Originally posted by forkedknightI reckon it would be impossible in that case.
I just noticed that in the instructions, it is the person arranging the cards that also gets to choose which card to describe.
Provided that constraint, Dejection's method, as well as a probably nearly infinite number of other methods, would work just fine.
I think the problem is a little more interesting if someone other than the arranger gets to choose the card.
Originally posted by luskinYes impossible.
I reckon it would be impossible in that case.
Using 4 random cards to describe a 5th random card! You cannot rely on having any particular suit or any particular number so have only the ranking order to use.
4 cards can be arranged in 24 differnt ways so only 24 cards can be described.
If you have cards that have up and down then it is possible. (Allows 16x24 permutations)
2 edits
Originally posted by doodinthemoodYes you get to choose which card to give back and which card to use as the "indicator". So, you give back the Qh and use the Jh to indicate the suit. Use the other 3 cards to code a difference of 1.
You get to pick the card to be shown. Don't choose the 4S. Choose the JH, then use dejection's method. It works fine.
oh, and you have to agree to take the smaller one, so they know to add the difference rather than subtract.
In some cases, you can allow the person to pick what card is to be coded, as some cards (7 of diamonds, for instance) clearly "point" in a certain direction, so as well as having the 4! possibilities for ordering, you can multiply it by two for which way you've made a card point. Of course, there are cards (ace of diamonds, 2 of clubs) that don't go along with the idea.
Originally posted by heinzkatI wouldn't pick the 10 of spades out of the 5 cards 4H 4D 4C 4S and 10S, I would pick the 4 of spades.
Just wondering, how four '4's would be arranged to describe a 10 of spades?
The 10 goes first (telling Peter the suit) and the fact that three fours are also visible, tells him that a 4 is the rank. If I was only given 3 - 4s, I would pick one of the 4s, so as to not alert him in the same fashion.
Originally posted by forkedknightYour solution is good but ... "I then arrange the other four cards in a special way, and give those 4 cards all face down"
I think it would not be impossible. See my solution.
FACE DOWN
Which is why I said a pack which differentiates UP and DOWN allows more permutations.
Basically the problem is how to encode 1-48 with 4 cards. I think its done.
🙂
Originally posted by forkedknightArguable whether it is within the rules. It seems to me, that the words "and give those 4 cards all face down, and in a neat pile, to Peter", are deliberately intended to exclude the binary solution.
Alright, so try upright and sideways; same effect, still in the rules...
Originally posted by doodinthemoodVerbal code? Ever heard of Schnapps? I did this verbal trick with banx99. People are annoyed at it. For example, to encode "Edward" in Schnapps"
It's not much of a "trick". More, as it is posed here, a logical puzzle. Learn a verbal code with your friend Greg, and it will be much more impressive.
* denotes a click of the fingers
# denotes a pause
"**Alright***This is so easy***#*Right?#Don't hesitate now, I've finished."
XD
I wonder if anyone can figure it out?
Originally posted by Dejectiondude...
Verbal code? Ever heard of Schnapps? I did this verbal trick with banx99. People are annoyed at it. For example, to encode "Edward" in Schnapps"
* denotes a click of the fingers
# denotes a pause
"**Alright***This is so easy***#*Right?#Don't hesitate now, I've finished."
XD
I wonder if anyone can figure it out?
as if u wuld go about doin that on rhp...
it onli works properly wen u say it
else it gets v obvious