The New Sudoku Players' Forum

[udosuk]

As Cec brought the topic up, here is another trick I know of:

You need an assistant and an innocent volunteer from the audience to perform this.

You turn your back from the audience (if necessary could be blindfolded). Your assistant asks the volunteer to randomly draw or pick 5 cards from the deck. After your assistant is shown the values of the cards, the volunteer is asked to lay them face-down on the table. You assistant then points to one of them, and the volunteer has to turn over that card, and reads out loud what it is (e.g. ace of spade). This process is then repeated for 3 more times, until 1 last card is left. Then you read out the value of that card, and the volunteer turns it over, to the amazement of the audience it's exactly what you declared.

There is no secret signal or anything at all times between you and your assistant or anyone else. Your assistant remains silent during the whole process and stays 20 feet away from you.

[Hud]

Does the assistant ask for the values in such an order that the missing value is midway between two other values like between the first and last values?

[udosuk]

You're on the right line Hud (sort of), but the method is not that simple...

By "value of a card" I mean the suit (spade, heart, club, diamond) plus the rank (ace to king)...

Hud's method would work only if there are 3 cards with consecutive values (2,3,4 of heart or the aces of spade, heart, club)...

[Cec]

Hey! this trick is more intriguing than "mine" You say the assistant remains silent. Am I right in assuming "winking" by the assistant is not permitted? Also, can the trick be performed with a brand new pack of cards meaning the cards have not been secretely marked in some way.

Cec

[udosuk]

Remember you could even be blindfolded. The assistant doesn't wink to you (unless she has a secret crush on you ) and you wouldn't see it anyway...

The cards could certainly be brand new or dusty old... No secret marks are needed and you wouldn't see them anyway...

[Cec]

Sorry udosuk, I forgot you can be blind-folded. Hmm?... still thinking... I hope the "magician" will eventually reveal this secret

Cec

[udosuk]

The "magician" will reveal the secret when this thread goes to the 2nd page...

[Cec]

Just to clarify something. Who reads out the identity of the four cards as they are turned over... the volunteer or the assistant?

Cec

[udosuk]

Your assistant remains silent during the whole process...

Read carefully...

[Cec]

Oops! Sorry about that. I'm still thinking

Cec

[coloin]

The assistant chooses the order......

3C, JD,KS,3H..........therefore it is the 7 of Clubs

Of 5 cards there will be at least 2 of one suit - e.g. 3 clubs and 7 clubs

Ist card therefore chooses the suit

The order of the 2nd,3rd and 4th cards are chosen in code form.

This determines how many card values above the 1st card [max 6] the 5th card is.

Take the cards in sequences [CDHS] the sequence Lowest,Highest, Medium can be code for + 4, e.g J Diamonds, K Spades, 3 Hearts.

Code: Select all

LMH......+1
MHL......+2
HLM......+3
LHM......+4
MLM......+5
HML......+6

I cant say that I have done it...but I remember reading it somewhere.

C

[udosuk]

Yep... But someone could still write a detailed "walkthrough" for this... If nobody does I'll do it on the 2nd page (if it ever gets there)...

[udosuk]

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

Code: Select all

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)...

Code: Select all

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.

Code: Select all

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!

[Cec]

Thanks usoduk for the explanation on this intriguing card trick. It would take some practice but that's to be expected for a trick as good as this.

Cec

[udosuk]

It would take some practice...

Certainly much less practice than your "quick hands with diversion" tricks (for me anyway)...