Can you figure out this card trick?

Are all of the cards in the deck the same one? If the audience member does not look at any of the cards but the one that is handed back to him/her, no one would know if the rest of the cards are all the same, other than the magician and assistant, of course.

All the cards in the deck are identical. This is why no one looks at the cards, except the magician.

This is possible without having all of the cards to be the same. I assumed that the audience, participant, and assistant (not the magician) could see all five cards after they had been blindly picked.

There are 4*3*2 = 24 ways for the remaining four cards to be arranged. The magician can calculate the correct order of the cards, and can then figure out which permutation it is.

Since the deck has 52 cards with four that are being used for the permutation, the magician has to choose one of 48 cards, or 2*24. This means that there is a way to encode a simple yes-no answer by card choice alone.

Now, we need to see how many ways that 5 cards can be partitioned into at most 4 suits.
5
4:1
3:2
3:1:1
2:2:1
2:1:1:1
There are six ways.
Now, how many ways are there to partition 4 into 4?
4
3:1
2:2
2:1:1
1:1:1:1
There are five ways. Somehow, by choosing how to partition the four cards by removing one card, the assistant conveys some information as to which suit the cards are in.

How about this encoding scheme. This shows which four card partitions come from which five card ones.
2:1:1 <- 2:1:1:1, 3:1:1 (2)
4 <- 5 (1)
2:2 <- 2:2:1 (2)
3:1 <- 4:1, 3:2 (2)

Perfect.

This manages to encode all of the possible partitions of five cards into a partition of four cards while limiting the uncertainty to be no more than two.

If the cards are divided 2:1:1, then the missing card is either in the same suit as the two cards or the odd one out.
If the cards are all of the same suit, then the missing card is of the same suit.
If there are two cards of one suit and two of another, then the missing card is in one of the two other suits.
If there are three cards of one suit and one of another, then the card is in one of those two suits.
If the cards are all of different suits, then your assistant messed up because that should never happen.

This will always limit the cards to be within two suits, so there are 24 possible cards (remember that four are already known). This is enough so that knowing which of 24 permutations were used can tell which is the right card. I have not decided on an encoding scheme for 24 permutations into 24 cards, but it should be much easier to come up with.

Aliens.

@PhiNotPi , I appreciate your answer, but you are working too hard. One more hint. 5 cards and 4 suits means that at least two of the cards must be in the same suit.

I confess that I did not figure this one out. The method is very simple, but devilishly clever.

Since there must be at least two cards in the same suit, the card that is picked to stay must be one that shares a suit with another card. Then, a card in the same suit is given as the first card. This indicates the suit of the unknown card, becuase the first card will be the same suit as the unknown card. Then the permutation of the remaining three can narrow it down to two cards, becuase there are six possible permutations and twelve cards. I haven’t quite figured out what to do next.

I like my first method better.

You are almost there. Consider the numerical difference between the two cards of the same suit.

The assistant can choose the card A so that (A-B) mod 13 is the smallest possible, where B is the card that he didn’t pick. This answer will make it so that the unknown card is within 6 cards below the known card, wrapping around. If the two cards are a 10 and a 6, the assistant, by picking the 10 to show to the magician, tell the magician that the unknown card is a card 3–9, in the same suit. Then, the permutation can tell him which of these cards it is.

Don’t you think that is a pretty solution? An easy way to handle the permutations is to order the 3 cards as small, medium and large (s,m,l) and assign the numbers 1 to 6 as
1. s,m,l
2. s,l,m
3.m,s,l
4, m,l,s
5. l,s,m
6. l,m,s

I understand this is a math puzzle. But still.

I think it should be possible to do this trick without looking at the cards.

Let’s blindfold the magician, for starters. Then the assistant will put the cards face down on a surface in front of the magician, the magician will touch the cards, feeling for the most recently removed card, and give the answer.

How are they stacked: 7 options (for example 2 stacked on one another, slightly offset and askew from the third card on the bottom.)
How far from the magician: 2 options (less than 6 inches, more than a foot)
What primary orientation: 4 options (12 o’clock to 6, 1:30 to 7:30, 3 to 9, 4:30 to 10:30)