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# An interesting problem.

Hello guys! I am pretty new to the community. In fact, what brought me here is this tricky problem I found and I can't solve it!

The problem is the following:

In one class the professor brought 2 decks of cards each containing 52 standard cards. The one deck was blue backed and the other was red backed. He then gave the decks to his assistant after which the assistant shuffled them nicely. After that, the professor left the room. The assistant then chose 5 students at random and let each one of them take one card and give it to him. The students did exactly that.Then the professor came back inside and the assistant revealed 4 of the cards to the professor in a previously arranged order (regarding their faces of the cards) to the professor (the professor can see the backs of the cards as well). Prove that taking into account the order of the revealed cards the professor can guess correctly the 5th card's color (red or black), value (A,2,3,4,5...J,Q,K) and back color (red or blue).

And also a sub-problem to this one:

If the professor had one deck of cards of one back color (for example red), and the story was all the same except that the students took 4 cards and the assistant revealed 3 of the cards to the professor in a previously arranged order (regarding their faces), prove that the professor can correctly guess the color and value of the 4th card!

Thank you very much in advance, I really hope this community can help me! :)

3 years, 6 months ago

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It's important the the professor can see the front and backs of the cards, and hence he actually gets a lot more information.

There are 104 possible cards, of which he knows 4. So, given the 4 cards, he needs to guess what the 5th card is. Thus, we need to create a way for the assistant to map each of the 100 possibilities, to a unique arrangement of the 4 cards. We know that the assistant can easily permute the cards around, which gives us 24 possibilities. How can we create 100, given that we can play with the front and backs?

Staff - 3 years, 6 months ago

We can easily exceed 100 possibilities with the fact that we can alter the orders and the front backs. By altering the faces of the 4 cards, we get 16 possibilities. By altering order, we get 24. The product of both is then 16x24=384, which is way more information than 100. For the one deck card case, there are 49 possibilities on the forth card, but we can alter only 6x8=48 ways (same reasoning as above) to reveal the cards. This means the professor needs some other extra signal from the assistant to actually know the exact card, but he can still guess with near 100% regarding the card with the available information if he does not do so.

- 3 years, 6 months ago

Very interesting! Anyone got any ideas on this? I'm kinda clueless, not used to solving this kind of problems, not sure where to start or in which direction I should go :/

Also thank you very much for your efforts Calvin! I really appreciate it!!!

- 3 years, 6 months ago