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The Great Houdini is performing again! Houdini has built up lots of reputation by showing his great card guessing tricks in [predecessor puzzle-1] and in [predecessor puzzle-2]. Everybody wants to see the famous magician, and the magic show is sold for many weeks in advance. Today Houdini shows the following awesome trick:

  • A girl from the audience carefully shuffles a deck of $52$ cards ($13$ spades, $13$ hearts, $13$ clubs and $13$ diamonds), and hands the deck over to Houdini's assistant.
  • The assistant looks through the entire deck (but without changing the order of the cards and without revealing them to Houdini).
  • The assistant then repeats the following step $52$ times: He takes the top card and puts it face-down on the table. Houdini guesses the suit. Then the assistant flips the card over and reveals it to everybody (including Houdini). Then the step is repeated.

Houdini's goal is of course to make as many correct guesses as possible. To reach this goal, Houdini and his assistant apply a sophisticated cheating strategy:

The backsides of the cards are not perfectly symmetric. The assistant has two different ways of putting down every card, and hence may communicate to Houdini one bit of information per card.

Question: Is there a cheating strategy for Houdini and his assistant that guarantees Houdini to make (a) at least $27$ correct guesses (b) at least $32$ correct guesses (even in the worst possible case)?

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    $\begingroup$ I would appreciate an female assistant, so Houdini can even get more guests! $\endgroup$ – buffo Mar 18 '15 at 10:27
  • $\begingroup$ Does the answer have to be a mathematical proof? Cuz with 2 different ways combined when putting down every card, 1 could easily hints all of the 4 suits $\endgroup$ – Alex Mar 18 '15 at 14:25
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Answer:

a. Yes
b. Yes

The strategy:

A:
Houdini and his assistant can easily guess 26 cards as said in answers to puzzle 2. To prove that he can easily guess 27 cards here are the needed steps:
Houdini and his assistant agrees that each odd card will show the color of the next (even) card and each even card will give the suit. Like that they can easily guess 26 cards.
But last card shouldn't even count, because Houdini can just count previous 51 cards and see which suit is missing.
So the "real" last card to guess is card 51. 51 is odd, so normally it would provide him information about the color of the 52nd card, but because at that moment Houdini will know which 2 suits(in worst case, in best case they are same) are left, he and his assistant should also agree on the order of the suits(for example Spades > Clubs > Hearts > Diamonds).
So if there are Clubs and Diamonds left and 51st card is Clubs, his assistant can put he 51st card normal way(which will mean greater than). With that information Houdini can easily find which card it is(only C and D are left and 51st card is "bigger" than 52nd, so it should be C)
So Houdini will guess all even cards(2, 4, 6, 8, ... , 50, 52) and second last, which is odd(51). That sums up to 27 guaranteed guesses in total
B:
It is actually possible to guess 34 suits and here's the strategy:
Assistant will put the first card that will show if the next three cards have more black or red suits. So Houdini will most probably guess the first card wrong. But then he knows that out of cards 2, 3 and 4 there are at least 2 red(or black) ones. 2nd and 3rd card will show the suit of that card and 4th card will again tell him whether next 3 cards are mostly black or red. Following this strategy, Houdini will guess 2 out of 3 cards in each group, so in total he will guess at least 34 suits($(52-1)*2/3$).
Example:
Assistant puts the first card up, so Houdini knows that out of the next 3 cards at least 2 are black. He says any suit for the first card, but probably guesses wrong.
Then assistant puts 2nd card up, so Houdini knows that first black card out of those 3 is Spades. Now Houdini should say Spades until it shows up(either this card or next one will be Spades).
Then assistant puts the 3rd card. Now it's upside down, so next black card in that group of 3 is Clubs.
The last card of that group of 3 tells Houdini the dominant color of the next 3 cards.

Following this strategy, Houdini will only guess 1 out of 3 cards wrong(worst case scenario)

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  • $\begingroup$ For part B, what about the case where the next three cards are like, "H,S,C" or "S,H,C"? In other words, the 2 same color cards don't have to be the first 2 cards. $\endgroup$ – JLee Mar 18 '15 at 13:49
  • $\begingroup$ @JLee see example: Now Houdini should say Spades until it shows up (either this card or next one will be Spades) $\endgroup$ – Novarg Mar 18 '15 at 13:57
  • $\begingroup$ So, he missed the first one, and he will miss one of the remaining 3, and he will get 2 correct. This seems like 50%. $\endgroup$ – JLee Mar 18 '15 at 14:01
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    $\begingroup$ @JLee a deck of cards has 52 cards in it(also stated in the question). There are $(52-1)/3 = 17$ groups of 3 cards out of which he will only miss 1. So in total he will guess $17*2=34$ correct $\endgroup$ – Novarg Mar 18 '15 at 14:03
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This is a follow-up to Novarg's excellent answer. I would prefer it as a comment, but lack the reputation.

You can apply the logic from part A to part B. On the 49th card, instead of indicating the dominant color of the final 3 cards, indicate the color of the 50th. The 50th card then indicates its own suit within that color. This leaves us with two final cards, which can be correctly guessed on their own as in part A. In this way, the entire final group of 3 can be correct, achieving a total score of 35.

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