Card-players - a hats type puzzle.
Nick Wedd
terms of the puzzle - as stated in this posting - and then allows them
to confer and plan their strategy. Then she draws four random cards
(from four different packs), and fixes one to the forehead of each
player. Each player can see the other three cards, but not his own.
Each must then simultaneously, and without communicating in any way, try
to write down the suit of his own card.
If at least one of them succeeds in writing the suit of his own card,
all four are rewarded. If all four fail, all four are punished. What
strategy should they use to maximise the chance of reward?
Nick
(The packs used consist of equal numbers of cards of four suits: spades,
hearts, diamonds, and clubs.)
--
Nick Wedd ni...@maproom.co.uk
Matthew Russotto
Nick Wedd <ni...@maproom.co.uk> wrote:
>Four card players are sitting around a table. The director explains the
>terms of the puzzle - as stated in this posting - and then allows them
>to confer and plan their strategy. Then she draws four random cards
>(from four different packs), and fixes one to the forehead of each
>player. Each player can see the other three cards, but not his own.
>Each must then simultaneously, and without communicating in any way, try
>to write down the suit of his own card.
>
>If at least one of them succeeds in writing the suit of his own card,
>all four are rewarded. If all four fail, all four are punished. What
>strategy should they use to maximise the chance of reward?
If they all write a suit without regard to what they see, chance of
losing is 81/256.
Assuming no hidden channels of communication and no tricks like
wearable mirrors, I don't see a way to improve it. There are many
ways they could cheat, but most violate the terms of the puzzle.
One cheating method is for at least one of them to decide to write the card of
the person opposite him. The opposite person will start by writing an
"S" on his paper, while the signaller writes in the normal
direction. After writing the S, the opposite person peeks at the
signallers paper (which will be "H", "C", "D" or "S") and continues
writing the correct answer.
--
There's no such thing as a free lunch, but certain accounting practices can
result in a fully-depreciated one.
Nick Wedd
<russ...@grace.speakeasy.net> writes
>In article <oSgDWrKp...@maproom.demon.co.uk>,
>Nick Wedd <ni...@maproom.co.uk> wrote:
>>Four card players are sitting around a table. The director explains the
>>terms of the puzzle - as stated in this posting - and then allows them
>>to confer and plan their strategy. Then she draws four random cards
>>(from four different packs), and fixes one to the forehead of each
>>player. Each player can see the other three cards, but not his own.
>>Each must then simultaneously, and without communicating in any way, try
>>to write down the suit of his own card.
>>
>>If at least one of them succeeds in writing the suit of his own card,
>>all four are rewarded. If all four fail, all four are punished. What
>>strategy should they use to maximise the chance of reward?
>
>If they all write a suit without regard to what they see, chance of
>losing is 81/256.
Right.
>Assuming no hidden channels of communication and no tricks like
>wearable mirrors, I don't see a way to improve it. There are many
>ways they could cheat, but most violate the terms of the puzzle.
They can improve it without cheating.
>One cheating method is for at least one of them to decide to write the card of
>the person opposite him. The opposite person will start by writing an
>"S" on his paper, while the signaller writes in the normal
>direction. After writing the S, the opposite person peeks at the
>signallers paper (which will be "H", "C", "D" or "S") and continues
>writing the correct answer.
Nick
--
Nick Wedd ni...@maproom.co.uk
d...@d-cubed.co.uk
people are writing down?
d...@d-cubed.co.uk
To maximise their chance of reward they should each write down the name
of a card suit.
(If any one of them didn't, the chances of losing would increase to
27/64! If two of them didn't, it would increase to 9/16 and so on. So
to maximise their chances, none of them should cock it up!)
Nick Wedd
d...@d-cubed.co.uk writes
>Does "no communication" include not being allowed to look at what other
>people are writing down?
It does.
Once the first card has been attached to a player, no information of any
kind may pas from any player to any other. All they can do is look at
each others' cards. (This goes for both puzzles.)
sunn...@yahoo.com
last guess a suit he doesn't see. The probability that the last person
is correct increases from 1/4 to 13/49. This is independent of the
other people's guesses, so the chance of them winning is 1 - (3/4)^3 *
36/49 = 541/784 = .6901.
I don't think all the players can employ this strategy, because then
their guesses are no longer independent. For example, if four hearts
appear they will always be wrong.
Nick Wedd
sunn...@yahoo.com writes
>Slightly better: Have the first three people guess randomly and the
>last guess a suit he doesn't see. The probability that the last person
>is correct increases from 1/4 to 13/49. This is independent of the
>other people's guesses, so the chance of them winning is 1 - (3/4)^3 *
>36/49 = 541/784 = .6901.
This calculation assumes that all four cards are drawn from the same
pack. The original posting specifies that each is drawn from a
different pack.
Peter Smyth
"Nick Wedd" <ni...@maproom.co.uk> wrote in message
news:oSgDWrKp...@maproom.demon.co.uk...
> Four card players are sitting around a table. The director explains the
> terms of the puzzle - as stated in this posting - and then allows them to
> confer and plan their strategy. Then she draws four random cards (from
> four different packs), and fixes one to the forehead of each player. Each
> player can see the other three cards, but not his own. Each must then
> simultaneously, and without communicating in any way, try to write down
> the suit of his own card.
>
> If at least one of them succeeds in writing the suit of his own card, all
> four are rewarded. If all four fail, all four are punished. What
> strategy should they use to maximise the chance of reward?
>
> Nick
For example, give the suits number values spades=0, hearts=1, diamonds=2,
clubs=3. Player A picks suit (B+C+D) mod 4 where B is the suit showing on
player B etc. Player B can reason that A will be correct if and only if
player B's suit is (A-C-D) mod 4. So if A=Hearts C=Diamonds D=Spades player
B reasons that A will be correct iff B=Clubs. This means that there is no
point in B guessing clubs because if he is right A will have been right
anyway. So B should guess one of the other 3 suits according to some
predetermined formula that I haven't worked out yet. Then player C can
reason that if his card is one of two suits then either A or B will be right
so C guesses one of the other two suits. Finally D can eliminate three suits
by the same logic and guesses the remaining one.
Assuming suitable formulas exist for the second and third players choice it
should be possible to get 100% success by ensuring that one and only one of
the four players is always right.
Peter Smyth
Mike Williams
That sounds as if it might be a really good approach if it weren't for
the fact that they have to write their answers simultaneously, making it
necessary for any players to see someone else's answer before deciding
their own answer.
--
Mike Williams
Gentleman of Leisure
sunn...@yahoo.com
Player A guesses 0-B-C-D mod 4,
Player B guesses 1-A-C-D mod 4,
Player C guesses 2-A-B-D mod 4,
Player D guesses 3-A-B-C mod 4.
Exactly one of them will be correct.
dgates
Wow! After initially being certain that there was no way for them to
improve upon random choices, I'm now thinking you're right... But
it's so complicated!
Let me try it with two guys and dice. If that works, then I'll
believe that yours work.
Abe and Ben flip coins and stick them to their foreheads, then each
guess if they have a Heads or Tail stuck to their head.
By pure random choice, they'll win 3/4 of the time.
However, Abe tells Ben "I'll write whatever I see on YOUR forehead."
The coins are flipped and Ben sees Heads on Abe's forehead. Ben then
reasons "If I've got Heads on MY forehead, then we've already won. So
I'll just write Tails and cover that remaining possibility."
Okay, I'm convinced. These four guys can win every time, even though
they're drawing from four different packs! :-)
Mark P
>
> Wow! After initially being certain that there was no way for them to
> improve upon random choices, I'm now thinking you're right... But
> it's so complicated!
>
> Let me try it with two guys and dice. If that works, then I'll
> believe that yours work.
>
> Abe and Ben flip coins and stick them to their foreheads, then each
> guess if they have a Heads or Tail stuck to their head.
>
> By pure random choice, they'll win 3/4 of the time.
>
> However, Abe tells Ben "I'll write whatever I see on YOUR forehead."
>
> The coins are flipped and Ben sees Heads on Abe's forehead. Ben then
> reasons "If I've got Heads on MY forehead, then we've already won. So
> I'll just write Tails and cover that remaining possibility."
>
>
A nice idea to distill this down to the two player game (though I kind
of like the idea of a 52 player game where 1 person always guesses
*exactly* which card is on his head).
Another way to look at your example (which I think makes it simpler to
explain) is that Abe will always guess what he sees, and Ben will always
guess the opposite of what he sees. Then for two coins, if they're the
same, Abe will guess correctly, and if they're different, Ben will guess
correctly. All possibilities covered.
Nick, did you invent this puzzle or do you know its origin?
Ob puzzle: Suppose there are M players and each receives a random
symbol chosen (separately) from among N possibilities with M < N. Is
the "obvious" generalization of this strategy optimal (i.e., each
assumes the sum of all symbols is a different value mod N), or is there
a better strategy?
Mark
Nick Wedd
<n...@my.real.email> writes
>dgates wrote:
>> Wow! After initially being certain that there was no way for them
>>to
>> improve upon random choices, I'm now thinking you're right... But
>> it's so complicated!
>> Let me try it with two guys and dice. If that works, then I'll
>> believe that yours work.
>> Abe and Ben flip coins and stick them to their foreheads, then each
>> guess if they have a Heads or Tail stuck to their head.
>> By pure random choice, they'll win 3/4 of the time.
>> However, Abe tells Ben "I'll write whatever I see on YOUR forehead."
>> The coins are flipped and Ben sees Heads on Abe's forehead. Ben
>>then
>> reasons "If I've got Heads on MY forehead, then we've already won. So
>> I'll just write Tails and cover that remaining possibility."
>>
>
>A nice idea to distill this down to the two player game (though I kind
>of like the idea of a 52 player game where 1 person always guesses
>*exactly* which card is on his head).
>
>Another way to look at your example (which I think makes it simpler to
>explain) is that Abe will always guess what he sees, and Ben will
>always guess the opposite of what he sees. Then for two coins, if
>they're the same, Abe will guess correctly, and if they're different,
>Ben will guess correctly. All possibilities covered.
>
>Nick, did you invent this puzzle or do you know its origin?
I did not invent it. I heard it from someone who first asked me the
two-player version, and then asked me to extend it to three and more
players. I can ask him where he got it. I posted the four-player
version here, to match the other four-player puzzle that I posted at the
same time.
>Ob puzzle: Suppose there are M players and each receives a random
>symbol chosen (separately) from among N possibilities with M < N. Is
>the "obvious" generalization of this strategy optimal (i.e., each
>assumes the sum of all symbols is a different value mod N), or is there
>a better strategy?
>
>Mark
Nick
--
Nick Wedd ni...@maproom.co.uk
Nick Wedd
sunn...@yahoo.com writes
Correct, well solved!
dgates
wrote:
I wonder if the puzzle wouldn't actually be easier (in terms of
telling it to anyone outside of rec.puzzles) if it were 6 guys rolling
dice.
For one thing, that eliminates the need for the "4 separate decks"
rule.
Also, maybe having actual numbers on their head would prompt the
thinking of "Well, the 6 dice have to add up to SOMETHING..."
although, even then, I still think that the "mod 6" thinking might
never enter people's heads.
Nah, if I ever tell this to a regular human being, it'll have to be
two dice :-(
Nis Jorgensen
wrote:
>Four card players are sitting around a table. The director explains the
D
S
H
C
I
P
E
L
A
A
E
U
M
D
A
B
O
E
R
S
N
S
S
D
S
Each player will make a different assumption about the distribution of
cards, and fill in his own suit to satisfy that assumption
The first player will assume that there is an even number of
clubs+diamonds, and an even number of clubs+hearts. Thus if she sees
CCS, she chooses S, CDD -> C, etc
The second player will assume that there is an even number of
clubs+diamonds, and an odd number of clubs+hearts
The third player will assume an odd number of clubs+diamonds and an
even number of clubs+hearts
The fourth player assumes odd clubs+diamonds and an odd number of
clubs+hearts.
It is easy to see that each assumption leads to one specific choice of
your own suit, and that exactly one of the assumptions will be valid.
Thus the players allways win.
/Nis
Nick Atty
wrote:
>I did not invent it. I heard it from someone who first asked me the
>two-player version, and then asked me to extend it to three and more
>players. I can ask him where he got it. I posted the four-player
>version here, to match the other four-player puzzle that I posted at the
>same time.
It really is quite astonishingly counter-intuitive. I still can't quite
get my head around it (I can understand *how* it works, and even *why*,
but it still feels impossible).
--
On-line canal route planner: http://www.canalplan.org.uk
(Waterways World site of the month, April 2001)
dgates
wrote:
Is this spoiler space some kind of puzzle in itself? :-)
I can see that it starts with DSHC for Diamonds Clubs Hearts Spades,
but I can't see what follows...
Nis Jorgensen
On Sat, 21 May 2005 10:39:13 -0700, dgates <dga...@spamlinkline.com>
wrote:
>On Sat, 21 May 2005 14:12:41 +0200, Nis Jorgensen <n...@superlativ.dk>
>wrote:
>>>(The packs used consist of equal numbers of cards of four suits: spades,
>>>hearts, diamonds, and clubs.)
>>
>>D
>>S
>>H
>>C
>>I
>>P
>>E
>>L
>>A
>>A
>>A
>>U
>>M
>>D
>>R
>>B
>>O
>>E
>>T
>>S
>>N
>>S
>>S
>>D
>>S
>
>Is this spoiler space some kind of puzzle in itself? :-)
Well, I didn't consider it a puzzle, really. Just me trying to be a
little bit creative.
>I can see that it starts with DSHC for Diamonds Clubs Hearts Spades,
>but I can't see what follows...
You are on the right track (although DSCH is NOT Diamonds Clubs Hearts
Spades). There was a mistake in my original version(corrected in the
quote above), but I don't think that made it much harder to "solve".
/Nis
Martin DeMello
>
> It really is quite astonishingly counter-intuitive. I still can't quite
> get my head around it (I can understand *how* it works, and even *why*,
> but it still feels impossible).
Here's one route to intuition:
Assume that there is a single card in a hat. Let each man have a single
guess. The strategy is then, obviously, for the first man to say
"clubs", the second "diamonds" etc - four possibilities, four men, and
you're guaranteed that one of them will get it right.
Now label clubs=0, hearts=1, diamonds=2, spades=3. Toss four cards into
a hat, and ask what the sum of the cards is, modulo 4. Again, the
strategy is for person 1 to say "0", person 2 to say "1", etc.
Then notice that, having decided on a guess for the sum modulo 4, and
looking at three cards, the fourth card is uniquely determined. The men
aren't guessing at their individual cards, they're systematically
covering the various values of sum modulo 4. One of them has to be
right, and when he is, his card gets correctly guessed almost as a
byproduct.
martin
Mark P
>
> Ob puzzle: Suppose there are M players and each receives a random
> symbol chosen (separately) from among N possibilities with M < N. Is
> the "obvious" generalization of this strategy optimal (i.e., each
> assumes the sum of all symbols is a different value mod N), or is there
> a better strategy?
>
Seeing no takers for this one I'll just spill it...
Yes, this is the optimal strategy for M < N. Any individual player,
regardless of his strategy, will be correct with probability 1/N. Then
the best the group can do is to have no overlap in correctness so that
at least one is correct with probability M/N. Since each player picks a
different value for the sum mod N, there cannot be any overlap, and they
will have a correct answer with probability M/N, which is the best possible.
dgates
That's not bad... although I was more easily to find faith by
reducing the four card suits to the two sides of a coin, and the four
guys to two guys. :-)
Nick Atty
<nos...@nandj.freeserve.co.uk> wrote:
>On Fri, 20 May 2005 10:30:28 +0100, Nick Wedd <ni...@maproom.co.uk>
>wrote:
>
>>I did not invent it. I heard it from someone who first asked me the
>>two-player version, and then asked me to extend it to three and more
>>players. I can ask him where he got it. I posted the four-player
>>version here, to match the other four-player puzzle that I posted at the
>>same time.
>
>It really is quite astonishingly counter-intuitive. I still can't quite
>get my head around it (I can understand *how* it works, and even *why*,
>but it still feels impossible).
Thinking about it a bit more, I've worked out why it feels so wrong.
When you hear the answer you think that it somehow breaks the rules of
chance, because the strategy is increasing the chances of being right.
But you aren't - this is because of the way it is scored.
Think of the two coins version. Suppose you each had a strategy of
always guessing heads. That would score the average of 1 point per
game and you'd only lose one in game in four. What this strategy does
is - essentially - take the extra point from the HH case and give it to
the TT case. Average score remains the same, but you always "win".
Good - that's slightly less jaw-dropping!
dgates
wrote:
So...
DSHCIPELAAAUMDRBOETSNSSDS
I see it as a SERIES of puzzles...
DSHC
Diamonds Spades Hearts Clubs
IPEL
Ingrid Pitt, Elsa Lanchester
AAAU
Alcoholics Anonymous, Alcoholics Unanimous
MDRB
Michael Douglas, Rhett Butler
OETSN
one eleven thirteen seven nine
SSDS
Sharon Stone, Dr. Seuss
But I don't know what it all adds up to! :-)
dgates
<nos...@nandj.freeserve.co.uk> wrote:
Interesting. Perhaps we humans are not used to thinking so
collaboratively.
You could muck it up further by saying that the four players are
strangers and the rules are as follows: If they all win, they each
get $50,000.
If only some win, then the winners get $1,000,000 and the losers are
killed and disposed of.
I haven't fully worked out the details, but it would be interesting to
see how the collaboration works now.
Ed Murphy
What happens if none of them win? I think that's crucial to figuring
out how the collaboration would work.
Ed Murphy
> Great idea. It must be that A+B+C+D = 0,1,2,3 mod 4.
>
> Player A guesses 0-B-C-D mod 4,
> Player B guesses 1-A-C-D mod 4,
> Player C guesses 2-A-B-D mod 4,
> Player D guesses 3-A-B-C mod 4.
Or, equivalently but perhaps easier to understand:
Player A guesses W such that W+B+C+D = 0 mod 4,
Player B guesses X such that A+X+C+D = 1 mod 4,
Player C guesses Y such that A+B+Y+D = 2 mod 4,
Player D guesses Z such that A+B+C+Z = 3 mod 4.
Ed Murphy
B doesn't need to know the value of A's choice (B+C+D mod 4). He just
needs to know which of his own choices would have the same truth value
as A's choice (A-C-D mod 4), and pick one of the others, say (A-C-D+1
mod 4).
Similarly:
C needs to pick something with a different truth value from A and B,
say (A-B-D+2 mod 4).
D needs to pick something with a different truth value from A and B
and C, say (A-B-C+3 mod 4).