The game
This is a one player game. $n$ numbers are generated uniformly randomly from $0$ to $1$. These numbers are then written on $n$ tiles. One number per tile. You may only look at one tile. You must then guess which tile has the largest number on it.
Quick example
The numbers $0.782$, $0.432$ and $0.882$ are generated randomly. Each wrote respectively on tiles $1, 2$ and $3$. You peek at tile $1$ and see $0.782$ and guess this as the largest tile. You are wrong. You loose.
Solution for n=2
Naturally in this simple case one peaks at any tile and if it greater than $\frac{1}{2} $ you guess that tile and if it is less you then guess the other tile. Let P denote the random variable that is the value on the tile you peak.
$\mathbb{P}[$You win ] = $\int_{0}^1 \mathbb{P}[$You win | $P = p ] f(p) dp $
$f(p) =1 $ clearly. Hence splitting the integral over zero to one half and one half to one yields three quarters.
My question
General formula for the Probability of winning for a given $n>2$
I want to know when one should "stick". I know the expected value of the maximum of $k$ tiles is $\frac{k}{k+1}$. Should we only stick if our peaked tile is greater than that? Intuitively I feel like no. Is there some recursion we could use?
Thanks :)
If the tile you choose has the number $p$, the probability that it is highest is $p^{n-1}$. If you stick, the probability of winning is $p^{n-1}$. If you switch, the probability of winning is $$\frac{1-p^{n-1}}{n-1}$$ This is the probability that you haven't chosen the highest, times the probability of choosing the highest, given that it is one of the other $n-1$.
You should stick when $$p^{n-1}>\frac{1-p^{n-1}}{n-1}$$
This simplifies to $$p>\frac1{\sqrt[n-1]n}$$