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This is just for curiosity's sake, but I found this problem, and I don’t know how to solve it. This is not the regular Monty Hall Problem!

You're a participant in yet another version of Let's Make a Deal! Each competitor must select from three doors to discover a reward. The reward is placed behind one of the three doors, and its location is determined uniformly randomly over the three doors. Every competitor picks independently, and multiple or none might select the right door. The winning door is disclosed after each turn.

You possess an advantage the others lack: you're aware that the reward will not be behind the same door as the last round. What is the probability that you will have chosen the correct door more than each other player by the end of the game?

Now this may just be a poorly worded problem, but here is what I'm confused about:

  • Does "have chosen the correct door more than each other player" mean for all players, you have chosen the correct door more times than they have? I would assume not, since it doesn’t specify a number of players. I figured that this meant that for "have chosen the correct door more than some other player"
  • How do we condition the number of rounds? Initially, my instinct was to say we find the answer over infinitely many rounds, but, using Gambler's Ruin (or just intuition), we get the trivial result that $$\lim_{N \to \infty}\frac{1-\left(\frac{0.\bar 3}{0.5}\right)^\frac{N}{2}}{1 - \left(\frac{0.\bar 3}{0.5}\right)^N} = 1$$ (this is the formula for winning Gambler's Ruin on an interval of [0, N] starting at N/2 with probability 0.5 that we win and 1/3 of the other player winning from Blitzstein & Hwang.)
  • Otherwise, I don't see a way to solve this problem without assuming a number of rounds. So what’s the right approach to this problem?
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1 Answer 1

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Let us assume you have $K$ other players and $N$ rounds. Each player has a $1/3$ probability in each round of guessing the correct door; this means that, over $N$ rounds, the number of correct doors a player chooses $\sim \text{Binomial}(N,1/3)$. With $K$ players, the cumulative distribution function of the maximum number of correct doors $P(n;N,1/3,K)$ is:

$$P(n;N,1/3, K) = P_{\text{Binomial}}(n;N, 1/3)^K$$

and we calculate the probability mass function by:

$$p(n; N, 1/3, K) = P(n; N, 1/3, K) - P(n-1; N, 1/3, K)$$

Now for us; the distribution of our number of correct doors, say $x$, is evidently $\text{Binomial}(N,1/2)$, since we will not pick the (wrong) door that was the correct door in the previous round, leaving us with two possibilities.

So the question comes down to: what is the probability that $x > n$?

$$p(x>n) = \sum_{i=0}^{N-1}p(i;N,1/3,K)\sum_{j=i+1}^Np_{\text{Binomial}}(j;N,1/2)$$

Now for a little simulation check:

K <- 3
N <- 4
tmp <- 0
for (i in 0:(N-1)) {
  tmp <- tmp + (pbinom(i, N, 1/3)^K - pbinom(i-1, N, 1/3)^K) * (1 - pbinom(i, N, 1/2))
}

# Simulation
wins <- 0
for (i in 1:10000) {
  max_other <- max(rbinom(K, N, 1/3))
  our_score <- rbinom(1, N, 1/2)
  wins <- wins + (our_score > max_other)
}

with result:

> tmp
[1] 0.3157604
> wins / 10000
[1] 0.308
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