Learn / Library / Brainteasers

Guess only when sure

BrainteasersHard

$6$ prisoners each get a red or a blue hat, all $2^{6} = 64$ patterns possible. Each prisoner sees the other $5$ hats but not his own. At the same moment, without communicating, each does the following: if he sees at least $4$ red hats he guesses blue; if he sees at least $4$ blue hats he guesses red; otherwise he passes. The prisoners are freed if at least one of them guesses and every guess made is correct. For how many of the $64$ patterns are they freed?

Checking an answer needs JavaScript. The hint and the solution open without it.

Hint

Count by the number of red hats in the whole pattern, and check what a red-hatted and a blue-hatted prisoner each see.

Solution

Let the pattern have $k$ red hats. A red prisoner sees $k - 1$ red hats and a blue prisoner sees $k$ red hats, and the rule says guess blue on $4$ or more red, guess red on $1$ or fewer red.
For $k = 0$ and $k = 1$ a blue prisoner guesses red and is wrong, and for $k = 5$ and $k = 6$ a red prisoner guesses blue and is wrong.
For $k = 3$ every prisoner sees $2$ or $3$ red hats and passes, so nobody guesses and they are not freed.
For $k = 2$ the red prisoners see $1$ red hat and guess red correctly while the blue ones pass, and for $k = 4$ the blue prisoners see $4$ red hats and guess blue correctly while the red ones pass.
The freed patterns number $(2 6) + (4 6) = 30$ .

Answer: $30$

Next question Back to the library