47 replies to this topic

[Rob_G]

Spoiler for Simple 62.5%

I'm not done thinking about this (nor have I really tested the ideas I posted previously), but this came to me in the shower (I'm convinced a large portion of the world's great epiphanies happen there).

The first two cards are throwaways, and the orientation tells what suit the other person will guess for cards 3-26. The suit will be the one that shows up most often (worst case, they all happen 6 times).

For the orientations of 3-26, they give an extra bit to the information for 27-50. So you can get all of those.

Lastly, the final two cards are simple to get as plasmid pointed out earlier.

In the case of a very large, but finite, deck, this would give at least a quarter of the first half, and all of the second half. So worst case is 50%/4 + 50% = 62.5%.

In the case of a regular deck the total is 0*2 + 24/4 + 24 + 2 = 32 cards guaranteed correct = 61.54%

Spoiler for What if

the two throwaways told you what to guess for up and down 3-26? Wouldn't that guarantee at the least half of those cards? Making 0*2+24/2+24+2=38 guaranteed correct or 73.07%?

[plainglazed]

think the first two cards could tell what to choose for one of up or down. two more cards could tell for the other. and you're right, then 12 of the next twenty four could be known. the problem tho arises with the balance of the cards because you have gleaned no more information.

Spoiler for another way to look at EventHorizon's response...

...is to say all the odd cards greater than 2 and less than 50 (24 cards) will all be guessed the suit indicated by the first two cards. so the direction of the airplane on those odd cards has no bearing whatsoever on what the second logician chooses and can instead be the first bit of info defining the even cards (the exposed orientation of the plane on the even cards being the second bit).

tho i may have misunderstood what you were saying.

[Rob_G]

Ok so I started working backwards. Here is what I'm thinking. I've only considered it for the last two (100%) and three (four) (75%) cards.

Spoiler for

Lets assign a number value to each suit.
4-spades
3-hearts
2-clubs
1-diamonds
Two cards is trivial we know the last two suits. If they are the same no problem. If not the last card's orientation tells us if the card has the lower or higher value. (up for hearts instead of clubs)
For the last three (four) cards there are 4 possibilities
1. All the same suit-trivial
2. Two different suits-arrows indicate if you have the higher or lower of the remaining suits on top (first two cards same direction up/up for highest suit of the remaining)
3. Three different suits and the first card is of higher or lower value than the rest-same as 2.(first two cards same direction up/up for highest suit of the remaining)
4. Three different suits and the first card has a value in between the other two-first two cards are opposite indicating middle suit and the last two are the same as the rest.

Maybe this can be expanded, I don't know, but it's what I came up with.

[plasmid]

Edit: miscounted, going back to make some corrections

Edited by plasmid, 23 May 2012 - 05:45 AM.

[plasmid]

Here's a way to get 2/3 of the cards right in an arbitrarily large deck, or 35/52 (67.3%) of a standard deck of playing cards.

Spoiler for

For each set of six cards:

The first card will give information about which of the suit pairs should be guessed for the next five cards. If it's up, then the guesser should guess either diamonds or spades. If it's down, then guess either hearts or clubs.

For each of the other five cards: if the card is up then guess whichever red suit was specified (either diamonds or hearts), and if the card is down then guess whichever black suit was specified (either spades or clubs). You will get at least 3 of the 5 cards correct.

If you got more than three out of the five cards correct, then great, you're already getting at least 2/3 correct for that set of six! If you didn't (meaning that 2/5 of those suits were not the suits specified for guessing), then look at the orientations of the two cards that were not in the predictable suits. Since they were going to be missed anyway, they are oriented to give you two bits of information - and those two bits of information will tell you the suit of the first card in the next set of six. So ultimately, you'll end up getting at least 3/5 correct from the first set of six and you'll get the first card correct out of the next set of six, so you're still on track to get 2/3 correct.

In a real deck, 6*8=48, so using this strategy with the first 49 cards (since sometimes this strategy will lead you to predict the card after the set of six under consideration) will guarantee that 4*8=32 of those are correct. Remember that the orientation of card 49 isn't specified by that strategy for sets of six so it can still be either up or down, so the orientation of cards 49 and 50 can tell you the suit of card 50, and you can get 51 and 52 by counting cards and using the strategy outlined before. That gives you a total of 35.

[plainglazed]

well it seems i've been pigeonholed as an overzealous OP. had adapted this puzzle from another which used the first two cards to define what the second logician guessed on half of those remaining plus the half then definable. with the fifty-two playing card format, figured one could do a little better by combining plasmid's described endplay and including the second card in what EventHorizon described as cards 3-26 guaranteeing seven of those twenty five. did not think that thirty three would necessarily be best. did expect it would take some long complicated algorithm to do better rather than such a simple and elegant solution. (though bushindo has not yet chimed in...) brilliant plasmid, kudos!

[bushindo]

well it seems i've been pigeonholed as an overzealous OP. had adapted this puzzle from another which used the first two cards to define what the second logician guessed on half of those remaining plus the half then definable. with the fifty-two playing card format, figured one could do a little better by combining plasmid's described endplay and including the second card in what EventHorizon described as cards 3-26 guaranteeing seven of those twenty five. did not think that thirty three would necessarily be best. did expect it would take some long complicated algorithm to do better rather than such a simple and elegant solution. (though bushindo has not yet chimed in...) brilliant plasmid, kudos!

This is a superb puzzle, plaingazed. My attempts at a solution were frustrated early due to some tunnel vision, but I did gain a lot from watching the evolution of the solution. I am in particular humbled and inspired by the plasmid's creativity and ingenuity. Thank you both for this educational experience.

[EventHorizon]

Spoiler for Simpler 35

The first card tells you the suit color that shows up most in the next three cards, and their respective orientations tell you the suit. If all three are the same color... let it ride until there's a suit of a different color. Once there is a card not predicted correctly, that tells you the color of the suit that shows up most in the next group of three (or more).

Worst case, there is always one missed in the group of three. This will leave a similar ending condition to that of plasmid's 35. Just let the bit of info from the last group tell the color for the 50th card, and it's orientation tell you it's suit.

I'm working on improving the worst case of this one, but it may not prove fruitful.

[plainglazed]

well, that puts me squarly in third place. am also curious if there is a way to eek out another (or more) mixing techniques or otherwise. very nicely figured EventHorizon. even i could manage being the assistant with this scheme. cheers

[CaptainEd]

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