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The Three Hat Problem

michael posted more than 13 years ago | from the how-mathematicians-occupy-their-time dept.

Science 325

jeffsenter 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.

cancel ×

325 comments

Re:Better solution (1)

Caine (784) | more than 13 years ago | (#302819)

There's 50% chance that Player3's guess is wrong, so you only win 50% of the time (or less).

Re:The requirements are unclear (1)

Caine (784) | more than 13 years ago | (#302820)

It's clearly stated that they must answer at the same time.

Some clarifications to the puzzle: (5)

DG (989) | more than 13 years ago | (#302823)

It seems some people aren't reading the article very well (on Slashdot? Horrors!)

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 differently-coloured 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)

The darn things turn up everywhere, don't they? :-)

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)

> Based on my experiments with a chunk of donut
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)

The monty hall problem is a great one... you can play it here [edmonton.ab.ca] .

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)

Yes, right here: http://www.bluehat.org [bluehat.org]

(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)

Haha, funny to see how people are taking this problem entirely too seriously, and forgetting their childhoods at the same time.

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)

I think a very similar puzzle was asked and solved in an earlier Ask Slashdot, here. [slashdot.org]

Sorry, article was archived (1)

monsted (6709) | more than 13 years ago | (#302840)

You'll have to search the whole thread [slashdot.org] for Another puzzle.

The requirements are unclear (2)

pen (7191) | more than 13 years ago | (#302841)

When I read the problem at first, I assumed that all of the players had to either guess at the same time or write down their answers. I didn't think that they were allowed to answer in succession. Then again, that's one of the keys to the puzzle.

--

Re:Answer is here (2)

sandler (9145) | more than 13 years ago | (#302842)

Sounds pretty difficult to pull off without any communication.

Research? (4)

sandler (9145) | more than 13 years ago | (#302843)

The problem has even spread to the Caribbean. At a workshop at a research institute in Barbados, one hardy group of theoretical computer scientists stayed up late one rum- soaked night, playing a drinking game based on the puzzle.

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)

When I read how the 3-hat-problem goes, the solution immediately popped up in my head (a person who sees two different hats passes, one who sees two similiar hats guesses the other color).
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)

If you're ever kidnapped by aliens and forced to play this game in return for your life, figure out what the names of the two colors are, and assign "down" to the one that comes first alphabetically, such as "blue" in the standard example, and "up" to the other, "red" in the standard example.

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 80-foot satellite dish in your ass regardless of the outcome.


-

Re:Sorry, article was archived (1)

Kimble (17437) | more than 13 years ago | (#302854)

...or, if you add the CID, #61, to the end of the URL, you'll go straight to that post. [slashdot.org]

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)

If any player guesses wrong (as opposed to passing), they all lose.

--

Re:Better solution (1)

kramer (19951) | more than 13 years ago | (#302867)

The solution works quite nicely, it just doesn't take into account that the guessing is simultaneous.

Re:How to cheat (1)

Ralph Wiggam (22354) | more than 13 years ago | (#302868)

It's only called cheating if you get caught. If you don't get caught it's called winning.

-B

Re:How to cheat (1)

Ralph Wiggam (22354) | more than 13 years ago | (#302869)

It's only called cheating if you get caught. If you don't get caught it's called winning.

-B

Re:How to cheat (1)

Ralph Wiggam (22354) | more than 13 years ago | (#302870)

Doh!

Re:50% (1)

mistered (28404) | more than 13 years ago | (#302872)

No, you can get a 75% chance of winning. Read the article again - the rules say each player has to guess at the same time (or pass). It's the passing that improves the odds. Because only the player that see two hats of the same colour is guessing, they've got a greater chance of getting it right.

Re:another interesting problem (1)

mistered (28404) | more than 13 years ago | (#302873)

What's so interesting about it? In a vacuum they would fall at the same speed, but there's a lot more air resistance on the feather than the ball so the ball hits first.

Re:What I don't get about the Monty Hall Problem (1)

mistered (28404) | more than 13 years ago | (#302874)

Before opening any doors, you have a 1/3 chance of picking the prize, and there's a 2/3 chance the prize is in one of the other doors. One of those doors is opened, and it doesn't have the prize. Now there's a 2/3 chance it's in the door you didn't pick. It relies on the host's knowledge of where the prize is.

Re:Some clarifications to the puzzle: (2)

mistered (28404) | more than 13 years ago | (#302876)

Ohh, so close but you missed one point:

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)

To get the answer with more no more that one response more than there are number of hats...

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)

Darn, should have read the full article first. The problem that I initially heard allowed for communication.

My answer then changes back to an answer that I first gave a co-worker. 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)

Probability is a funny, funny thing...

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)

If you had RTFA, you would know that there is a strategy where the group has a 75% chance of success.
--
Obfuscated e-mail addresses won't stop sadistic 12-year-old ACs.

Re:75% of the time? (1)

PurpleBob (63566) | more than 13 years ago | (#302904)

Yes. The players must all answer simultaneously and are not allowed to communicate.
--
Obfuscated e-mail addresses won't stop sadistic 12-year-old ACs.

Re:This defies random odds (2)

UnknownSoldier (67820) | more than 13 years ago | (#302912)

> Probability is a funny, funny thing...

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)

They do have to answer at the same time. The reason that it works at 75% is because of probability. Here are the possible outcomes:

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)

I think your solution generalizes to any number of players.

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)

am I missing something here?

Yeah, all players have to guess simultaneously.

Re:Better solution (2)

spazimodo (97579) | more than 13 years ago | (#302949)

uh, the player 3 guessing red thing reduces your chance of success to %50. remember, they all have to guess at the same time.

-Spazimodo

Fsck the millennium, we want it now.

Re:The requirements are unclear (2)

Fjord (99230) | more than 13 years ago | (#302954)

The odd thing is that the solution I got for answering in succession is as likely as when they all answer at the same time (being 75%). I'm still now sure why my same time answer works. I'm working on that now.

Re:Answer is here (1)

mobets (101759) | more than 13 years ago | (#302957)

um... no comunication

___

Re:Hmmmm (1)

BradleyUffner (103496) | more than 13 years ago | (#302958)

you can still always win if you pass when you see 2 hats of the same color. even if they are ALL the same color. ex. I see 2 red, so i pass, next person sees 2 red, BUT they also know that i passed. Now they know that i saw 2 hats of the same color. If 2 people pass then they know that all the hats are the same color.
=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\ =\=\=\=\

Re:This defies random odds (1)

BradleyUffner (103496) | more than 13 years ago | (#302959)

you can still win 100% of the time. (click my userinfo and read my other post to see how)
=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\=\ =\=\=\

Re:This defies random odds (2)

snorb (109422) | more than 13 years ago | (#302967)

Much like the Monty Hall problem, at first this seems to defy the rules of probability, but it doesn't. The key points:
  • The group has agreed to a strategy ahead of time.
  • Only one needs to be right - the rest of the group can pass.
  • If they are all wrong, they are not penalized any more than if only one of them were wrong.
50% of the time each individual's answer is wrong (as expected) but the group is wrong only 25% of the time.

another interesting problem (1)

krokodil (110356) | more than 13 years ago | (#302968)

The Playboy has a nice article about lead ball and birds feather
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)

yep, you are missing something, thats communication and no communication is allowed. afaics they discuss strategy beforehand, they go into their separate chaning rooms where a randomly coloured hat gets placed on their head. they walk out, look at each other and then answer at the same time, the answer being red, blue or pass.

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)

this doesn;t work cos the maths is backwards.

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)

the obvious answer seems to be the lead ball but it depends on the conditions. the feather would fall slower due to air resistance overcoming it's weight and so it's terminal velocity would be slower than that of the lead ball wihch is nice and streamlined.

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)

Not sure you read the rules of the puzzle correctly. If any of the players guess incorrectly, they all lose. So, player 3 will be wrong about 50% of the time, never mind players 1 and 2.

The Red Hat problem (5)

SpanishInquisition (127269) | more than 13 years ago | (#302983)

Why does their x.0 releases always suck?
--

Read the problem more closely (1)

jshowlett (134148) | more than 13 years ago | (#302985)

The rules of the game are about the collective guesses of the team (at least one right guess and no wrong guesses), not individual guesses. And players can pass. 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 Trial: 1 2 3 4 P1 X + - - P2 X - + - P3 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.

(oops)Re:Read the problem more closely (1)

jshowlett (134148) | more than 13 years ago | (#302986)

The rules of the game are about the collective guesses of the team (at least one right guess and no wrong guesses), not individual guesses. And players can pass.

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)

If you only have one opportunity to pass or choose a color, I would say that there would be a problem. However, if they keep saying, "Come on, anybody have a guess?" if everybody passes (i.e. there are rounds of passing or guessing until somebody makes a guess) then it would go something like this for a group of seven:

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)

It goes down. The block in the boat displaces the same mass of water as its own mass. The block on the lake floor displaces the same volume of water as it own volume. Concrete is denser than water, so the sunk block displaces less water than the floating one.

Remember "Mastermind" (1)

jsin (141879) | more than 13 years ago | (#302991)

How funny, my wife just go back into the game "Mastermind" where you pick four colored pegs and the other player gets ten tries to guess the colors, with a minimal amount of feedback in the form of red and white pegs that indicate if you have chosen just a correct color (white) or correct color and position (red).

Blue hat? (3)

ritlane (147638) | more than 13 years ago | (#302994)

Wait... I'm confused...

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)

Although it's not near as complex, as a mind such as mind can't handle such things, it is kind of cool.

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)

Switching your guess does not benefit you. What am I missing here people?
Mathematically, you're correct.

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)

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.

If you see all hats the same colour, then invert the color you see and guess that.
CAN'T... HANDLE... MULTIPLE... SPELLINGS... MUST... IDENTIFY... AUTHOR'S COUNTRY... Gaaak!

Re:Play monte hall! (1)

been42 (160065) | more than 13 years ago | (#303003)

Is the boat in the lake? Does he throw the block into the lake? Did he check to make sure nobody is swimming underwater around the boat? Assuming he's in the water, throwing the block into the water, wouldn't the level stay the same? Based on my experiments with a chunk of donut, my coffee cup, and a tiny paper boat (an experiment about as scientific as the Pepsi Challenge), that's my guess.

Hrmm, lets see (1)

revelation0 (164235) | more than 13 years ago | (#303005)

How about this (I'm not positive about the additional rules of the game, so this may not be a viable answer) but here is what should happen -

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)

How come the problems I encounter at work aren't nearly this interesting! My problems are typically as complex as, "Gee, should the GUI button say OK or Okay?".

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)

Fred Durst of Limp Bizkit know about this blantant Tradmark infringement>>>

Re:75% of the time? (1)

dannywyatt (175432) | more than 13 years ago | (#303017)

am I missing something here?

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)

I've come up with what (I think) is a better solution than the one presented in the article. Feel free to poke holes in it if I'm wrong.

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)

Not sure you read the rules of the puzzle correctly.

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)

everyone take off their hats and look at them :)

Oh wait .... that defeats the question.

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)

If Player 3 always guesses red, then he will be wrong 50% of the time, and since, if one person guesses wrong, the whole team is wrong, this gives a 50% success rate.

What I don't get about the Monty Hall Problem (1)

CoreyG (208821) | more than 13 years ago | (#303045)

Perhaps somebody could enlighten me about the Monty Hall problem, aside from it being offtopic.

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 .5. Switching your guess does not benefit you. What am I missing here people?

non-JavaScript version (1)

brlewis (214632) | more than 13 years ago | (#303052)

You can also play an open-source non-JavaScript Monty Hall [webappcabaret.com] simulation and see the aggregate results for everyone who's played.

For the Bridge Players out there (1)

dave cutler (222400) | more than 13 years ago | (#303056)

In addition to its similarity to the Monte Hall problem, the 3 hat problem bears a remarkable similarity to the "restricted choice" problem in bridge http://www.rpbridge.net/4b73.htm. All are applications of conditional probability arguments.

Re:Better solution (1)

dave cutler (222400) | more than 13 years ago | (#303057)

Nonsense!

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!!!

Re:Better solution (1)

idontrtfm (235021) | more than 13 years ago | (#303067)

Wrong.

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)

First the link works perfectly. Thank god.

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)

Probability is a best guess, when you don't have all the facts. Nothing funny or strange about it all.

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)

What a silly problem. The color of the hats are determined my independant coin tosses. No communication is allowed. Other that cheating and viewing your own hat in some way, the maximum chance is 50%. Coin toss. Old statistics problem.

This is like a word problem you would get in third grade where the wording of the problem would be such that you could be tricked. I am amazed that any mathematician would waste their time analyzing it.

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)

The link given didn't require any registration when I went to read the article. I only wished they had explained how the hamming codes were used. IIRC hamming codes depend on ordering to do their thing.

Wow (1)

Evil Adrian (253301) | more than 13 years ago | (#303079)

Slow news day, huh guys? :-P
---

Re:75% of the time? (1)

ocbwilg (259828) | more than 13 years ago | (#303081)

I don't see why the article claims they can win only 75% of the time....am I missing something here?

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)

1 2 3

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)

if I'm losing money, but not losing as much as I thought I was, and do plan to make more eventually, I must tbe wearing a RedHat

Duh... (1)

Dancin_Santa (265275) | more than 13 years ago | (#303084)

More fun [greylabyrinth.com]

Dancin Santa

Re:Too simple to be true? (1)

abrett (308268) | more than 13 years ago | (#303087)

This is fairly simple to work out for 3 hats, but the article states that a similar theory applies for n hats where n is one less than a power of two(eg. 3, 7, 15 etc). I'm guessing things get less obvious when you start dealing with more hats (hence the article)...

My favourite color is.... (1)

JohnnyKnoxville (311956) | more than 13 years ago | (#303089)

red!... no wait... blue! er, no, red...

Re:Better solution (1)

Anemophilous Coward (312040) | more than 13 years ago | (#303090)

This solution would be fine...if the rules of the puzzle allowed someone to be wrong. In reading the rules you'll see: "The group shares a hypothetical $3 million prize if at least one player guesses correctly and no players guess incorrectly."

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)

you are an idiot.

'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)

Player 3 allways guesses red ? Then the probability of success is at most 50%. there must be _no_ incorrect guess. It would be not that hard to invent a 12.5% solution for 1 incorrect from 2^3 total possibilities :-)

heard about this from a friend recently (1)

JiffyPop (318506) | more than 13 years ago | (#303110)

i don't do the nyt reg, but this sounds like the problem when a row of prisoners is lined up and can see all of the people in front of them (and their hats). they can discuss a plan the night before, but they can not say anything but a color once the hats are put on. the gaurds also tell them what the possible colors are. also, everyone can hear the answers from the people behind them. even when it is expanded to n people and m colors (with mn) only one person will die with the best algorithm! it took me about ten minutes to figure out after i was told what the best possible outcome was...

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 n-2 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)

that doesn't necessarily work for ANY number (greater than 4 or otherwise). ALL OF THE HATS CAN BE THE SAME COLOR!!! think about it. you want to garauntee that people get it correct, not have a certain probability of it...

Re:This defies random odds (1)

JiffyPop (318506) | more than 13 years ago | (#303112)

first the obligatory IANAMM (i am not a math major).

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 n-k 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 k-1 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)

That solution is against the "no communication rule". No signals allowed.

Hmmmm (5)

BillyGoatThree (324006) | more than 13 years ago | (#303115)

"Three-fourths of the time, two of the players will have hats of the same color and the third player's hat will be the opposite color. The group can win every time this happens by using the following strategy: Once the game starts, each player looks at the other two players' hats. If the two hats are different colors, he passes. If they are the same color, the player guesses his own hat is the opposite color."

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)

Yeah, but it would really suck if the aliens read this post too......

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 np-complete 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)

say that again

Re:50% (1)

lbarbato (410651) | more than 13 years ago | (#303127)

That is true if you had no other information. However, you do have other information in the form of knowing the odds of a particular set of hats being chosen, and that is what is being used here.

75% of the time? (1)

plluke (412415) | more than 13 years ago | (#303133)

What if the strategy were as follows: Each person goes into the room and decides to wait 30 seconds before saying anything. Then whoever sees that the other two people are wearing the same colored hats, they will say "pass". If two of them say pass, the third one will know the color of his own hat. This covers the case of three hats being the same color. If the hats are dispersed in a 1-2 relationship, then that one person will say pass...nobody else will and the next two people can guess by looking at each other. This seems like a pretty obvious solution to get a 100% chance of winning and is basically the first thing that popped into my head...I don't see why the article claims they can win only 75% of the time....am I missing something here?

So what is the strategy for larger teams? (2)

Hilary Rosen (415151) | more than 13 years ago | (#303137)

If there are seven players, what is my strategy? Do I guess Red if I see 4 blue hats? If I have to see 6 blue hats then we will pass almost 90% of the time. And why isn't this under the Red Hat topic?
--
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