47 replies to this topic

[EventHorizon]

EventHorizon, I'm not grasping the method. In particular, I can't tell if the "group of three (or more)" is always at least three, or if it resets upon color change. In particular, how would you point the backs for the opening sequence, and what would suits would the observer call?

SHDCSDHCSCHD

Spoiler for Explanation

The first group of three is HDC. Since H and D are red, that's what the orientation of the first card (S) will say (whichever orientation is designated for red, lets say up).

The guesser will get H (lets say H is up) and D (down) correct. Since C cannot be guessed correctly (wrong color), it's orientation tells the guesser what the color will be for the next three (SDH). Again, it's red (up). So the guesser will incorrectly guess H on C, but that's fine. It tells the guesser to use it as the color instead.

The orientation on S (incorrectly guessed as D, signalling it's orientation is the color indicator) gives the color of black (down) for CSC, while D (down) and H (up) are guessed correctly.

Since all three cards are black in CSC, all three will be guessed correctly (up,down,up). The guesser will just keep using black until one is not a black suit. The very next card is red (H), so it's orientation gives the color for the group of three of D??.

U-UDU-DDU-UDU?-? are the orientations.
?-OOX-XOO-OOOX-? are if they are guessed correctly.
?-HDH-DDH-CSC?-? are the suits guessed.
The hyphen marks the groups (usually three unless the first 3 are correctly guessed).



So the "three or more" quip did confuse things a little bit. It was just there to say that if all three cards in the group of three were the same color, there was no card to give a color for the next group of three. So the guesser would just keep guessing that color until one was missed to give the color for the following three cards.

Hopefully that clears things up. Let me know if you have further questions.

[CaptainEd]

Yes, thanks. What a marvelous solution!

[plainglazed]

was reviewing some older threads for inspiration or possible extension puzzles when a comment EventHorizon made got me thinking anew on this one.

Spoiler for It's nice to be correct, but mistakes allow learning...

see spoiler title

a little more specifically - can we gain extra bits of info from a revealed card that is different then expected. have a solution for 36 and would not be surprised if that could be bested by one or two. pretty interesting approach and totally different from the two 35 card solutions if anyone is interested.

[plainglazed]

Spoiler for somewhat of a HINT

it is possible to predict the suits of 36 out of 52 cards after correctly predicting less than 25% of the first quarter of the deck

[phaze]

Is it possible to examine the back of every card before they begin turning them over?

[plainglazed]

hey hey phaze - very glad for the question and interest. to answer though, no. the only info the guessing logician has to go by are all previously displayed cards and the direction of the plane on those cards as well as the direction of the plane on the top of the remaining face down deck.

Spoiler for a slight modification on my previous HINT

it is possible that no more than three of the first thirteen cards guessed are guaranteed to be correct (in one system giving 36 correct predictions, anyway).

[plainglazed]

well, when i said

can we gain extra bits of info from a revealed card that is different then expected. have a solution for 36 and would not be surprised if that could be bested by one or two. pretty interesting approach and totally different from the two 35 card solutions if anyone is interested.

i wasnt totally accurate. in my 36 card solution, i did use a principle used in the 35 card solutions; at least initially. after that the prior solutions methodologies are indeed different.

[plainglazed]

Had bumped this thread over a month ago after an extended period with a previous optimal answer. It's about to go off of the front page of the forum and even though there is at least one person following this topic (that you plasmid?), think it's about time to offer another solution. Besides, suspect this solution can be bested.

Spoiler for a solution with 36 correct suits predicted

the direction of the plane on the first two cards indicates the suit to be guessed on cards 2-14. Out of those thirteen cards at least four must be correct (pigeon hole principle). The direction of the plane on cards 3-14 along with the direction of the plane on cards 15-26 could give twelve more correct suit predictions. One could intentionally miss any one of those twelve cards three different ways. Now three of those twelve (cards 15-26) could be missed in ₁₂C₃ or 220 different combinations and those three cards could be missed 3³ or 27 different ways for a total of 5940 permutations which is greater than 2¹². In other words, every 12 digit binary number could be represented thus one can correctly predict 9 out of cards 15-26 and again have 12 bits of extra information. So 4 out of 14 + 9 out of 12 + 9 out of 12 + 12 out of 12 gives 34 out of 50 correctly predicted with the last two predicted as previously determined.

Think that logic is sound. 37 anyone?

[EventHorizon]

Had bumped this thread over a month ago after an extended period with a previous optimal answer. It's about to go off of the front page of the forum and even though there is at least one person following this topic (that you plasmid?), think it's about time to offer another solution. Besides, suspect this solution can be bested.

Spoiler for a solution with 36 correct suits predicted

the direction of the plane on the first two cards indicates the suit to be guessed on cards 2-14. Out of those thirteen cards at least four must be correct (pigeon hole principle). The direction of the plane on cards 3-14 along with the direction of the plane on cards 15-26 could give twelve more correct suit predictions. One could intentionally miss any one of those twelve cards three different ways. Now three of those twelve (cards 15-26) could be missed in ₁₂C₃ or 220 different combinations and those three cards could be missed 3³ or 27 different ways for a total of 5940 permutations which is greater than 2¹². In other words, every 12 digit binary number could be represented thus one can correctly predict 9 out of cards 15-26 and again have 12 bits of extra information. So 4 out of 14 + 9 out of 12 + 9 out of 12 + 12 out of 12 gives 34 out of 50 correctly predicted with the last two predicted as previously determined.

Think that logic is sound. 37 anyone?

Spoiler for 37

Very similar to the one you gave.

First two cards give the suit to guess for cards 2-10. You can get 3 of those cards by the pigeon-hole principle.

Cards 3-10 give their bits to let you guess cards 11-18. You can miss 2 cards, 8c2 = 28 pairs in 3^2 different ways = 252.... not quite 2^8. However, you could also just miss 1 card intentionally which gives 8c1 * 3 extra possibilities for a total of 276. That's the extra 8 bits of info needed for the next group.

Just like yours, you'll get the last group all correct and the final two.

3 of 10 + 6 of 8 + 6 of 8 + 6 of 8 + 6 of 8 + 8 of 8 + 2
= 3 + 6 + 6 + 6 + 6 + 8 + 2
= 3 + 24 + 10 = 37

[EventHorizon]

Spoiler for What if it was an infinite deck?

We've mentioned this before, and plainglazed's new algorithm gives 75% (9 of 12, same as my modification 6 of 8).

I decided to see what would happen if we used plainglazed technique, including my addition of using all the possibilities with less missed cards, and see what happens as the cards in the groupings increases. It turns out that you can get....

80% with 80 cards per grouping.
90% with 1640 cards per grouping.
95% with 2820 cards per grouping.
96% with 3375 cards per grouping.
97% with 4267 cards per grouping.
98% with 5950 cards per grouping.
99% with 10700 cards per grouping.
99.5% with 19399 cards per grouping.
99.9% with 79961 cards per grouping.
99.99% with 633144 cards per grouping.

So with an infinite deck, you can get basically all of them.