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Terrible Wizard Inner circle 1973 Posts 
Are there any decent puzzles using playing cards?


jimgerrish Inner circle East Orange, NJ 2881 Posts 
Spellbinder's "Missing Card Puzzle" might be considered "decent" when properly dressed up by a presentation. It's in The Wizards' Journal #25.
Jim Gerrish
http://www.wizkidzinc.com http://www.magicnook.com Home of The Wizards' Journals: http://magicnook.com/wizardsTOC.htm 

Terrible Wizard Inner circle 1973 Posts 
Cool . Thanks for the info.


landmark Inner circle within a triangle 4670 Posts 
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.
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

Terrible Wizard Inner circle 1973 Posts 
Cheers landmark. I'll think about those,


wulfiesmith Special user Beverley, UK 781 Posts 
An interesting post ... I'll think about those too


ipe Regular user 174 Posts 
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 Regular user 174 Posts 
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 Regular user 174 Posts 
This one can be presented as logic puzzle, even if I strongly suggest to present it as magic trick.
You are blindfolded. 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 Inner circle New Jersey 1740 Posts 
Are the face up cards all together or are they randomly distributed throughout the deck?


ipe Regular user 174 Posts 
Quote:
On Oct 31, 2018, 0pus wrote: Face up cards are randomly distributed in the deck. The deck can be shuffled while you are blinded.
What would a real mindreader do?


murf Loyal user San Antonio, TX 251 Posts 
"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 Loyal user San Antonio, TX 251 Posts 
Quote:
On Oct 31, 2018, ipe wrote: Turn over 17 cards as you deal into two piles. Murf 

Scott Cram Inner circle 2676 Posts 
Quote:
On Oct 31, 2018, ipe wrote: Turn over the six and the blueback 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 bluebacked 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 redbacked card? The proposition says nothing about ONLY evennumbered cards being red backed. So, the face of the redbacked 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 Inner circle 2676 Posts 
Quote:
On Oct 31, 2018, ipe wrote: This puzzle reminds me of some of a routine from Bob Hummer's "HalfADozen Hummers", but in that one, exactly half of the cards must be faceup. Obviously, that doesn't work here. Quote:
On Nov 3, 2018, murf wrote: 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 faceup cards. How? If the number of faceup cards in the 17 cards you take is U1 (U for Up, 1 for first pile), the number of facedown cards in the 17 cards you take is D1 (D for Down, 1 for first pile), and the number of faceup 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 facedown 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 faceup cards. 

landmark Inner circle within a triangle 4670 Posts 
Nice explanation, Scott. I think the solution there is a bit Miraskilish, tooi.e. the number of red cards in one pile equals the number of black cards in the other pile.
Gerald Deutsch's Perverse Magic: The First Sixteen Years available now.
All proceeds to nonprofit charity Open Heart Magic. You can read my daily blog at Musings, Memories, and Magic 

ipe Regular user 174 Posts 
Quote:
On Nov 3, 2018, murf wrote: Murf, Scott's reply is the correct solution. You have to check the blackback card too. In this kind of problems, it is useful to remember this (logical) equivalence: if A then B = if notB then nonA So, "if a card shows an even number on its face, then its back is red" is equal to "if it is a notred (i.e. black) back, then it is a noteven (i.e. odd) number". Therefore, we should check the even card and the blueback card.
What would a real mindreader do?


ipe Regular user 174 Posts 
Quote:
On Nov 3, 2018, Scott Cram wrote: Correct and well explained. 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 Loyal user San Antonio, TX 251 Posts 
Quote:
On Nov 4, 2018, ipe wrote: 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 notB then nonA"! Just shows what happens when you try to answer without thinking things through. But I DID manage to get the discussion rolling.... Murf 

ipe Regular user 174 Posts 
@Murf
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.
What would a real mindreader do?


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