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Alice and Bob have two positive integers, x and y respectively, glued to their foreheads, so that each can read the other’s number but not their own. They also know that |x − y| = 1. The following conversation is overheard between Alice and Bob:

Alice: I don’t know my number.

Bob: I don’t know my number.

Alice: I don’t know my number.

Bob: I don’t know my number.

...

They say this to each other 200 times. And then we overhear them say: Alice: I know my number! Bob: I know my number too!

Can you explain the conversation and figure out the numbers x and y. You may assume that Alice and Bob each know that the other is extremely smart — so each is confident that if the other has sufficient information to deduce the answer then he/she definitely will.

paul garrett
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Jason Lindquist
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  • I /think/ the solution is Alice has 3 and Bob has 2. So when Alice sees Bob's number she knows she must have either 1 or 3. But she eventually figures out that if she had 1 then Bob must have been able to immediately deduce that he has 2 (since the numbers are positive integers so must be $\gt0$). Alice therefore deduces that she must have 3. – Mufasa Sep 15 '13 at 14:20
  • @Mufasa, a single denial by Bob would then be sufficient. Alice is smart, and knows that Bob is smart, and that Bob knows that Alice is smart and knows that Alice knows that Bob is smart et cetera :-). See here – Jyrki Lahtonen Sep 15 '13 at 14:33

2 Answers2

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Let us denote Alice's number as $a$ and Bob's as $b$. So, we have:

  • Alice: I don’t know my number. $\Rightarrow$ Bob concludes what Alice already knows: $b > 1$, because if $b = 1$, Alice would deduce hers was $2$, i.e., she'd know the answer.

  • Bob: I don’t know my number. $\Rightarrow$ Alice concludes what Bob already knows: $a > 1$, because if $a = 1$, Bob would deduce his was $2$, i.e., he'd know the answer.

Now both of them know that their numbers are greater than $1$.

  • Alice: I don’t know my number. $\Rightarrow$ Bob concludes what Alice already knows: $b > 2$, because if $b = 2$, Alice would deduce hers was $1$ or $3$, but she already knows it's not $1$, i.e., she'd know the answer.

  • Bob: I don’t know my number. $\Rightarrow$ Alice concludes what Bob already knows: $a > 2$, because if $a = 2$, Bob would deduce his was $1$ or $3$, but he already knows it's not $1$, i.e., he'd know the answer.

Now both of them know that their numbers are greater than $2$.

And so on.

Vedran Šego
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Hint

Alice's first turn.

  • Suppose Alice sees $y=1$. Then Alice knows that $(x,y)=(2,1)$ is the only possibility and says she knows.
  • Suppose Alice sees $y\ge 2$. Then Alice says she doesn't know.

Now it's Bob's turn.

  • If Alice said she knows, Bob knows that $(x,y)=(2,1)$ so we are finished.
  • If $x=1$ then Bob knows that $(x,y)=(1,2)$ and says so.
  • If $x=2$ then Bob knows that $(x,y)=(2,3)$ because otherwise Alice would have seen $y=1$ and said so. He announces that he knows.
  • If $x \ge 3$ Bob says he doesn't know.

Now it's Alice's second turn.

  • If Bob said he knows, Alice can work out from what she sees what the pair is. (Either $(1,2)$ or $(2,3)$.)
  • Otherwise, $x\ge 3$ and $y\ge 2$. If $y=2$ Alice announces she knows. Otherwise, we have a situation where it is Alice's turn and she knows $x,y\ge 3$. This bears some similarity to the original problem.

This reminds me very much of the blue-eyed islanders.

not all wrong
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