The Three Hat Problem
michael posted more than 13 years ago  from the howmathematiciansoccupytheirtime dept.
325jeffsenter writes: "The NYTimes has a nice article on the three hat problem, which has recently become quite popular among mathematicians. Three people are given either a red or a blue hat to wear. The goal is to have someone guess the correct color of his/her own hat with no person guessing incorrectly." Read the article for the rules of the puzzle. This problem is quite comparable to the Monty Hall problem, where people initially think that they can't do better than chance, but then realize that there is an extra source of information which can be tapped  either the host's knowledge of which door has the prize, or in this case, the fact that which player makes a guess can be determined after the game has started, that is, based on information available about the hats. Think about it  it's an interesting puzzle.
Re:Better solution (1)
Caine (784)  more than 13 years ago  (#302819)
Re:The requirements are unclear (1)
Caine (784)  more than 13 years ago  (#302820)
Some clarifications to the puzzle: (5)
DG (989)  more than 13 years ago  (#302823)
Here's some key points:
1) All guesses must be correct. If any of the 3 players guess and get it wrong, everybody loses.
2) Guesses are simultanious, not sequential  ie, you write down your guess, and all 3 guesses are passed to the host, who then reads them
3) There are 3 hats, but two colours. This means that out of 12 possible combinations, there are 2 that are "all hats same colour"  so 1/6th of the time. Thus, if you see 2 hats of the same colour, the probability that your hat is the same color is 1/6  which in turn implies that guessing that your hat is a different colour will be correct 5/6th of the time.
Thus, if you see any two differentlycoloured hats, you pass. If you see all hats the same colour, then invert the color you see and guess that. This starts at 5/6ths correct with 3 hats, and gets better for larger numbers of hats not a power of 2.
Hamming codes (1)
aleksey (1519)  more than 13 years ago  (#302824)
It's always great to see theoretical work evolve from recreational activities.
Re:Play monte hall! (1)
armb (5151)  more than 13 years ago  (#302835)
There's your problem  donuts float better than concrete blocks. Throwing something that floats overboard will leave the boat at the same level.
Throwing something that sinks won't.

Play monte hall! (2)
Booker (6173)  more than 13 years ago  (#302836)
Oh, and by the way, you should always switch.
Here's another one I like, but in a differetn (physics) vein  a man is in a boat holding a cement block. He throws it overboard. Does the lake level go up, go down, or stay the same?

Re:Blue hat? (2)
KlomDark (6370)  more than 13 years ago  (#302837)
(It was extremely tempting to turn that into a goatlink, but I didn't :) )
It's a Dr Seuss April Fools joke (4)
KlomDark (6370)  more than 13 years ago  (#302838)
This is all taken from the Dr Seuss' book "One Fish. Two Fish. Red Fish. Blue Fish [amazon.com] ", just changed to be One Hat Two Hat Red Hat Blue Hat.
Truely LMAO!
Answer, I think (spoiler) (1)
monsted (6709)  more than 13 years ago  (#302839)
Sorry, article was archived (1)
monsted (6709)  more than 13 years ago  (#302840)
The requirements are unclear (2)
pen (7191)  more than 13 years ago  (#302841)

Re:Answer is here (2)
sandler (9145)  more than 13 years ago  (#302842)
Research? (4)
sandler (9145)  more than 13 years ago  (#302843)
Sounds fun! How can I get a job at said "research" institute??
Answer obvious to me (1)
hoss_33 (10488)  more than 13 years ago  (#302845)
This needs a bunch of professional mathematicians thinking for days?
Not so obvious with a lot of hats, though.
How to cheat (5)
Syberghost (10557)  more than 13 years ago  (#302847)
Then when it's time to look at your fellow players, pick the one farthest to your right and look at his chest for down, or his hat for up. Point your whole face.
Then glance with your eyes at the others. If even one of them has read this post, you're in good shape. If they both have, you're free.
Unless the aliens are just shitting you, and intend to implant an 80foot satellite dish in your ass regardless of the outcome.

Re:Sorry, article was archived (1)
Kimble (17437)  more than 13 years ago  (#302854)
That problem's intersting, but not quite the same, though  in the 3 hat problem, everyone guesses simultaneously. Plus, only one person has to be right for the group to win, as long as everyone else passes. Plus, you can't win all of the time, but you can win a lot more of the time than you'd think.

How many classes do you have to take
Re:Better solution (2)
Mike Schiraldi (18296)  more than 13 years ago  (#302857)

Re:Better solution (1)
kramer (19951)  more than 13 years ago  (#302867)
Re:How to cheat (1)
Ralph Wiggam (22354)  more than 13 years ago  (#302868)
B
Re:How to cheat (1)
Ralph Wiggam (22354)  more than 13 years ago  (#302869)
B
Re:How to cheat (1)
Ralph Wiggam (22354)  more than 13 years ago  (#302870)
Re:50% (1)
mistered (28404)  more than 13 years ago  (#302872)
Re:another interesting problem (1)
mistered (28404)  more than 13 years ago  (#302873)
Re:What I don't get about the Monty Hall Problem (1)
mistered (28404)  more than 13 years ago  (#302874)
Re:Some clarifications to the puzzle: (2)
mistered (28404)  more than 13 years ago  (#302876)
There are 3 hats, but two colours. This means that out of 12 possible combinations
There are three hats, and each has two possible colours. Thus there are 2 * 2 * 2 = 8 possible combinations, not 12, and there's a 2/8 chance of them being all the same colour.
Answer is here (1)
gbr (31010)  more than 13 years ago  (#302879)
1. Assume 3 hats (works with any number)
2. First person looks at other peoples hats, and says "I see two red hats, therefor mine must be blue"
3. Second person looks at all other hats, and says "I see two red hats, and person 1 saw two red hats, therefor my hat is red"
4. Third person looks and sees two red hats, and repeats person two's phrase.
5. First person hears all the phrases, and concludes that his guess is wrong, and changes it to red.
Works for any combination of red and blue hats, and any number of people.
Re:Answer is here (1)
gbr (31010)  more than 13 years ago  (#302880)
My answer then changes back to an answer that I first gave a coworker. Random probability of the coin toss is 50/50.
Look at all the other hats, and then conclude that yours is in the monirity of what you saw. Works for 4+ hats, but not for 3 hats.
ie:
Re:This defies random odds (1)
Saige (53303)  more than 13 years ago  (#302891)
You'd think that the best you can do is 50% correct, since there's that chance for your hat being blue or red.
However, if you look at the possibilities for three people, the hats can be :
rrr
rrb
rbr
brr
bbr
brb
rbb
bbb
If you count, 6 out of the 8 possibilities have two hats of one color, and one hat of the other. Therefore, as the article said, you improve your odds of guessing correctly if you see two hats of the same color. It's all a matter of perspective, as to whether you look at the individual random item, or the full set of random items.
I agree, at first appearance it does seem to defy what we're taught about odds/probability and the like. I won't pretend to be quite comfortable with it...

Re:50% (1)
PurpleBob (63566)  more than 13 years ago  (#302903)

Obfuscated email addresses won't stop sadistic 12yearold ACs.
Re:75% of the time? (1)
PurpleBob (63566)  more than 13 years ago  (#302904)

Obfuscated email addresses won't stop sadistic 12yearold ACs.
Re:This defies random odds (2)
UnknownSoldier (67820)  more than 13 years ago  (#302912)
Probability is a best guess, when you don't have all the facts. Nothing funny or strange about it all.
Re:The requirements are unclear (1)
FreshGroundPepper (70119)  more than 13 years ago  (#302913)
RRR Lose
RRB Win
RBR Win
RBB Win
BRR Win
BRB Win
BBR Win
BBB Lose
The solution is that the only person to answer is the one that sees two hats of the same color on the other person. In 6 of the 8 cases that means that only 1 person says anything (and they are correct). In the 2 losing cases, they all say that they have the incorrect hat color. They can all answer at the same time and still have a 75% success rate.
FGP
Perhaps the best solution. (2)
dsplat (73054)  more than 13 years ago  (#302915)
Player 1 guesses Red only if he sees all Blue hats. This strategy is only triggered in two cases regardless of the number of players. It will always be wrong in half of those two cases, and it is the only guess that ever takes place.
Each subsequent player ignores the colors of the hats of the players who preceded him. He is only concerned with the colors of the hats of the player who will follow. If he sees only Blue hats among them, he states that his own hat is Red.
This strategy works because if the first player passed, the rest of the players know that he saw at least one Red hat. If a player sees no Red hats among the players that follow him, and he is not the first player, then he knows that the players who passed to him saw the Red hat on his head.
Since there must be a single guess to differentiate between two cases to start the inference, there is one losing case. We can't improve the odds to a sure thing. However, we have just reduced them to one losing case regardless of the number of players. Therefore, the best strategy loses still loses 1 time in 2^n cases.
If the players believe that the contest is rigged, Player 1 can randomly select the color of his guess. This brings the worst case in a rigged contest from 0% back up to 50%. I don't think there is a way to improve on that.
Re:75% of the time? (1)
Christopher Whitt (74084)  more than 13 years ago  (#302917)
Yeah, all players have to guess simultaneously.
Re:Better solution (2)
spazimodo (97579)  more than 13 years ago  (#302949)
Spazimodo
Fsck the millennium, we want it now.
Re:The requirements are unclear (2)
Fjord (99230)  more than 13 years ago  (#302954)
Re:Answer is here (1)
mobets (101759)  more than 13 years ago  (#302957)
___
Re:Hmmmm (1)
BradleyUffner (103496)  more than 13 years ago  (#302958)
=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=
Re:This defies random odds (1)
BradleyUffner (103496)  more than 13 years ago  (#302959)
=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=
Re:This defies random odds (2)
snorb (109422)  more than 13 years ago  (#302967)
another interesting problem (1)
krokodil (110356)  more than 13 years ago  (#302968)
problem which become quite popular among physicists.
You have lead ball and birds feather and drop them down from high
tower. Which one will reach ground first?
Think about it  it's an interesting puzzle.
Re:75% of the time? (1)
fyonn (115426)  more than 13 years ago  (#302973)
this waiting and then answering depending on what you see stuff is still communication as far as I am concerned
dave
Re:I have read one similar to this (1)
fyonn (115426)  more than 13 years ago  (#302974)
they initially paid $10 each for the room, or $30 in total. the woman came and gave them back a dollar each so that the new figures are that they paid $9 each or $27 on total. however of that $27 dollars, the room cost $25 and she kept $2. you're adding instead of subtracting.
all I can say is that I hope they good decent room service
dave
Re:another interesting problem (1)
fyonn (115426)  more than 13 years ago  (#302975)
are they dropping through an atmosphere or a vacuum, if the latter both would hit at the same time.
of course if the lead ball was shot and the feather was attached to a pidgeon then the feather would hit first
what are the flaws in my argument here?
Re:Better solution (1)
klieber (124032)  more than 13 years ago  (#302981)
The Red Hat problem (5)
SpanishInquisition (127269)  more than 13 years ago  (#302983)

Read the problem more closely (1)
jshowlett (134148)  more than 13 years ago  (#302985)
(oops)Re:Read the problem more closely (1)
jshowlett (134148)  more than 13 years ago  (#302986)
The correct strategy can cause all three players to guess wrong together 25% of the time, while the remaining 75% of the time one player will guess correctly and two will pass, something like this:
X=Wrong, +=Right, =Pass
TrialNo: 1 2 3 4
Player1: X +  
Player2: X  + 
Player3: X   +
The number of right guesses is the same as the number of wrong guesses, but they still win 75% of the time. This is because the strategy helps tell the players when they ought to pass instead of trying to guess at all.
Re:So what is the strategy for larger teams? (2)
Sir Tristam (139543)  more than 13 years ago  (#302989)
The first round, if anybody saw six hats of a color, they would guess the other color; otherwise, they would pass. If all the hats are the same color, you lose right here, otherwise you're going to win. If there are six hats of one color and one of the other, you win right here.
The second round, everybody knows that nobody saw six hats of a color. (Ahh... but does this count as communications? Hmmm...) So, anybody who sees five hats of a given color guesses the other color, while everybody else passes. If you have five hats of one color and two of the other, the two with the other color will guess correctly, and you win.
The third round, everybody knows that nobody saw five hats of a color. Anybody who sees four hats of a given color guesses the other color. This is guaranteed to be the last round, as the three in the minority guess their hat color correctly.
Even numbers should be fun. Having four hats, two of each color, would result in winning in the second round with everybody guessing their hat color correctly.
Solution (2)
gowen (141411)  more than 13 years ago  (#302990)
Remember "Mastermind" (1)
jsin (141879)  more than 13 years ago  (#302991)
Blue hat? (3)
ritlane (147638)  more than 13 years ago  (#302994)
There is something other than Red Hat?
Lane [rit.edu]
I have read one similar to this (2)
woody_jay (149371)  more than 13 years ago  (#302998)
Three guys come want to stay in a hotel and pay equal amount each to share a hotel room. When they ask the woman behind the counter what the price is for a single room, she replies that it is 30 dollars. This is to their delight as they will be able to split that very evenly between the three of them. They each give her $10 and are happily on their way to thier room. About an hour later, the woman's boss comes in and asks if anything happened while he was out. She tells him the story of the three men to which he responds be telling her that they paid too much and that the price of the room should have been $25. He tells her to take $5 and give it back to them. In the process of returning the money, she remebers how they all wanted to pay an equal amount. So to make it easy on them, she gave them each $1 back, and kept the remaining $2 for herself.
This is where it get's tricky. They each paid $9 which multiplied by 3 is $27. She kept $2 and when added on to the $27 that they paid is $29. Where did the 30th dollar go?
I love that one.
Re:What I don't get about the Monty Hall Problem (1)
lorian69 (150342)  more than 13 years ago  (#302999)
In reality, when faced with this situation and a possible prize, people will tend to second guess themselves, somehow deciding that the other choice might be more likely to be correct since they didn't remove it... no logical sense, but logic doesn't always play a big part in a swift, nervous decision.
Re:Some clarifications to the puzzle: (1)
lorian69 (150342)  more than 13 years ago  (#303000)
Re:Play monte hall! (1)
been42 (160065)  more than 13 years ago  (#303003)
Hrmm, lets see (1)
revelation0 (164235)  more than 13 years ago  (#303005)
The three people walk into the room. The first person to notice that the other two are wearing different color hats passes. Then the person to their left looks to the other person and guesses the opposite color hat. If no one passes, someone just blurt out the color of the others hats.
That should get it about every time
Revelations 0:0  The begining of the end.
Real world problems (2)
zettabyte (165173)  more than 13 years ago  (#303006)
No wonder custom business software is often over engineered. We're bored to death with mundane business logic!
Does, (1)
chowpalace (166596)  more than 13 years ago  (#303008)
Re:75% of the time? (1)
dannywyatt (175432)  more than 13 years ago  (#303017)
Yes, they have to answer simultaneously. Answering in succession would constitute communication between players, which is disallowed.
Better solution (1)
Erasmus Darwin (183180)  more than 13 years ago  (#303026)
In the article, in the 3 hat case, they've got a 75% chance of winning (their solution generated a loss if all 3 hats were the same color). I've managed to come up with an 87.5% chance of winning, instead:
Player 3's strategy will be to always guess 'red'.
Player 2's strategy will be to guess 'red' only if Player 3's hat is 'blue'. Otherwise, Player 2 passes, knowing Player 3 will win.
Player 1's strategy will be to guess 'red' only if Player 2 and Player 3 both have 'blue'. Otherwise, Player 1 passes, knowing Player 2 or Player 3 will win it.
The only time this solution will fail is the case where all three hats are blue. This occurs 1 out of 8 times (12.5%).
In the 15 player case mentioned later on, the article claims they've got a situation that works 15 out of 16 times (93.75%). Using my method above, it should work 99.99% of the time.
In the immortal words of Joel Hodgson, "What do you think, sirs?"
Re:Better solution (2)
Erasmus Darwin (183180)  more than 13 years ago  (#303029)
That was, indeed, my problem. I didn't realize that they had to guess simultaneously. Had the players been able to guess in order, mine would've worked. (And still would've tied back to the hamming code concept.)
Owel, back to the drawing board.
It's simple ...... (1)
sawb (187496)  more than 13 years ago  (#303030)
Oh wait
This defies random odds (1)
Ergo2000 (203269)  more than 13 years ago  (#303038)
It's been a while since I've played with finite math, and a lot of it's been forgotten, however this seems to defy the nature of random odds. i.e. Am I incorrect or is this based upon the supposition that by looking at the other draws you can base conclusions on your own (which of course defies the nature of random odds and is a common fallacy).
Re:This defies random odds (1)
Ergo2000 (203269)  more than 13 years ago  (#303039)
Ahhh...
Re:The requirements are unclear (1)
Ergo2000 (203269)  more than 13 years ago  (#303040)
This confused me as well however one part states "Once they have had a chance to look at the other hats, the players must simultaneously guess the color of their own hats or pass" so they are guessing simultaneously. I think they should reword "pass" to "give no answer" as pass implies successive answering.
Re:Better solution (1)
mapMonkey (207912)  more than 13 years ago  (#303043)
What I don't get about the Monty Hall Problem (1)
CoreyG (208821)  more than 13 years ago  (#303045)
According to my understanding of the problem, you have 3 choices with only one correct choice. After you choose, one of the 3 choices(not yours) is shown to be incorrect and removed. You now have the option of changing your guess. The option of changing your guess completely redefines the problem! The probability of you picking the correct answer is no longer based out of 3 choices, but out of 2! Since an incorrect choice was removed from the 3, that leaves a correct choice and an incorrect choice. The 3rd choice is now completely and utterly irrelevant and devoid of the problem. It never existed! It is not a possible choice! The problem, once an incorrect choice is removed, is now based on 2 possible choices. One is right, one is wrong. The probability of you having originally picked the correct door is
nonJavaScript version (1)
brlewis (214632)  more than 13 years ago  (#303052)
For the Bridge Players out there (1)
dave cutler (222400)  more than 13 years ago  (#303056)
Re:Better solution (1)
dave cutler (222400)  more than 13 years ago  (#303057)
Your solution fails whenever player 3's hat is blue. This happens 50% of the time.
Monty Python, not Monte hall (1)
Exedore (223159)  more than 13 years ago  (#303058)
So if the concrete block weighs as much as a duck...
A witch!!! Burn it!!!
Good wording at Grey Labyrinth (1)
Jodiamonds (226053)  more than 13 years ago  (#303059)
Re:Better solution (1)
idontrtfm (235021)  more than 13 years ago  (#303067)
Player 3 always guesses.
50% of the time he is wrong.
The choices of the others only increase the probability that someone else will be wrong further decreasing the possibility that the team wins.
Too simple to be true? (1)
ishrat (235467)  more than 13 years ago  (#303068)
Next the article is an interesting read and so is the solution to the problem, but I still have this doubt whether it was not one of the April fool jokes played around the world, or else the application of the theory of probability should have been simple enough. But then I am no mathematician and what may seem obvious to me may not be so obvious afterall.
Re:This defies random odds (1)
shyster (245228)  more than 13 years ago  (#303074)
No it's not. I think that would be called an estimate or best guess. Probability [dictionary.com] is the likelihood that something will occur. In this case, the likelihood that your best guess would be correct.
That being said, this doesn't defy random odds because it's not random. By introducing the twist where you can see the other's hats, and only 1 has to make a guess, then you can pull little tricks like having the one with the most complete information (2 hats of the same color) guessing.
Re:50% (2)
shyster (245228)  more than 13 years ago  (#303075)
Let me guess...I bet you got alot of those tricky word problems wrong, didn't you? But you were always the first one done!
What do you guys think my odds of getting that one right are? =)
Umm No reg required (1)
discovercomics (246851)  more than 13 years ago  (#303077)
Wow (1)
Evil Adrian (253301)  more than 13 years ago  (#303079)

Re:75% of the time? (1)
ocbwilg (259828)  more than 13 years ago  (#303081)
Yes, you are. They have to guess simultaneously.
Re:What I don't get about the Monty Hall Problem (1)
Proud Geek (260376)  more than 13 years ago  (#303082)
p(1) = 1/3
p(2) = 1/3
p(3) = 1/3
select 1:
p(1) = 1/3
p(2U3) = p(2) + p(3) = 2/3
remove 2:
p(1) = 1/3
p'(2) = 0
p'(3) = p(2U3)  p'(2) = 2/3  0 = 2/3
What you are probably missing is that although you resolve the uncertainty associated with 2, you don't change the probability associated with the prize being behind 2 or three, versus the probability of the prize being behind 1.
I just check my checkbook balance (3)
typical geek (261980)  more than 13 years ago  (#303083)
Duh... (1)
Dancin_Santa (265275)  more than 13 years ago  (#303084)
Dancin Santa
Re:Too simple to be true? (1)
abrett (308268)  more than 13 years ago  (#303087)
My favourite color is.... (1)
JohnnyKnoxville (311956)  more than 13 years ago  (#303089)
Re:Better solution (1)
Anemophilous Coward (312040)  more than 13 years ago  (#303090)
From what I see of your strategy, you base the success rate on only one person guessing correctly while the other two may be right or wrong. If Player 3 is always guessing red, but really has blue which causes Player 2 to guess red, they all still lose since one of them (player 3) guessed incorrectly.
Re:What I don't get about the Monty Hall Problem (1)
3am (314579)  more than 13 years ago  (#303100)
'Mathematically, you're correct' blah, blah, blah...
dope. coders trying to do math. in fact he is wrong, and you are wrong. write a program that imlements the Monty Hall problem, and then do it 500 times. then talk about mathematics.
glad you could correct an error that thousands of mathematicians have been overlooking for the last 50 years... idiot...
Re:Better solution (1)
snehumak (316134)  more than 13 years ago  (#303106)
heard about this from a friend recently (1)
JiffyPop (318506)  more than 13 years ago  (#303110)
THE ANSWER: (SPOILER!!)
1) each color is given a number (alphabetically to remove the ambiguity) from 1 to m
2) the first person adds up all of the colors in front of him modulo m. he will likely die (especially with a lot of colors)
3) the second person adds up all of the people in front of him modulo m. he then subtracts this from the color the first person said. (if it is negative he adds m) this gives him the color on his head.
4) the nth person subtracts the sum of all of the previous n2 people's answers modulo m from the first persons response, and adds m if necessary, to get the color on his head!
this is one of the coolest math problems i have ever heard. does anyone have some more alone the same vein?
not an answer (1)
JiffyPop (318506)  more than 13 years ago  (#303111)
Re:This defies random odds (1)
JiffyPop (318506)  more than 13 years ago  (#303112)
no, this does not break any rules (obviously, since it does work). the key is to think about where the players can get their information. the kth person can see the colors on the nk hats in from of him (but not the ones of the people that preceed him). everyone but the first person also has the knowledge of what the sum of all colors modulo m is, and what the colors of the k1 people between them and the first person. this is enough information to reconstruct the color of their own hat, so only the first person has the possibility of dying. he dies a heros death, however, and gives all of the other players the extra information that they need to save themselves.
Re:How to cheat (1)
madfgurtbn (321041)  more than 13 years ago  (#303114)
Hmmmm (5)
BillyGoatThree (324006)  more than 13 years ago  (#303115)
I was going to post a solution like this (before reading the rest of the article, duh) but then I thought it didn't work. That's because I was taking a single point of view (which is Timothy's point). I should have gone ahead and done it....
There are 8 possible universes. The algorithm works as follows:
RRR = every player sees two reds, every player says "blue"  LOSS
BBB = every player sees two blues, every player says "red"  LOSS
RRB = the reds see conflict and pass, the blue sees two reds and guesses "blue"  WIN
RBR = the reds see conflict and pass, the blue sees two reds and guesses "blue"  WIN
BRR = the reds see conflict and pass, the blue sees two reds and guesses "blue"  WIN
RBB = the blues see conflict and pass, the red sees two blues and guesses "red"  WIN
BBR = the blues see conflict and pass, the red sees two blues and guesses "red"  WIN
BRB = the blues see conflict and pass, the red sees two blues and guesses "red"  WIN
That's 6 wins in 8 plays or 3 wins in 4 plays. I'm not going to try to extend this to further cases.

Re:How to cheat (1)
Kormath (344334)  more than 13 years ago  (#303120)
Re:heard about this from a friend recently (1)
bark76 (410275)  more than 13 years ago  (#303122)
THE ANSWER: (SPOILER!!)
Nope, sorry but your answer doesn't work. If one person gets it wrong, then they lose. Each person either passes or guess right in order for them to win (and they can't all pass either, at least one person has to guess right). The article never explained how that 75% solution works for groups of 7, 15, etc, anyone been able to find any links related to this problem that might explain this? The 3 person version is easy, all you have to look at are 2 people and they're either the same or not, but with more than 2 people it doesn't stay easy. I'd like to see some of the other solutions too if they're available online.
With your solution, the chances of winning all rest solely on the first person (1 in m at best), everyone else may as well pass after the first one guesses. The best solution must give the first person a chance to pass.
I wonder if there is a npcomplete solution to this problem?
Re:So what is the strategy for larger teams? (1)
bark76 (410275)  more than 13 years ago  (#303123)
players must simultaneously guess the color of their own hats or pass
Sorry, the article says nothing about multiple rounds, no one can be wrong and at least one person must be right.
Re:How to cheat (1)
bark76 (410275)  more than 13 years ago  (#303124)
Re:50% (1)
lbarbato (410651)  more than 13 years ago  (#303127)
75% of the time? (1)
plluke (412415)  more than 13 years ago  (#303133)
So what is the strategy for larger teams? (2)
Hilary Rosen (415151)  more than 13 years ago  (#303137)
