Player survives if generated number < previous number. Probability of Alice winning the game Alice, Bob and Carol are playing a game. Each round they press a number, which generates a random number independently uniformly distributed between $[0,1]$. If the number generated by this player is smaller than the smallest number so far, the player survives. Otherwise they get eliminated. They play in turns, i.e. Alice goes first, then Bob, then Carol, then Alice again etc.
What's the probability that Alice eventually wins this game?
Consider we independently sample $n$ times, the probability that $x_1$ is the smallest is $\frac{1}{n}$, that's the probability that Alice survives after $n$th round.
But how do we find the probability that Alice out-survive the rest of the players?
If there are two players, Alice and Bob, the answer is trivial. But somehow from two to three makes all the difference..

Let $a$ be the number of surviving rounds before the first eliminating round, and $b$ be the number of surviving rounds between the two eliminating rounds. Thus the game ends in $a + b + 2$ rounds. We consider the probability of this situation happening.
If we generate $a + b + 2$ random numbers, then their order take equal probability $\frac 1{(a + b + 2)!}$ for each possible permutation. Among all $(a + b + 2)!$ such permutations, only $a(a + b + 1)$ of them correspond to the situation above, namely the $(a + 1)$-th number can be inserted anywhere before the smallest of the first $a$ numbers, and the $(a + b + 2)$-th number can be inserted anywhere before the smallest of the first $(a + b + 1)$ numbers.
This means that the probability that one player gets eliminated on the $(a + 1)$-th round and another player gets eliminated on the $(a + b + 2)$-th round is equal to $\frac{a(a + b + 1)}{(a + b + 2)!}$.
It is clear that A survives to the end if and only if we are in one of the following two cases:

*

*$a \equiv 1 \pmod 3, b \equiv 0 \pmod 2$;

*$a \equiv 2 \pmod 3, b \equiv 1 \pmod 2$.

Therefore the final answer is equal to $$\sum_{a \equiv 1 \pmod 3}\sum_{b \equiv 0 \pmod 2} \frac{a(a + b + 1)}{(a + b + 2)!} + \sum_{a \equiv 2 \pmod 3}\sum_{b \equiv 1 \pmod 2} \frac{a(a + b + 1)}{(a + b + 2)!}.$$
It is approximately equal to $0.46649280488530$.
