Okay, in case people didn't understand coloin's reply, here is the trick explained in details.

The trick relies on 2 things, which are both (implicitly) decided by the assistant:

1. which card among the 5 to be the secret card for you to "guess"

2. the sequential order of the first 4 cards for the volunteer to read out

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`Observation #1:`

Of any set of 5 cards (no joker), at least 2 of them must be of the same suit.

Proof:

There are only 4 suits (spades, hearts, diamonds, clubs). Enough said.

Therefore, once the assistant are shown the 5 cards, she can immediately pick out a pair of same suit cards (in case of 3/4/5 cards having the same suit, she just randomly choose 2 of them)...

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`Observation #2:`

In any pair of same suit cards, one card is always 1 to 6 ranks "higher" than the other.

Proof:

In "higher" we are talking about a "circular" relationship (not unlike the modular arithmetic).

Rank the card as the following: 9,10,J,Q,K,A,2,3,4,5,6,7,8,9,10,J,Q,K,A...

You'll see every card has 6 cards that are "higher" than it and 6 cards that are "lower".

E.g. For the "J", the 6 higher cards are Q,K,A,2,3,4, the 6 lower cards are 5,6,7,8,9,10.

Here the assistant will pick the "higher" card of the pair of same suit card to be the

secret (5th) card, and the "lower" card to be the

1st card to be read out by the volunteer.

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`Observation #3:`

Of any (distinct) set of 3 cards, they can be arranged in a total of 6 different ways.

Proof:

Let's decide a priority order for the ranks and suits:

A > K > Q > J > 10 > 9 > 8 > 7 > 6 > 5 > 4 > 3 > 2

spades > hearts > diamonds > clubs

When we compare 2 cards, the ranks will decide which one is "larger", and in case of a tie the suits will then be compared...

Hence, "A of spades" is the largest card, followed by "A of hearts", "A of diamonds"...

The smallest card of all would be "2 of clubs", preceded by "2 of diamonds", "2 of hearts"...

Now, for the set of 3 cards, label the largest one "L", the middle one "M", the smallest one "S".

E.g. You have 4 of diamonds, Q of hearts, and J of spades. So L="Q of H", M="J of S", S="4 of D".

It's easy to see there are only 6 different ways to arrange them:

S-M-L

S-L-M

M-S-L

M-L-S

L-S-M

L-M-S

The assistant, having already decided the

1st and

5th cards, has 3 cards left. In her mind she labels those 3 cards as described above, and then arranges them according to how many

ranks the 5th card is "higher" than the 1st...

If 1 rank higher, she arranges them as S-M-L.

If 2 ranks higher, she arranges them as S-L-M.

If 3 ranks higher, she arranges them as M-S-L.

If 4 ranks higher, she arranges them as M-L-S.

If 5 ranks higher, she arranges them as L-S-M.

If 6 ranks higher, she arranges them as L-M-S.

And this arrangement decides the order of the 2nd, 3rd & 4th cards for the volunteer to read out...

Example:

The 5 cards drawn by the volunteer are "A of spades", "Q of hearts", "J of diamonds", "5 of spades", "2 of spades".

The assistant sees 3 spades. She decides to use the "A & 5" as the 1st and 5th card respectively. ("A & 2" or "2 & 5" are also usable, it's a free choice for the assistant.)

Noticing 5 is

4 ranks higher than A, the assistant knows she has to arrange the remaining 3 cards as "M-L-S", i.e. "J of diamonds", "Q of hearts", "2 of spades".

So, she signals the volunteer to read out these 4 cards in order: "A of S", "J of D", "Q of H", "2 of S".

From the 1st card, you know the 5th card must be a spade. From the order of the 2nd-4th cards, you know it is 4 ranks higher than A, i.e. a 5. Therefore without hesitation you declare the 5th card to be "5 of spades", which is of course correct!