Assuming Peter and You have agreed on a method beforehand (but have no hidden markings or outside communication), explain in detail how this feat is always possible and how the four face-up cards unequivocally reveal the fifth.
Solution:
Among any five cards, at least two cards must have the same suit, because there are only four suits. Choose two cards of the same suit. One of them will be hidden, and the other will be shown as the first card in the pile.
Use the rank order:
A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K, then back to A.
Between two cards of the same suit, one direction around the cycle is always between 1 and 6 steps. Choose the visible card so that the hidden card is 1 to 6 ranks ahead of it.
The first visible card tells Peter the suit of the hidden card. The order of the other three visible cards tells Peter how many ranks to count forward.
Beforehand, Peter and the magician agree on a fixed sorting order for cards, for example:
Clubs, Diamonds, Hearts, Spades, and within each suit A, 2, 3, ..., K.
After the key card is placed first, sort the remaining three cards by that agreed order and call them A, B, and C.
Use this code:
ABC = 1
ACB = 2
BAC = 3
BCA = 4
CAB = 5
CBA = 6
So the magician puts the key card first, then arranges the other three cards in one of those six orders to encode the distance.
Example:
Suppose the five cards are:
♣9, ♥K, ♦4, ♣3, ♠A
The two clubs are ♣3 and ♣9.
Choose ♣3 as the key card and hide ♣9, because counting forward from 3 to 9 gives:
4, 5, 6, 7, 8, 9
That is 6 steps.
The remaining cards are:
♥K, ♦4, ♠A
Sorted by the agreed order, they are:
A = ♦4
B = ♥K
C = ♠A
Distance 6 is encoded by CBA, so the magician gives Peter the four cards in this order:
♣3, ♠A, ♥K, ♦4
Peter sees ♣3 first, so he knows the hidden card is a club. He sees the remaining three cards are in CBA order, so the distance is 6. He counts six ranks forward from 3 and gets 9.
Therefore Peter announces:
Nine of Clubs
The trick works every time because the first card gives the suit, and the six possible orders of the remaining three cards give the rank distance from 1 to 6.
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