Name the hidden card
Let's take on the deck of cards today and see where it takes us to: You and another intelligent friend of yours offer to perform this trick together in front of an audience. There is a deck of 52 cards in front of you. Assume the deck is neat with 13 of each type. The deck is shuffled and your friend asks the audience (one of them) to pick 5 cards. Your friend then takes those 5 cards and sees the cards that the audience has picked without showing it to you. Then your friend will pick one card from that 5 cards and hand it back to the audience. Then your friend neatly piles the remaining 4 cards in an order and places it on the table downwards and you start to pick them up. After you pick all the 4 cards, you are able to tell for sure what the 5th card with the audience is. How is that possible? Can you do the same experiment with 4 cards? Rule out all silly assumptions like "ESP between you and your friend, somehow your friend signals to you what it is, etc.," It is purely logicalLevel: Hard
posted by Karthik Nagaraj at
2:58 PM
5 Comments:
Thanks! Also try to get the answer. Its interesting. Spread the word amongst your friends!
I haven't thought about the second bit (the possibility with 3+1 cards), but here's a solution to how it can be done with 5 cards (4+1):
Among the 5 cards, atleast 2 have to be of the same suit. Assuming you are the guesser, your friend gives the audience one of those 2 cards, and shows you the other one as the first card among his 4. You now know which suit the audience's card belongs to.
How your partner chooses which one of the 2 similar-suited cards to give back to the audience is simple. If you consider all the 13 cards in a suit in a cyclic order or priority (let's say -A-1-2-3-4-...-J-Q-K-A- ), it's easy to see that any two cards will have a distinct number of cards in the arcs separating them on the circle. Your partner can choose the first card in a clockwise minor arc, so you now have to determine the clockwise offset of the other card and you are home. This offset can lie in one out of the next six positions on the circle, clockwise. Any of these six positions can be indicated by an appropriate permutation of the remaining three cards your partner has to show you, if you both have agreed on a global ordering of all the 52 cards in the pack.
--Abracadabra.
Ahem. The solution to the second part of the problem (the possibility with 3+1 cards) occurs to you with a little bit of thinking. If the problem is rephrased, it asks whether a one-one map exists from the set of unordered sets of 4 cards into the set of ordered 3-tuples. What the magician does is a simple inversion of the map, to determine the fourth (and hence all four cards).
The number of elements in the first set (the domain) is 52C4 = 52*51*50*49/24, whereas the number of elements in the second set (the codomain) is 52P3 = 52*51*50, which is lesser than 52C4. Hence, there is no way of establishing a one-one map from the larger first set into the second smaller set, proving that the card trick cannot be performed with a total of 4 cards.
--Abracadabra.
Abracadabra,
I think you have cracked both the parts of the solution. Excellent. So time for a new one!
I guess this puzzle is completed then. The basic idea in this puzzle is that your partner and you have to agree upon certain rules and that your partner helps you out by handing out the cards in a pre-defined fashion. Now, amongst 5 cards that are picked up, there is bound to be more than one suite's card in the 5. Your partner would choose the lowest clockwise offset suite card (among the 2 cards that are of the same suite) to give it back to the audience. So the first card that is shown to you will help you identify the suite. The next 3 cards are shown in a particular order that reveals the offset of the 5th card on the same clockwise distribution of the cards.
Obvoisuly now with 4 cards, there cannot exist a similar suite card with 100% guarantee, this test cannot be performed with 4 cards.
It will be nice if anonymous would put in his name next time. Congrats on getting the puzzle right!
This puzzle is SOLVED!
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