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I have a question about the finite analog of the puzzle proposed here involving mathematicians guessing the contents of boxes.

Specifically, suppose there are $k$ unopened boxes each containing a single symbol from an alphabet of $r$ symbols, where $k$ and $r$ are possibly finite nonzero cardinals. A mathematician wants to guess the contents of an unopened box by opening some proper subset (or subsets, if he wants) of the boxes, examining those symbols, and then selecting an unopened box to guess.

Denote the mathematician's guessing strategy (which may of course be randomized) by $\sigma$. For an initial box configuration $C$, let $P_\sigma(C)$ be the probability that the mathematician guesses correctly when using strategy $\sigma$ on configuration $C$. Let $\pi_\sigma$ be the infimum of $P_\sigma(C)$ over all initial box configurations $C$.

When $k=\omega$, it was shown that for any $0\le p \lt 1$, there is a strategy $\sigma$ satisfying $\pi_\sigma\ge p$. (Actually, this was shown for $r=2^{\aleph_0}$, but the same argument works for arbitrary $r$. Compare as well this question , where Eric Naslund showed that $\omega$ mathematicians who independently try the game can guarantee that there is at most one wrong guess between them).

Can someone please explain a simple reason that, for finite $k$ and $r$, $\pi_\sigma\le 1/r$?

This is intuitively obvious, of course - but of course, it's also intuitively obvious when $k=\omega$.

I have a question about the finite analog of the puzzle proposed here involving mathematicians guessing the contents of boxes.

Specifically, suppose there are $k$ unopened boxes each containing a single symbol from an alphabet of $r$ symbols, where $k$ and $r$ are possibly finite nonzero cardinals. A mathematician wants to guess the contents of an unopened box by opening some proper subset (or subsets, if he wants) of the boxes, examining those symbols, and then selecting an unopened box to guess.

Denote the mathematician's guessing strategy (which may of course be randomized) by $\sigma$. For an initial box configuration $C$, let $P_\sigma(C)$ be the probability that the mathematician guesses correctly when using strategy $\sigma$ on configuration $C$. Let $\pi_\sigma$ be the infimum of $P_\sigma(C)$ over all initial box configurations $C$.

When $k=\omega$, it was shown that for any $0\le p \lt 1$, there is a strategy $\sigma$ satisfying $\pi_\sigma\ge p$. (Actually, this was shown for $r=2^{\aleph_0}$, but the same argument works for arbitrary $r$. Compare as well this question , where Eric Naslund showed that $\omega$ mathematicians who independently try the game can guarantee that there is at most one wrong guess between them).

Can someone please explain a simple reason that, for finite $k$ and $r$, $\pi_\sigma\le 1/r$?

This is intuitively obvious, of course - but of course, it's also intuitively obvious when $k=\omega$.

I have an elementary question about the finite analog of the puzzle proposed here involving mathematicians guessing the contents of boxes.

Specifically, suppose there are $k$ unopened boxes each containing a single symbol from an alphabet of $r$ symbols, where $k$ and $r$ are possibly finite nonzero cardinals. A mathematician wants to guess the contents of an unopened box by opening some proper subset (or subsets, if he wants) of the boxes, examining those symbols, and then selecting an unopened box to guess.

Denote the mathematician's guessing strategy (which may of course be randomized) by $\sigma$. For an initial box configuration $C$, let $P_\sigma(C)$ be the probability that the mathematician guesses correctly when using strategy $\sigma$ on configuration $C$. Let $\pi_\sigma$ be the infimum of $P_\sigma(C)$ over all initial box configurations $C$.

When $k=\omega$, it was shown that for any $0\le p \lt 1$, there is a strategy $\sigma$ satisfying $\pi_\sigma\ge p$. (Actually, this was shown for $r=2^{\aleph_0}$, but the same argument works for arbitrary $r$. Compare as well this question , where Eric Naslund showed that $\omega$ mathematicians who independently try the game can guarantee that there is at most one wrong guess between them).

Can someone please explain a simple reason that, for finite $k$ and $r$, $\pi_\sigma\le 1/r$? Why is this so obvious, exactly?

This is intuitively obvious, of course - but of course, it's also intuitively obvious when $k=\omega$.

I have a question about the finite analog of the puzzle proposed here involving mathematicians guessing the contents of boxes.

Specifically, suppose there are $k$ unopened boxes each containing a single symbol from an alphabet of $r$ symbols, where $k$ and $r$ are possibly finite nonzero cardinals. A mathematician wants to guess the contents of an unopened box by opening some proper subset (or subsets, if he wants) of the boxes, examining those symbols, and then selecting an unopened box to guess.

Denote the mathematician's guessing strategy (which may of course be randomized) by $\sigma$. For an initial box configuration $C$, let $P_\sigma(C)$ be the probability that the mathematician guesses correctly when using strategy $\sigma$ on configuration $C$. Let $\pi_\sigma$ be the infimum of $P_\sigma(C)$ over all initial box configurations $C$.

When $k=\omega$, it was shown that for any $0\le p \lt 1$, there is a strategy $\sigma$ satisfying $\pi_\sigma\ge p$. (Actually, this was shown for $r=2^{\aleph_0}$, but the same argument works for arbitrary $r$. Compare as well this question , where Eric Naslund showed that $\omega$ mathematicians who independently try the game can guarantee that there is at most one wrong guess between them).

Can someone please explain a simple reason that, for finite $k$ and $r$, $\pi_\sigma\le 1/r$?

This is intuitively obvious, of course - but of course, it's also intuitively obvious when $k=\omega$.

I have an elementary question about the finite analog of the puzzle proposed here involving mathematicians guessing the contents of boxes.

Specifically, suppose there are $k$ unopened boxes each containing a single symbol from an alphabet of $r$ symbols, where $k$ and $r$ are possibly finite nonzero cardinals. A mathematician wants to guess the contents of an unopened box by opening some proper subset(s) of the boxes, examining those symbols, and then selecting an unopened box to guess.

Let the mathematician's guessing strategy, which may of course be randomized, by $\sigma$. For an initial box configuration $C$, let $P_\sigma(C)$ be the probability that the mathematician guesses correctly when using strategy $\sigma$ on configuration $C$. Let $\pi_\sigma$ be the infimum of $P_\sigma(C)$ over all initial box configurations $C$.

When $k=\omega$, it was shown that for any $0\le p \lt 1$, there is a strategy $\sigma$ satisfying $\pi_\sigma\ge p$. (Actually, this was shown for $r=2^{\aleph_0}$, but the same argument works for arbitrary $r$. Compare as well this question , whose answer showed that $\omega$ mathematicians who independently try the game can guarantee that there is at most one wrong guess between them).

Can someone please explain a simple reason that, for finite $k$ and $r$, $\pi_\sigma\le 1/r$? Why is this so obvious, exactly?

This is intuitively obvious, of course - but of course, it's also intuitively obvious when $k=\omega$.

I have an elementary question about the finite analog of the puzzle proposed here involving mathematicians guessing the contents of boxes.

Specifically, suppose there are $k$ unopened boxes each containing a single symbol from an alphabet of $r$ symbols, where $k$ and $r$ are possibly finite nonzero cardinals. A mathematician wants to guess the contents of an unopened box by opening some proper subset  (or subsets, if he wants) of the boxes, examining those symbols, and then selecting an unopened box to guess.

Denote the mathematician's guessing strategy (which may of course be randomized) by $\sigma$. For an initial box configuration $C$, let $P_\sigma(C)$ be the probability that the mathematician guesses correctly when using strategy $\sigma$ on configuration $C$. Let $\pi_\sigma$ be the infimum of $P_\sigma(C)$ over all initial box configurations $C$.

When $k=\omega$, it was shown that for any $0\le p \lt 1$, there is a strategy $\sigma$ satisfying $\pi_\sigma\ge p$. (Actually, this was shown for $r=2^{\aleph_0}$, but the same argument works for arbitrary $r$. Compare as well this question , where Eric Naslund showed that $\omega$ mathematicians who independently try the game can guarantee that there is at most one wrong guess between them).

Can someone please explain a simple reason that, for finite $k$ and $r$, $\pi_\sigma\le 1/r$? Why is this so obvious, exactly?

This is intuitively obvious, of course - but of course, it's also intuitively obvious when $k=\omega$.