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Distinct numbers on hats

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The numbers $1$ , $2$ , $3$ , $4$ , $5$ , $6$ are written on $6$ cards. $5$ prisoners, numbered $1$ to $5$ , each draw one card without replacement and wear it as a hat; every ordered way of dealing $5$ cards is equally likely, $720$ in all. Each prisoner sees the other $4$ hats, so his own number is one of the $2$ numbers he does not see. A prisoner with an odd number guesses the larger of those $2$ numbers, and a prisoner with an even number guesses the smaller. For how many of the $720$ deals are at least $4$ of the $5$ guesses correct?

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Hint

Let $u$ be the card nobody wears. Every prisoner is choosing between his own card and $u$ .

Solution

Let $u$ be the card nobody wears. Each prisoner has $2$ candidates: are his own card and $u$ , so an odd prisoner is right exactly when his card is larger than $u$ , and an even prisoner when his card is smaller than $u$ .
If $u = 1$ , $u = 5$ or $u = 6$ , the number of right guesses is at most $3$ , $3$ and $2$ respectively, so no deal reaches $4$ .
For $u = 2$ the larger cards are $3$ , $4$ , $5$ , $6$ : all $3$ odd prisoners must wear larger cards and exactly $1$ of the $2$ even prisoners wears the single smaller card $1$ , which gives $2 \times 4! = 48$ deals with $4$ right guesses.
For $u = 3$ all $5$ guesses are right when the odd prisoners take $4$ , $5$ , $6$ and the even prisoners take $1$ , $2$ , which gives $3! \times 2! = 12$ deals; for $u = 4$ exactly $4$ are right when one of the $3$ odd prisoners wears a smaller card, which gives $3 \times 2! \times 3! = 36$ deals.
Adding the $3$ cases gives $48 + 12 + 36 = 96$ .

Answer: $96$

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