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Here is a magic trick I saw. My question is how the magician and his partner did it.

Given the simple French deck of cards, with $52$ cards. A person from the audience chooses randomly five cards from the deck and gives it to the partner (the partner works with the magician), without showing it to the magician. Then the partner (who sees the five cards) chooses four cards from the five cards, and gives it to the magician one by one (so the order of the four cards matters). From that the magician knows the fifth card.

The partner and the magician can’t communicate during the trick. How did they do it?

I thought that amoung the five cards there will be two with the same sign (Spades,Hearts,Diamonds or Clubs) and one of these two cards will be the fifth, and the other will be the first card to give to the magician...

Let $S$ be a set of all cards.

So we have a set $A$ of all ordered $4$-couples of the set $S$ and a set $B$ of all $5$ subsets of the set $S$.

Connect 4-couple in $A$ with a subset in $B$ iff all four card from that 4-couple are in that subset. Clearly each 4-couple is connected to $8$ 5-subsets and every 5 subset is connected to $5!= 120$ 4-couples.

This relation give us a bipartite graph $G=(A,B)$. Now this graph satisfies Hall matching condition. Take any subset $X$ in $A$. Then the set of neighbours $N(X)$ satisfies: $$ 120cdot |N(X)| geq 48cdot |X|implies |N(X)|geq |X| $$
So there exist a matching which saturate all vertices in $A$.

It's a cute one. Let's just assume the cards are numbered $1 dots 52$. We draw cards $a_1 < a_2 < a_3 < a_4 < a_5$, and hand out in this order $b_1, b_2, b_3, b_4$. The way we order this set $B=b_1, b_2, b_3, b_4$ already allows to transmit quite some information. Indeed, choosing the orders paramounts to choosing a permutation on the 4 elements of $B$. There are $4! = 4 cdot 3 cdot 2 cdot 1 = 24$ such permutations.

This is not enough still to transmit the $5$th which is in a set of size $52$. Well, we also know that the 5th card is different from those $4$ firsts ones: there is only $52-4 = 48$ possibilities.

I'll leave it to you to complete the strategy. Note that $48/24 = 2$, so we just need to find a trick to only half the search space. Note also that we have still not fixed how we choose the secret card among the set $A$. At last, note at least one of the two inequalities:

- $a_2 leq 25$

- $a_4 geq 58-25$.

There is in fact a solution which uses your idea: that some suit will be repeated twice.

Say that among the $13$ ranks within a suit, ordered $A, 2, 3, 4, 5, 6, 7, 8, 9, 10, J, Q, K$, each rank "beats" the next six ranks, wrapping around when we get to the end. For example, $A$ beats $2,3,4,5,6,7$, and $10$ beats $J, Q, K, A, 2, 3$.

If you have two cards of the same suit, exactly one of them beats the other one. So you should pass, in order:

1. A card of the same suit as the missing card, which beats the missing card. This leaves six possibilities for the missing card.


2. The remaining three cards, in an order that encodes which of the six possibilities it is.

For the second step, we should order all $52$ cards in the deck somehow; for instance, say that $clubsuit < diamondsuit < heartsuit < spadesuit$ and $A < 2 < 3 < dots < Q < K$. Then the order of the last three cards is one of "Low, Middle, High", "Low, High, Middle", and so on through "High, Middle, Low". Just remember some correspondence between these six possibilities and the values $+1, +2, +3, +4, +5, +6$, and add the value you get to the rank of the first card (wrapping around from $K$ to $A$ of the same suit).

For example, say that the correspondence we chose in the second step is

beginarrayccc
textLow & textMiddle & textHigh &+1 \
textLow & textHigh & textMiddle &+2 \
textMiddle & textLow & textHigh &+3 \
textMiddle & textHigh & textLow &+4 \
textHigh & textLow & textMiddle &+5 \
textHigh & textMiddle & textLow &+6
endarray

and you draw the cards $4clubsuit, 5spadesuit, 5diamondsuit, Aclubsuit, Jspadesuit$.

• We have two possibilities for the repeated suit, so let's choose $spadesuit$.


• In the cyclic order in that suit, $5spadesuit$ beats $Jspadesuit$, so the first card we pass is $5spadesuit$.


• We want to encode the offset $+6$, which is the ordering High, Middle, Low.


• So we pass that ordering: after $5spadesuit$ we pass $5diamondsuit, 4clubsuit, Aclubsuit$ in that order, because $5diamondsuit > 4clubsuit > Aclubsuit$.

For any 4 cards given, we have 24 possible ways of ordering then (4 choices for the first card, 3 choices for the second and so on).

With 4 cards out of the deck, we have 52-4 = 48 possible cards to be the fifth one.

So, it would be impossible to guess the number and the sign of the fifth card only by the order of the 4 cards given.

My guess is: the partner makes some binary non-verbal communication (e.g.: blinking his left or his right eye). Blinking his left eye means the sign is Spades or Hearts and his right eye means Diamonds or Clubs (it isn't 100% correct since we may have up to 26 cards from 2 signs, but it is just an example).

Thus, we are going to have 24 possible ways of ordering 4 cards blinking his left eye plus 24 possible ways of ordering 4 cards bliking his right eye. Summing them, we cover all the 48 possible cards left.

They probably developed some mnemonic to memorize and process it all very fast, but it is mathematically possible.