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The Magic Cafe Forum Index » » Puzzle me this... » » Puzzles with playing cards (0 Likes) Printer Friendly Version

Terrible Wizard
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Are there any decent puzzles using playing cards?
jimgerrish
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Spellbinder's "Missing Card Puzzle" might be considered "decent" when properly dressed up by a presentation. It's in The Wizards' Journal #25.
Terrible Wizard
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Cool Smile. Thanks for the info.
landmark
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1) Take the A,K,Q,J of each suit from the deck. The challenge is this: create a 4x4 square of those cards such that there are no two cards of the same value in any row, column or long diagonal.

2) Harder: create a 4x4 square of those same cards such that there are no two cards of the same value OR THE SAME SUIT in any row, column, or long diagonal.
"I use my five illusions to create the sense I'm useful to six."



You can read my daily blog at Musings, Memories, and Magic
Terrible Wizard
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Cheers landmark. I'll think about those, Smile
wulfiesmith
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An interesting post ... I'll think about those too
ipe
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There are four (normal) cards in a row on the table:

a three |a six |a blue back |a red back


You know that this sentence is true: if a card shows an even number on its face, then its back is red.

Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on its face, then its back is red?
What would a real mindreader do?
ipe
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Using exactly 9 cards, build a vertical structure for balancing a glass full of water on top. (No card can be placed flat on the table.)
What would a real mindreader do?
ipe
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This one can be presented as logic puzzle, even if I strongly suggest to present it as magic trick. Smile

You are blind-folded. Someone hands to you 52 playing cards and tell that there are 17 face up cards on that deck. You have to divide the deck in two piles, each pile with the same number of face up cards. No trick is allowed, only logical thinking.
What would a real mindreader do?
0pus
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Are the face up cards all together or are they randomly distributed throughout the deck?
ipe
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Quote:
On Oct 31, 2018, 0pus wrote:
Are the face up cards all together or are they randomly distributed throughout the deck?

Face up cards are randomly distributed in the deck. The deck can be shuffled while you are blinded.
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murf
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"There are four (normal) cards in a row on the table:

a three |a six |a blue back |a red back


You know that this sentence is true: if a card shows an even number on its face, then its back is red.

Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on its face, then its back is red?"

The six.

Murf
murf
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Quote:
On Oct 31, 2018, ipe wrote:
This one can be presented as logic puzzle, even if I strongly suggest to present it as magic trick. Smile

You are blind-folded. Someone hands to you 52 playing cards and tell that there are 17 face up cards on that deck. You have to divide the deck in two piles, each pile with the same number of face up cards. No trick is allowed, only logical thinking.


Turn over 17 cards as you deal into two piles.

Murf
Scott Cram
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Quote:
On Oct 31, 2018, ipe wrote:
There are four (normal) cards in a row on the table:

a three |a six |a blue back |a red back


You know that this sentence is true: if a card shows an even number on its face, then its back is red.

Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on its face, then its back is red?


Turn over the six and the blue-back card. Why?

The six has an even number on its face, so we need to make sure that the back is red in order for the statement to be true.

The blue-backed card needs to be turned over to make sure that it DOESN'T have an even number on the face, otherwise the rule is violated.

Why NOT turn over the 3? Because the color of the back doesn't affect the truth of the proposition. The back of the 3 is red? Great. The back of the 3 is blue, green, purple, or some other color? Great. Neither one matters.

Why NOT turn over the red-backed card? The proposition says nothing about ONLY even-numbered cards being red backed. So, the face of the red-backed card could be odd (in which case it has no relation to the rule) or even (it which case it's just a confirming instance of the rule), and neither case adds anything to our knowledge of the truth of the proposition.

---

Here's the same puzzle, but presented in a way that people grasp quicker.

Imagine you're at a party where people are drinking either beer or milk, and you have to make sure that no one under the age of 18 is drinking a beer.

You see 4 people with drinks:
The first person is drinking a beer, and may be over or under 18.
The second person is drinking milk, and may be over or under 18.
The third person is definitely over 18, and you can't be sure what they're drinking.
The fourth person is definitely under 18, and you can't be sure what they're drinking.

Which people do you need to investigate, in order to make sure that no one under 18 is drinking a beer?

Most people quickly and correctly determine that you need to investigate the first and fourth. The first person is definitely drinking a beer, so you need to check his ID to make sure that he or she is over 18. You need to check the drink of the fourth person, because you know they're under 18, and you need to make sure they're not drinking beer.

The second person is drinking milk, so the rule doesn't apply. The third person is known to be over 18, so it doesn't matter whether they're drinking beer or milk.
Scott Cram
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Quote:
On Oct 31, 2018, ipe wrote:
This one can be presented as logic puzzle, even if I strongly suggest to present it as magic trick. Smile

You are blind-folded. Someone hands to you 52 playing cards and tell that there are 17 face up cards on that deck. You have to divide the deck in two piles, each pile with the same number of face up cards. No trick is allowed, only logical thinking.


This puzzle reminds me of some of a routine from Bob Hummer's "Half-A-Dozen Hummers", but in that one, exactly half of the cards must be face-up. Obviously, that doesn't work here.

Quote:
On Nov 3, 2018, murf wrote:

Turn over 17 cards as you deal into two piles.

Murf


Close, Murf, but not quite.

What you need to do is take any 17 cards, and turn over all 17 cards. Both piles will now contain the same number of face-up cards. How?

If the number of face-up cards in the 17 cards you take is U1 (U for Up, 1 for first pile), the number of face-down cards in the 17 cards you take is D1 (D for Down, 1 for first pile), and the number of face-up cards left behind in the stack of 35 if U2 (U for Up, 2 for second pile), we have:
U1 + D1 = U1 + U2

Note that we're not concerned at all about the number of face-down cards in the second pile.

Cancel out U1 on both sides of that equation, and you'll see that D1 = U2. This means that you can simply turn the pile of 17 over, and you'll have 2 piles of cards with the same number of face-up cards.
landmark
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Nice explanation, Scott. I think the solution there is a bit Miraskil-ish, too--i.e. the number of red cards in one pile equals the number of black cards in the other pile.
"I use my five illusions to create the sense I'm useful to six."



You can read my daily blog at Musings, Memories, and Magic
ipe
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Quote:
On Nov 3, 2018, murf wrote:
"There are four (normal) cards in a row on the table:

a three |a six |a blue back |a red back


You know that this sentence is true: if a card shows an even number on its face, then its back is red.

Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on its face, then its back is red?"

The six.

Murf



Murf,

Scott's reply is the correct solution. You have to check the black-back card too.

In this kind of problems, it is useful to remember this (logical) equivalence:


if A then B = if not-B then non-A


So, "if a card shows an even number on its face, then its back is red" is equal to "if it is a not-red (i.e. black) back, then it is a not-even (i.e. odd) number". Therefore, we should check the even card and the blue-back card.
What would a real mindreader do?
ipe
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Quote:
On Nov 3, 2018, Scott Cram wrote:
What you need to do is take any 17 cards, and turn over all 17 cards. Both piles will now contain the same number of face-up cards. How?

If the number of face-up cards in the 17 cards you take is U1 (U for Up, 1 for first pile), the number of face-down cards in the 17 cards you take is D1 (D for Down, 1 for first pile), and the number of face-up cards left behind in the stack of 35 if U2 (U for Up, 2 for second pile), we have:
U1 + D1 = U1 + U2

Note that we're not concerned at all about the number of face-down cards in the second pile.

Cancel out U1 on both sides of that equation, and you'll see that D1 = U2. This means that you can simply turn the pile of 17 over, and you'll have 2 piles of cards with the same number of face-up cards.


Correct and well explained. Smile

Of course 17 is an arbitrary number here, the trick can be generalized with both the number of card in the deck and the number of face up cards as big as you like.
What would a real mindreader do?
murf
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Quote:
On Nov 4, 2018, ipe wrote:
Quote:
On Nov 3, 2018, murf wrote:
"There are four (normal) cards in a row on the table:

a three |a six |a blue back |a red back


You know that this sentence is true: if a card shows an even number on its face, then its back is red.

Which card(s) must you turn over in order to test the truth of the proposition that if a card shows an even number on its face, then its back is red?"

The six.

Murf



Murf,

Scott's reply is the correct solution. You have to check the black-back card too.

In this kind of problems, it is useful to remember this (logical) equivalence:


if A then B = if not-B then non-A


So, "if a card shows an even number on its face, then its back is red" is equal to "if it is a not-red (i.e. black) back, then it is a not-even (i.e. odd) number". Therefore, we should check the even card and the blue-back card.


Being a software engineer who has been involved in computers for over 50 years, you'd think I might rememeber things that I often use, like "if A then B = if not-B then non-A"! Just shows what happens when you try to answer without thinking things through. But I DID manage to get the discussion rolling....

Murf
ipe
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@Murf Smile Smile


Anyway, replying to the original poster, you can present the famous Monty Hall problem (https://en.wikipedia.org/wiki/Monty_Hall_problem) using playing cards. Smile
What would a real mindreader do?
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