47 replies to this topic

[Time Out]

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?

[plainglazed]

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.

[plasmid]

Ok, just a tiny bit better:

Spoiler for

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.

[Time Out]

Spoiler for ?Suit predictions. Would perhaps

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

[bonanova]

Ok, just a tiny bit better:

Spoiler for

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.

Spoiler for If

they're different suits.

[plainglazed]

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 - .

[EventHorizon]

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

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!

[EventHorizon]

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%

[plasmid]

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

Spoiler for

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, 22 May 2012 - 12:01 AM.

[plainglazed]

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