然后表示反对：“这不公平！”
请问后手玩家获胜的概率是多少？
将你的答案保留10位小数，即用0.abcdefghij的形式提交。