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# A Challenged Flight Deck

## Question

On the return journey from the Arctic Circle of Logicians Fresh Air/Fresh Ideas annual seminar, each pair of logicians received a complimentary pack of fifty two standard playing cards. These particular playing cards featured an image of the mighty de Havilland C-6 Twin Otter; the very aircraft they were currently shivering in. As such, each face down card would have one of two different orientations. The plane would either be pointing straight up against a sunless grey sky in an apparent death stall or, because of the slight tilt of its wings, would appear to be stuck in a sustained tailspin. Due to the logicians nervous desire for any distraction from the umhh, cold; they developed the following challenge: If a pair works together where only one knows the order of a shuffled deck and pre-arranges the direction of the plane on each card (without changing the order of the deck) and the other predicts the suit of each card, one at a time, from the evenly stacked face down deck, what scheme yields the maximum guaranteed correct number of suit predictions?

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On the return journey from the Arctic Circle of Logicians Fresh Air/Fresh Ideas annual seminar, each pair of logicians received a complimentary pack of fifty two standard playing cards. These particular playing cards featured an image of the mighty de Havilland C-6 Twin Otter; the very aircraft they were currently shivering in. As such, each face down card would have one of two different orientations. The plane would either be pointing straight up against a sunless grey sky in an apparent death stall or, because of the slight tilt of its wings, would appear to be stuck in a sustained tailspin. Due to the logicians nervous desire for any distraction from the umhh, cold; they developed the following challenge: If a pair works together where only one knows the order of a shuffled deck and pre-arranges the direction of the plane on each card (without changing the order of the deck) and the other predicts the suit of each card, one at a time, from the evenly stacked face down deck, what scheme yields the maximum guaranteed correct number of suit predictions?

I'd some some clarification. In the bold part, does it mean that the second logician predicts a card, the card is revealed, and then the logician makes the next guess, and so on?

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Do the card designers have some personal vendetta against the manufacturers of the de Havilland C-6 Twin Otter? (I'm imagining cards all with different pictures of de Havilland planes suffering catastrophic structural failures). Edited by Morningstar
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I'd some some clarification. In the bold part, does it mean that the second logician predicts a card, the card is revealed, and then the logician makes the next guess, and so on?

that's correct. the first logician arranges the orientation of the cards without changing the order in some predetermined system then places the evenly stacked deck face down. the second logician predicts the suit of the top card, that card is revealed, then the back of the next card and its orientation is revealed, the suit of that card is guessed, etc.

had struggled with the wording. hope that helps?

Do the card designers have some personal vendetta against the manufacturers of the de Havilland C-6 Twin Otter? (I'm imagining cards all with different pictures of de Havilland planes suffering catastrophic structural failures).

nah, the same image is one each card. orienting the cards plane up or down just gives a "bit" of info to the other logician who does not know the order of the cards.

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[spoiler='

?Can the hand used to reveal the suite be changed

']

Agree that a particular orientation means red suit and the other means black suit.

After a guess is made the hand used to turn the card indicates the suit of nex cardt, say

left hand is club/heart and right is diamond/spade.

Opening gambit could be signalled by an open palm gesture left/right as above.

Think this would work with mittens on too.

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hmm, guess that again works making a max of 51? let's say the only info the second logician receives is the orientation of the top card of the remaining deck, the cards already revealed, and the orientation of all previous top cards.

EDIT: oops, had missed your play for determining the first card. also failed to mention my doubts as to the ability of even the wiliest of logicians to flip over a playing card from a stacked deck using either hand - wearing mittens!

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I can think of a strategy that will guarantee that at least half of the cards' suits will be correctly predicted fairly easily. I'll ponder a little further and see if I can do any better.

Edited by plasmid
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nah, the same image is one each card. orienting the cards plane up or down just gives a "bit" of info to the other logician who does not know the order of the cards.

I know. I was being facetious. Sorry. Anyway, I have an idea, but it's not very efficient. I'm working on it.

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I can think of a strategy that will guarantee that at least half of the cards' suits will be correctly predicted fairly easily. I'll ponder a little further and see if I can do any better.

yeah, think you can do at least a little better. good to know you're on the case.

I know. I was being facetious. Sorry. Anyway, I have an idea, but it's not very efficient. I'm working on it.

suspected as much but didn't want to just leave you hangin' (@ 15,000')

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A slight tweak of my initial approach would on average do a little better than getting half the suits right, but still can't guarantee more than half. For the sake of getting at least one strategy out to start things off...

Set up the deck so that if a card is of one particular suit, say spades, it's oriented with the plane right-side-up. Any other suit and the plane is upside-down. The guesser will always guess spades if it's right-side-up and be guaranteed to get all of those, and will always guess one pre-determined other suit if it's upside-down, say hearts, and get all of those suit right as well.

On average you can do a little better by modifying the strategy: Keep a running tab of how many times each suit has already shown up - or more importantly, how many of each suit are still left in the deck - and use that to base the card orientation and guesses on. If the next card is of the most abundant suit left, then it gets oriented with a right-side-up plane and guessed correctly. Otherwise, it's oriented with an upside-down plane and the guesser guesses the second most abundant suit left. If there are multiple suits tied for most abundant, they would have to agree on a precedence of suits ahead of time to know which suit will be denoted by a right-side-up plane (say spades > hearts > diamonds > clubs) and the right-side-up suit would be whichever one is highest on the precedence list of those that are tied for most abundant.

The second strategy would lead to more cards being correctly predicted on average, but would not guarantee more than half.

Edited by plasmid
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Firstly,hope that the mittens quip came across as was meant, a bit of fun. Sorry if it didn't.

Second a question...

Can the orientation of the revealed cards be changed, to convey information to the guessing logician?

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hey plasmid - all of what you say is true. will reiterate that the guarantee part is intended, just in case. good work, it will come. you'll appreciate the answer.

Firstly,hope that the mittens quip came across as was meant, a bit of fun. Sorry if it didn't.

of course it did, should have added the appropriate smiley. was just attempting to extend the fun. can be hard to get the tone right in this forum format but always go with the assumption that it's all about the fun, myself.

Second a question...

Can the orientation of the revealed cards be changed, to convey information to the guessing logician?

you bring up an excellent point. please assume the faces of all cards are bilaterally symmetric and reveal no additional info nor can their orientation be changed. hope that doesn't throw anyone who may have been taking that tack.

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Ok, just a tiny bit better:

Assign numbers to the suits: spades=0, hearts=1, diamonds=2, clubs=3. Plane orientations are assigned a number: plane up=1, plane down=0. The player who orients the cards will indicate the suit of every other card by using the orientation of two cards. Take the orientation of the first card of the pair as the first digit in a binary number, and the second card of the pair as the second digit of the binary number, and that binary number indicates the suit of the second card. For example, if the second card in the deck is a diamond (=2, or in binary 10) the first card's plane will be pointing up and the second's will be pointing down. Using that strategy, every other card will be predicted correctly.

Use that strategy for the first 50 cards. Once you're down to the last two cards in the deck, they players will need to have counted cards, and they should know the suits of the last two cards. For the 51st card, the code changes to be: plane-up means the 51st card is the higher numbered suit of the two remaining suits, and plane-down means the 51st card is the lower numbered of the two remaining suits. They'll be able to get both of the last two cards correct, for a total of 27/52.

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"the next card is not a spade" for example , be acceptable.

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Ok, just a tiny bit better:

Assign numbers to the suits: spades=0, hearts=1, diamonds=2, clubs=3. Plane orientations are assigned a number: plane up=1, plane down=0. The player who orients the cards will indicate the suit of every other card by using the orientation of two cards. Take the orientation of the first card of the pair as the first digit in a binary number, and the second card of the pair as the second digit of the binary number, and that binary number indicates the suit of the second card. For example, if the second card in the deck is a diamond (=2, or in binary 10) the first card's plane will be pointing up and the second's will be pointing down. Using that strategy, every other card will be predicted correctly.

Use that strategy for the first 50 cards. Once you're down to the last two cards in the deck, they players will need to have counted cards, and they should know the suits of the last two cards. For the 51st card, the code changes to be: plane-up means the 51st card is the higher numbered suit of the two remaining suits, and plane-down means the 51st card is the lower numbered of the two remaining suits. They'll be able to get both of the last two cards correct, for a total of 27/52.

they're different suits.

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well done, plasmid. now we're making progress. however, you can still guarantee more correct predictions. and bonanova's comment adds some customarily additional clarity. @ Time Out - tho am most often a New Word Riddles Forum contributor, this one is straight forward Math/Logic so the answer to your question would be no - not this time out - .

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If your bit of information lets the other person get the right suit, no further information can be gleamed. However, if neither value will get the right answer, you can pass other types of information instead.

So, I haven't gotten a polished answer to give yet, but here's some of the things I've been playing with in regard to this puzzle.

So far, two different solutions for 50% (ignoring endgame) have been given. One, you select two suits and tell which of those two the next card is. Since you can only get cards of those two suits right, that's 50%. The other approach gets every other card right by encoding the suit in two card's orientations. Every other card, 50% again.

Here's a rudimentary combination of the two: Choose two suits and use the bit to choose between those two suits, but with the cards you cannot give the right answer, take every pair of wrong answers to give the suit of cards counting back from the end (so the guy needs to keep track of when the information is put into effect).

But this removes the use of the orientations of the cards at the end. So to improve this, the cards you get wrong tell you (counting backwards from third to last card (last two are easy as previously pointed out)), whether the card will be possible to guess with the two initial suits. If not, you choose from between the other two.

This however, still only guarantees about 50%. You could have all the cards you cannot predict first.

Another option is perhaps that the information you give is to change the two suits you choose between. With two wrong answers gives the chance to change one of the suits, or you can have three answers to set both to something (ie, there are only 6 possible pairs given 4 suits). Perhaps if you get one wrong on purpose that you could have gotten means to change both to the other two suits.

The approach I'm going to test out initially is a combination of the last idea and the combination of the two 50% ideas. Use the first 2,3, or 4 errors to change the suits you choose from, and the rest of the wrong answers to help with identifying suits of the cards counting back from the end (possibly getting one wrong on purpose if it improves things to change what I'm choosing between). I'm thinking I can get something approaching 75%, but don't quote me on that yet.

You could also use the first unpredictable card's orientation to choose between different strategies based on how the deck is shuffled. Or even the first N wrong answers (N chosen beforehand) to give a different set sequence of "predicable-with-initial-suits vs not" that will maximize the score by matching the shuffled deck to the the 2^N options. This approach could even be broken up into groups of X cards, where the wrong answers (however many there are) let you choose the sequence for the next X cards (perhaps not choosing the very best one to allow more information for the next group of cards).

That's all I've got to share as of yet, hopefully more to come. Great puzzle plainglazed!

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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%

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Here's one implementation of the sort of approach EventHorizon was talking about

Pick one suit that will be "easy", one suit that will be "sort of easy", and two suits that will be "tricky".

For every "easy" card in the deck, the card will be oriented plane-up. ANY time the player sees a card oriented plane-up, they will predict the "easy" suit. These cards will always be predicted correctly. End of story (and hence the name).

For the "sort of easy" suit, the card orienter will always orient it plane-down. These will always end up being predicted correctly. However, a plane-down orientation does NOT mean that it should always be predicted as the "sort of easy" suit because of the following exception.

For the first "tricky" card in the deck, the player will of course make an incorrect prediction regardless of whether it was plane-up or plane-down. The first time the player makes an incorrect prediction, they will realize that the plane orientation was not telling them the suit of the card they just turned over, it was telling them the suit of the next "tricky" card that they will be able to predict. If it was plane-up, that means they should look for a signal to predict "tricky suit #1", and if it was plane-down, they should look for a signal to predict "tricky suit #2". The signal that they should predict a tricky suit is: the next time they see a card oriented plane-up that turned out NOT to be the "easy" suit, they will know that the next card oriented plane-down will be a "tricky" suit.

From the card-orienting person's perspective, whenever they come to a "tricky" suit card, they will need to look further in the deck and find the first point where there are two cards that are each in a "tricky" suit (not necessarily both in the same suit) and do NOT have a "sort of easy" card between them. Then the card orienter makes the first of those "tricky" suit cards plane-up, and the second plane-down.

So in order to get a "tricky" suit predicted correctly, there will have to be at least three cards involved: one "tricky" suit card that is incorrectly guessed but which will tell the player the suit of the next tricky card that they will be able to predict, one "tricky" card oriented plane-up to let them know that the next plane-down card will be one of the tricky suits, and the correctly predicted "tricky" card that will be oriented plane-down. There are 26 "tricky" cards in the deck, so at most 26/3 = 8 could be predicted correctly. However, if there are "sort of easy" cards that prevent the second and third "tricky" cards from being consecutive, then the card that would have been the second card of that three-card series would instead have to be oriented plane-down (indicating that the next plane-down card will NOT be a tricky suit) instead of plane-up and the player would have to keep waiting for the signal to predict a tricky suit. In a worst-case scenario, all 13 of the "sort of easy" cards would interrupt a series like that, so there would effectively be only 13 "tricky" cards available to encode tricky suits, and 13/3 = 4 of them would be predicted correctly.

It seems horribly inefficient to only be able to guarantee 4 of the tricky cards, so I wouldn't be too surprised if there turns out to be a better approach.

Edit: just saw that EventHorizon posted a different approach as I was typing which gives 32 guaranteed predictions, better than the 30 with this strategy.

Edited by plasmid
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ooh, nice going plasmid and EventHorizon - one can still guarantee just a little more...

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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%

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%?

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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.

...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.

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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.

Lets assign a number value to each suit.

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.

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Edit: miscounted, going back to make some corrections

Edited by plasmid
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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.

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.

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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!

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