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The Three Hat Problem 325

Posted by michael
from the how-mathematicians-occupy-their-time dept.
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.
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The Three Hat Problem

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  • by Anonymous Coward
    The real power of math is abstraction. That's why we all have to deal with it.
  • by Anonymous Coward
    You are a fucking idiot. Read the puzzle again. Simultaneous answers. NO communication. Now go fucking crawl back into your hole and die.
  • There's 50% chance that Player3's guess is wrong, so you only win 50% of the time (or less).
  • It's clearly stated that they must answer at the same time.
  • by DG (989)
    Oops, how embarrasing. :(

    Thanks for catching that.

  • A decent bit of logical deduction, my dear Watson.

    Of course, you _could_ have just followed the link to the website in my identity block - but then that would have been cheating. ;)

  • by DG (989) on Tuesday April 10, 2001 @06:32AM (#302823) Homepage Journal
    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.

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

    It's always great to see theoretical work evolve from recreational activities.
  • I beleive the optimal solution to the 5hat problem yields a win 22/32 times (or rather 11/16).

    Devide the 32 possible combinations into 3 groups:
    1) All hats same (2 possible)
    2) 4/1 split (10 possible)
    3) 3/2 split (20 possible)

    Now, to get group 3 right look at the other 4. If same number of red/blue then pass. If there is a majority guess the oposite.

    This will make you get all the situations in group 2 wrong. It is possible to salvage the last two in group one though. If all you see is the same, guess what you see.

    Are there any better solutions?

    -Harry

    PS. I beleive that for any group where there are are a power of 2 members (1,2,4,8,etc) it is impossible to get above 50%. Can anyone confirm?

  • Guestimate how long it will take me to do a task, based on past performance and how much was budgeted.
    Arithmetic (Multiplication, Division, Simple Estimation)

    Made an estimate of what sprinkler I needed for my lawn and type of grass, taking into account water coverage and projected rainfall.
    Arithmetic (Subtraction, Division)

    Determined what operations to make to get a bit structure in the format I desired, over a system with endian differences (using Fortran - yuck!)
    Boolean Algebra

    For fun, calculated how many managers it would take to run a company, based on each manager having a maximum of 4 people under them (still working on those formulas - I leave it as an excerise for the reader to determine why I decided on 4, or you could go join the Discordians and find out for yourself).
    Sumnation Math (1st sem. calculus)

    Calculated my wife's reading rate, to determine when she would be done with the Harry Potter book
    Arithmetic (Multiplication, Division)

    Cut down a recipe for 10 to a recipe for 2 1/2.
    Arithmetic (simple division).

    Determined whether it would be a better idea to make an extra car payment, house payment, get a CD, or invest in a mutual fund
    Arithmetic (Simple Comparisons)

    Tried to figure out your age, based on how little math you have had (19? 20?)
    No math just a flame.

    It's been a slow math week, too. In my job (systems programmer), I've used logic, binary arithmetic, calculus, trig, geometry, statistics, and other flavors of math.
    I see no trig. No geometry. No statistics. All most all is just arithmetic. And all but your just for fun problem was high school math.

    I'm gonna have to agree for the first guy dude. For 90% of the people out there (prolly more really) a solid understanding of High School level math is all it really takes.

    -Harry
  • Interesting. I'm not convinced that this is completely true though. Well, it is in the "real" world, but not in the "math" world that this problem lives in.

    In the real world it would be trivial to ignore player 4, and have 1-3 operate just like in the 3hat game.

    In the math world this can't really be done. There is no way to decide which player to ignore.

    Basically the way I'm looking at the problem (and the way I think the problem is supposed to be looked at) the only information you have to base a decision on is the number of red/blue on everyone else's head. You can't differentiate between them. Am I making sense? I hope so.

    -Harry
  • Or am I missing something? I seem to have a strategy that will work for all odd-numbers-of-players games, with a failure only when all the hats are the same color. I assume that when the article talked about the optimal strategy being unknown it was referring to the case of an even number of players.

    The strategy is as follows:
    Look at the other hats, and see whether there's one color that's a minority (if you see a tie, keep your mouth shut; i.e. always pass). Then, if you see m hats showing the minorty, pass until round m+1 at which point you guess that your hat has the minority color.

    An example with nine players. Say players p1 through p6 have R hats, and p7, p8 and p9 have B hats:

    Walking in, players p1 through p6 each see five R hats and three B hats, and so decide that if the game lasts to round four, they will guess "B".

    Players p7, p8, and p9 each see six R hats and two B hats, and so decide to guess "B" in round three, if the game gets that far.

    Then, everyone passes for rounds one and two. In round three, the three players with B hats all guess "B", and the game is over (with the team having won).

    This strategy of course results in losing the game if all the hats are the same color, or if there are an even number of players and the number of R hats equals the number of B hats.
  • > What am I missing here people?

    The basic idea you're missing is that when you made your initial pick you chose from a different set of doors than you're presented with when you get to do your switch-or-stay choice.

    First of all, clear your mind of the notion that simply because there are two options, both of them must be equally likely to lead to reward. The real world rarely works like that, and in fact you should never assume that even theoretical problems work that way unless its so stated. As a trivial example, suppose that we play a game in which we walk up to a random person on the street and ask them for their signature. Would you like to win when they sign with their right hand or with their left?

    So what is stated: the prize is behind a randomly chosen door, and before the game is equally likely to be behind any one of the three. The host chooses randomly when he can. You are in control of your own strategy.

    Now, let's take a diversion and play a game similar to Monty Hall, but with one important difference: the host chooses one of the two doors regardless of what may be behind it and just places a big X on the door marking it off-limits. You can then switch or not. Now, then, the universes play out this way:

    case A: (1/3 chance) You choose initially the correct door: no matter which door the host Xes out, you lose if you switch and win if you stay.
    case B: (1/3 chance) You initially choose incorrectly and the host Xes out the door with the prize. No matter what you do, you lose. Sometimes life is like that.
    case C: (1/3 chance) You initially choose incorrectly and the host Xes out the other empty door. You win if you switch, but lose if you stay.

    It is important to note that the probabilities for the different cases are NOT computed by simply using 1/3 since we have three cases, but rather from the statement of the problem: we _know_ that the location of the prize is distributed uniformly, and therefore have a 1/3 chance of guessing correctly initially (case A). In the 2/3 of the time that we guess incorrectly initially, half of the time the host will X out the prize door (so 2/3 * 1/2 == 1/3 chance for case B) and half the time the host will X out another empty door (2/3 * 1/2 == 1/3 chance for case C)

    So we see that the strategy analysis works out like this: if you switch, you win 1/3 of the time and lose 2/3 of the time. If you stay, same odds. Therefore, there's no advantage to either strategy.

    Now, how does this differ from the regular Monty Hall game? The difference is that the host doesn't choose randomly, and only Xes out a door without the prize. This then redistributes the probabilities between cases B and C. Specifically, case B never happens, so we have case A when we guess correctly initially (1/3 chance), and case C when we guess incorrectly initially.

    It is important to note that the reason changing the rules and eliminating case B doesn't cause the probability of case A to go up is that the probability of case A doesn't depend at all on the host's action.

    For example, let's go back to the evil game above and assume that our host doesn't like us much at all and therefore if the host can X out the door with the prize on it, he will do so 2/3 of the time. Then, the distributions become:

    case A: (1/3 chance) win on stay.
    case B: (2/3 * 2/3 == 4/9 chance) always lose.
    case C: (2/3 * 1/3 == 1/9 chance) win on switch.

    Then, we have that if we stay we will win 1/3 of the time, whereas if we switch we will win only 1/9 of the time.

    It is useful to work with other Monty Hall variants (for example, if the host prefers to open the left-most door you didn't choose when given a choice, or a game where the host may choose to open the door which you initially chose, bumping your choice to the next door) and to then check one's reasoning with a math prof. or computer simulation. Going through exercises like this can really help one appreciate some of probability's finer points.
  • The vast number of comments posted to discuss various solutions to and aspects of this puzzle shows that mathematical reasoning is useful, or at least interesting, if only for mental recreation and exercise.

    The mathematical reasoning I mention is not to be confused with school classes labelled ``math''. As you have found out, they are not very interesting.

    I can easily complain about English classes like you do about Mathematics classes. Whereas you ask why we are taught angle bisection and not cheque book balancing, I can equally ask why we are made to know Shakespear and not practice on writing résumés.

    Some people argue that literature is intellectually interesting and important to an advance civilization, whether it is directly useful or not. The same can be said about mathematics. Provided it is taught and learned right. The problem is that mathematics is not conveyed right by schools.

    Please give mathematics a second consideration. Do not look at it as something that should be used -- any area is useful only to a small niché. Look at it as a possible pastime, an interesting art, an interesting science, a way of thinking (the insistence on precision and proofs is certainly a viable philosophy), and a framework for solving puzzles.

  • Persons A,B,C...

    Person A looks and if B is Red and C is Blue,
    then Person A flips a coin and says red or blue.

    Next, person B looks. If C is Blue, person B guesses Blue.

    Now, person C guesses red.

    If it gets by person A, then there are only 3 possibilities for B&C. By getting by B, there's only one left.

    There's a 1/4 chance that A will have to guess, so a 1/8 chance that he will guess and guess wrong. If B or C guess, they'll be right.

    So, you can do it with only a 1/8 chance of doing it wrong.

    I know this doesn't follow the "everybody guess at the same time rule," but I didn't see that in the NYT article...
  • If even one of them has read this post, you're in good shape. If they both have, you're free.

    Great! What if the aliens read this post? Now they will make us wear masks.
    __
  • Does somebody know where "Red hat" got their name from?
    __
  • It sounds like you were crunched for time as a kid. There was plenty of time when I was a little kid to read Seuss and other more educational books. My parents never had to introduce me to books, I just picked up what I wanted too. I think most other people had the choice as kids to read many things.
  • > 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.

    --
  • 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?

    ---

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

    (It was extremely tempting to turn that into a goatlink, but I didn't :) )

  • by KlomDark (6370) on Tuesday April 10, 2001 @06:10AM (#302838) Homepage Journal
    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!

  • I think a very similar puzzle was asked and solved in an earlier Ask Slashdot, here. [slashdot.org]
  • You'll have to search the whole thread [slashdot.org] for Another puzzle.
  • 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.

    --

  • Sounds pretty difficult to pull off without any communication.
  • by sandler (9145) on Tuesday April 10, 2001 @05:41AM (#302843) Homepage
    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??

  • So other than "the people who developed computer microchips", and developers (who have to analyse algorithms), and network admin (who have to optimize networks), and people who build houses (who make estimates based on simple plans), and buyers for large companies who make purchases based on economic indicators and polls, and may be chemists, yah, you're right.

    You see a bad buyer might use someone elses formula, but a better buyer will understand that model and adjusted it to better suit her particular environment.

    We don't normally train folks to be below average, therefore everyone (named above, and possible just a few more) needs to be trained in math, not just counting.

    Joe
  • 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.
  • As author of the original post, I'd like to give you the recognition you deserve here, because the moderation system will surely miss it:

    In my opinion, your response is funnier than my original post.

    -
  • by Syberghost (10557) <syberghost&syberghost,com> on Tuesday April 10, 2001 @05:38AM (#302847) Homepage
    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.


    -
  • I'm sorry, was I supposed to be impressed?
  • Correctly observed. The first person to guess will always have atleast a 25% chances to fail. If he is going to relay enough information to the other guessers.

    The simultaneous guess solution is perfect since it overlaps the wrong guesses while spreading out the correct guesses.

    Schematics (W=Wrong guess, C = Correct guess, - = pass):

    r r r W W W
    b b b W W W
    r b b C - -
    b r r C - -
    b r b - C -
    b b r - - C
    r r b - - C
    r b r - C -

  • Actually, no. This doesn't work since with this algorithm you will have to do a guess when you see hats with different color (you can't pass since that would mean they were the same color according to your reasoning). This guess would make 2/8 = 25% wrong directly.
  • Everyone: If everyone has the same color guess that you have the opposite color.

    Everyone but p1: If p1 has the opposite color of the rest guess that you have the same color as p1.
    -----
    Extend rules for higher order solutions. Following is a schematic of how I reached the above rules.

    r r r r W W W W
    r r b b - C - -
    r b r r - C - -
    r b b b C W W W
    r b r b - - C -
    r b b r - - - C
    r r r b - - - C
    r r b r - - C -
    b r r r C W W W
    b r b b - C - -
    b b r r - C - -
    b b b b W W W W
    b b r b - - C -
    b b b r - - - C
    b r r b - - - C
    b r b r - - C -
  • Oops, you asked for the 7 player solution. Ok, I am not 100% sure but it should be like this:

    Everyone: If everyone has the same color guess that you have the opposite color.

    Everyone but p(x): If p(x) has the opposite color of the rest guess that you have the same color as p(x).

  • It says they pass or guess simultaneously, and that they win if someone guesses correctly without anyone else guessing incorrectly.

    So the condition for winning does *not* have to occur. If everyone passed, they'd lose.

  • ...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

  • There's some nice psychologie going on here. It's similar to this 30$, three guys bla thing.
    Our mind seems to prefer "shortcuts" based on similar numbers.
    The Monty Hall experiement can made clear to everyone changing just quantities.
    Let's say you have 1000 doors, with 999 false choices and one winner.
    You choose one and then the host shows 998 false doors. Now it's easy to see that indeed changing choice is the better decision and everyone whom I explained it this way immediatly got it.
    It's just that 2 is so damn near to 3 what confuses people.

  • Thus, if you see 2 hats of the same colour, the probability that your hat is the same color is 1/6

    No, it's 1/2.

    --

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

    --

  • Okay, look. Here are the eight possible outcomes:

    000
    001
    010
    011
    100
    101
    110
    111

    Let's give the people names:
    ABC
    DEF
    GHI
    JKL
    MNO
    PQR
    STU
    VWX

    Now, we're only talking people who see two hats of the same color. That leaves
    ABC
    F
    H
    J
    M
    Q
    U
    VWX

    There are 12 people here. Six of them are wearing a hat that's the same color as the ones they see. Six are not.

    1/2.

    QED

    --

  • I don't disagree with that. Note that the post that started this all was:

    Thus, if you see 2 hats of the same colour, the probability that your hat is the same color is 1/6

    I (correctly) pointed out that the probability was actually 1/2. That's what the previous few posts have been in reference to.

    --

  • You're completely wrong buddy.

    Here's my explanation of the Monty Hall problem that so many of you think is wrong.

    When you initially choose a door, probability of getting a prize is 1/3 (3 doors, one prize, hence 1 in 3 chance). The chance of picking a goat is 2/3.

    After a host opens one of the two doors, he does so with FULL knowledge! He will not open a prize door by tossing a coin, he will always open a door with a goat. This is EXTREMELY important clue to the puzzle!

    Now, after he open a door with a goat, he does so with full knowledge AND the door that he opens is one of the TWO doors that you didn't choose (i.e. he will not open a door that you chose). Once he does that, things change BIG time.

    The probability of your current door having the prize is still 1/3! Nothing changed since you picked the door among three other doors. The door that's left (DoorL), however, now has a chance of 2/3 of being the door with a prize! Why? Well... initially it (DoorL) had a chance of being the prize with P=1/3, the door that was opened (DoorG) had a chance of P=1/3 but now the chance of the DoorG having the prize is P=0! So, the chance of DoorL having the prize now must be P=2/3.

    Q.E.D.
  • When you initially choose a door, probability of getting a prize is 1/3 (3 doors, one prize, hence 1 in 3 chance). The chance of picking a goat is 2/3.

    After a host opens one of the two doors, he does so with FULL knowledge! He will not open a prize door by tossing a coin, he will always open a door with a goat. This is EXTREMELY important clue to the puzzle!

    Now, after he open a door with a goat, he does so with full knowledge AND the door that he opens is one of the TWO doors that you didn't choose (i.e. he will not open a door that you chose). Once he does that, things change BIG time.

    The probability of your current door having the prize is still 1/3! Nothing changed since you picked the door among three other doors. The door that's left (DoorL), however, now has a chance of 2/3 of being the door with a prize! Why? Well... initially it (DoorL) had a chance of being the prize with P=1/3, the door that was opened (DoorG) had a chance of P=1/3 but now the chance of the DoorG having the prize is P=0! So, the chance of DoorL having the prize now must be P=2/3.

    Q.E.D.
  • Heh... what's the point of him opening a door with a prize? That'd be a pretty stupid show IMHO.
  • If the rules are the players must guess simultaneously, but not necessarily at a specific time, then we can bend the rules slightly and win 87.5% of the time with 3 players.

    Here's my strategy.

    When the 3 hats are distributed, one of the players yells "NOW". then, each player that can see 2 RED hats guesses BLUE. Otherwise all players pass. If all hats are RED, we lose at this point, otherwise we will eventually win.

    Then, if all players pass, we have new information. That information is: "There is no more than 1 RED Hat". So, we co-ordinate another guess. Someone yells "NOW" again. Then, everyone that can see one RED hat correctly guesses "BLUE".

    If all players pass again, new information is obtained: "There are NO red hats". Then, after someone co-ordinates the guesses again, all players can guess "BLUE" and be right.

    The generalised solution has a failure rate of 1/2^N, where N is the number of players. For 15 players, the chance of failure with this strategy is 1/32768, far better than the 1/16 for the simpler strategy in the article.

    This problem is similar to the "Three Philosophers" problem that was mentioned in Scientific American some years ago. In this one, three philosophers are aleep under a tree, and a bird soils the foreheads of all of them. When they wake, they all start laughing at each other. Then one suddenly stops laughing. Why?

    --
  • You don't use any advanced mathematics? Within 1 year of graduating, I had to apply Newton's Method to a problem I was dealing with. Half of all SQL I do comes from principles I learned in Set Theory. And I won't even go into the analytical thinking gained through the mathematics courses I did.

    Grow up, please

  • The problem was "How do you swap two numbers without an additional variable?"

    a and b being the same variable isn't in the problem domain.

    Simple nuff.

  • Solution in Pseudo C

    a := a + b;

    b := a - b;

    a := a - b;
  • The solution works quite nicely, it just doesn't take into account that the guessing is simultaneous.
  • 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.

  • You may be right, but in all honesty...

    I have always hated Dr. Seuss...

    I'll never forget the way my teachers in grade school (around 1st-2nd grade) treated me: They wanted me to read the stupid Dr. Seuss books, and other kid books - rather than let me read the ones I could see I wanted to read - Science Experiment books, Alfred Morgan electricity books, books on space and technology, computers, etc. - Telling me those were for the "older" children, only (like, WTF? I might learn something? What the hell am I going to learn about "Green eggs and ham, Sam I am"? How to read? I already knew how to do that, unlike my lamer peers!)...

    Thank the gods my parents had some sense, and bought me both a science and a regular encyclopedia for me before I turned 6 years old...

    That, and plenty of Lego...

    Worldcom [worldcom.com] - Generation Duh!
  • Not crunched for time at all...

    I could just see what utterly useless pap Dr. Seuss was. The first book I remember being "introduced" to was a science encyclopedia my parents got in the mail - they knew I liked building toys and such, so they showed it to me, and I read that thing nearly cover-to-cover, so they got me more like it (always asking my input on whether it was something I would like to read). Later they got a set of encyclopedias (Brittanica - not a cheap set to get), mainly for school work. Any other kinds of book wanted to read, I could pick out from the bookstore or elsewhere.

    Don't get me wrong - my reading interests weren't all reference, nor did I read only "adult" oriented books - I read my share of children's novels. I just enjoyed the kind of books that had more than 10-15 pages, and more words than pictures. For those books that had pictures, I vastly preferred pictures that at least looked like a real setting, rather than a complete candy-coated strange-ass drug-induced fantasy land (and if you look at Dr. Seuss in this light as an adult, perhaps that is why some adults are fascinated by his work...perhaps).

    Worldcom [worldcom.com] - Generation Duh!
  • Sue me - I suck at spelling - I think it is rather funny that you point this out - I have had the sig running for quite a long time, yet you are the first to point it out.

    I will change it immediately, thanks for the update...

    Worldcom [worldcom.com] - Generation Duh!
  • by cr0sh (43134)
    I just didn't like Dr. Seuss books...

    Worldcom [worldcom.com] - Generation Duh!
  • The first thing I drew in kindergarten was a picture of a UFO. What is strange to me about this is that I don't have any recollection of seeing any images of "UFOs" before this, either on TV or in a book or somesuch...

    Worldcom [worldcom.com] - Generation Duh!
  • It's easy. Just become a "theoretical" computer scientist.

    I'm buying my pseudo-ticket today!
  • I'm neither a total moron NOR a troll.

    For the people who developed computer microchips, then the advanced math is VERY relevent. However, unless you find yourself under that particular umbrella, things like advanced calculus are surprisingly abstract. I'll even go so far as to say that grade 12 is getting a little .. iffy. I'm not ungrateful. I just speak my mind.

    yeah, hate what you don't understand ...
    And, for the record, I have always gotten approximately 95% in ALL of my math courses, so I DO understand the material.

    ------------
    CitizenC
  • by norton_I (64015) <hobbes@utrek.dhs.org> on Tuesday April 10, 2001 @07:07AM (#302910)
    They talk about this at the bottom of the article. It would appear to break the laws of random odds, since the other players hats give you no information about your own. And in fact, if you look carefully, you only guess right 50% of the time. The trick is, when one person guesses wrong, all three do, yet it only counts as one failed trial.

    Of the 8 possible hat combinations, 6 of them will have exactly one person answer correctly, and the other 2 will have all 3 people answer incorrectly.
  • > 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.
  • 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.
  • Of course my face, although not my hat, is red. The word simultaneously does appear in the rules to describe how the guesses must happen. This solution only works if the guesses are sequential. It is interesting only in that it demonstrates that as long as the only communication is a choice of guessing or passing there will always be at least one losing case.
  • Grashopper, calculus is only the beginning to true enlightenment, and will only reach the first layer of abstraction with it. You need much more mathematics, at the university level, to understand its importance. Mathematics will be critical in engineering, physics, finance (no, finance isn't just addition :-P), etc. You just showed your ignorance, sir.
    --
  • by e271828 (89234) on Tuesday April 10, 2001 @12:49PM (#302927)
    Nobody seems to have posted this yet, and since I've wasted a few hours on this already, I figured I should share. Here's the solution for the general case where the number of players is n=2^k-1; it involves Hamming codes.

    The general case is perhaps best illustrated by example for the case of 7 players. Here's how it goes:

    First, number the players 1 through 7. Now think of a player with a red hat as being assigned a binary zero, and a player with a blue hat as being assigned a binary 1. Then an assignment of hats to the players can be associated with a length 7 binary sequence: As an example, if all the players have red hats, this correspond to the sequence 0000000 (seven zeros), and if all but the last player have red hats, we get the sequence 0000001. We'll call length 7 binary sequences words; every hat assignment has an associated word, and vice versa. For future reference, denote the two words in the example above as c0 and c1 respectively.

    A brief digression on Hamming codes: A length 7 Hamming code is a certain (carefully chosen) subset of the 2^7=128 possible words. This subset consists of 16 words, called codewords. We need just one fact about Hamming codes to describe the players' strategy: [Fact 1] If any one bit in a codeword is flipped, the resulting word is not a codeword. For example: c0, the all zero word, is always a codeword in a Hamming code; c1, which differs from c0 in just one bit, cannot be a codeword.

    Suppose I am one of the players. Here's my strategy:

    By observing the hats of the other 6 players, I construct two words. The first, w0, is the word that would be associated with our hat assignments if my hat is red. The second, w1, is the word that would be associated with our hat assignments if my hat is blue. We have 3 possible cases:

    1. Neither w0 nor w1 is a codeword
    2. w0 is a codeword
    3. w1 is a codeword
    Cases 2 and 3 are mutually exclusive because of Fact 1. My strategy?
    In case 1, I say nothing, in case 2, I declare my hat to be blue, and in case 3, I declare my hat to be red.
    That's it. This strategy will result in the team winning every time the word associated with the hat assignment is not a codeword, which happens (128-16)/128=7/8 of the time.

    So, why does this work? To understand, we need one more property of Hamming codes: [Fact 2] For every word that is not a codeword, there is a unique bit, which if flipped, will produce a codeword. For instance, if the last bit in c1 is flipped, we get c0, the all zero codeword. Now suppose that the hat assignment is associated with a word that's not a codeword (which, as shown above, happens 7/8 of the time). To be specific, suppose the assignment corresponds to c1. Here, the 7th bit is the unique bit that can be flipped to get a codeword. Then, by Fact 2, players 1 through 6, after constructing their w0 and w1 words, will find themselves in Case 1 : neither word is a codeword. They'll keep quiet, while the 7th player will declare his hat to be blue (Case 2). Success!

    Of course, if the hat assignment corresponds to a codeword, then every player will speak up, and they'll all guess wrong! Example again: if every player is assigned a red hat (corresponding to c0), they'll all say they have blue hats.

    It's not hard to see how this generalizes for an arbitrary number of players n, where n=2^k-1. The players will succeed if and only if the associated word is not a codeword. A Hamming code of length n=2^k-1 has 2^(n-k) codewords, so success is achieved 1-2^-k of the time, which tends to 1 as k tends to infinity.

    There, I knew those grad courses in coding theory would come in useful sometime!

  • by frankie (91710) on Tuesday April 10, 2001 @09:24AM (#302929) Journal

    This puzzle comes from a paper by James Aspnes, Richard Beigel, Merrick Furst, and Steven Rudich [google.com]. Attribution: I got this information from my friends on Mattababy [mattababy.org].

    Date: Tue, 10 Apr 2001 10:58:43 -0400
    From: Steven Rudich
    Subject: New Yorkl Times fails to credit me.

    Today, the New York Times had a lively article on a brilliant puzzle going around mathematical circles.

    Just before doing my post-doc, Richard Beigel and I formulated formulated our voting puzzles. Subsequently, our puzzles were developed into a now classic paper in theoretical computer science (with M. Furst and J. Aspnes). There paper can be found at www.rudich.net/papers/voting.ps . My initial solution to the first voting puzzle was a coding theory solution that is strictly identical to the much touted solution in the NYT article. When giving talks on the subject, I mentioned that if my voters had the ability to abstain, that my coding solution was still optimal and that the voters would make no errors in a most situations. This is strictly identical to the hat puzzle in the NYT article. My talks were very popular and many of the people in the article attended these talks. Subsequently, Burt Enderton gave a simple solution to my puzzle that did not have the abstention interpretation and hence is not a solution to the hat puzzle. I liked Burt's solution and put it in our paper.

    In the last few months I have reminded people that *all* the proofs and connections to coding theory were developed by me. Nonetheless, these same people ignored me in the Times article.

    I have had many ideas stolen, but never featured in vivid detail in the Times.

    Steven
  • How about this solution which works 100 percent of the time? I don't see why this isn't a solution.

    It isn't a matter of who is right. The challenge is making sure no one is wrong. Reread the challenge.
  • 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.
  • 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.
  • The spirit of the puzzle means that you don't know a partner passing versus them guessing. I know the words don't say that exactly, but that's the boiled-down spirit of it. You could say they write the word pass on a piece of paper instead of writing the color guess.

  • A "win" is considered to be when the money is split among the players, not just a correct guess.

    Again, it's a team game, if any money is going to be shared. Because information is not perfect, wrong guesses must be made for the game to be played at all, it is your responsibility to guess incorrectly if the circumstances are correct, or else you will not be helping to consolidate all the wrong guesses into a concentrated play.

  • 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.
  • by SpanishInquisition (127269) on Tuesday April 10, 2001 @05:33AM (#302983) Homepage Journal
    Why does their x.0 releases always suck?
    --
  • 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.

  • 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.
  • by ritlane (147638) on Tuesday April 10, 2001 @05:38AM (#302994) Homepage
    Wait... I'm confused...

    There is something other than Red Hat?



    ---Lane [rit.edu]
  • 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.
  • 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!
  • "It's 0.75."

    The cases in which you see two hats of the same color are RR-R, RR-B, BB-R, and BB-B, where the last color in each case is your own hat. These cases have equal probability of occurring. The frequency with which your hat is the same color is 1/2 of these equally probable cases, and so the probability is 1/2.

    So, the answer to the question "If you see two hats of the same color, what is the probability your hat is the same color?" is 1/2.

    However, the answer to the question "If (at least) two hats are the same color, what is the probability all hats are the same color?", the probability is 1/4.

    This is because the equally probable cases in which two hats are the same color (as is always the case of course) are RRR, RRB, RBR, RBB, BRR, BRB, BBR, and BBB. Of those, the proportion that have all three the same color is 1/4, and hence the probability is 1/4.

  • Your comments are consistent with a misintepretation of the question. Specifically, your comments appear consistent with an attempt to demonstrate how the algorithm described in the New York Times article yields a 75% probability of success. However, the question addressed here has been a different problem. The question in this thread is, if you see two hats of the same color, what is the probability your hat is the same color?

    That probability is indeed 1/2.

    Perhaps you would like to reconsider your assertions that others were in error.

  • You three know that if the two hats you can see are of the same color, chances are that your hat is of a different color.

    This is a very old probabilistic fallacy. Since all the hats were chosen independently, the probability that your hat is a different colour from the other two is still 50%.

    The apparent contradiction is resolved in this problem because you don't always actually make a guess, and a "win" is based on three answers, not one. Take a look at my other [slashdot.org] post.

    Also, I was saying there is no way to guarantee a win - in other words, have zero probability of failure.

    --

  • Am I incorrect or is this based upon the supposition that by looking at the other draws you can base conclusions on your own

    You would be correct if a single person were being asked to guess either red or blue for his own hat.

    But, this is not the case. Three people are being asked to give simultaneously one of three answers: red, blue, or pass. Each person can see the hats of the other two, and finally, they have an agreed strategy for guessing.

    The best way to think of it (IMHO) is this: the guessing strategy is completely mechanical, so it's fair to think of it as part of the randomized process, rather than the response to one.

    --

  • Someone suggested that if the players are allowed to give their answers in sequence, and subsequent players can know the others' responses, then you can guarantee a win.

    However, this is clearly untrue, because of the following reasoning:

    The first person to guess has access only to the colours of the other two hats. This gives no information about the colour of his own, since they were chosen independently. Any concrete guess will have a 50% chance of being correct. Thus, in order to guarantee a win, the first person must pass.

    Since the first person must pass regardless of what colours the other two have, the second person has no new information about his own hat. He, too, must pass.

    By the same reasoning, the third person also has no new information, and is therefore forced to guess, and incurs the wrath of the other two, 50% of the time.

    There is no way to guarantee a win except by allowing the players to collaborate in some way.

    --

  • 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.

  • I'm gonna have to agree for the first guy dude. For 90% of the people out there (prolly more really) a solid understanding of High School level math is all it really takes.
    I agree, although I think that opportunities to use the likes of trig, geometry, statistics and calculus are there but people don't have the confidence in their knowledge and/or are too lazy to do it.
  • by JWhitlock (201845) <John-Whitlock AT ieee DOT org> on Tuesday April 10, 2001 @07:46AM (#303037)
    It's true, not everyone needs math. In fact, if most people don't learn math, then folks like me get paid more for our math knowledge.

    I used math over the last seven days to:

    Guestimate how long it will take me to do a task, based on past performance and how much was budgeted.

    Made an estimate of what sprinkler I needed for my lawn and type of grass, taking into account water coverage and projected rainfall.

    Determined what operations to make to get a bit structure in the format I desired, over a system with endian differences (using Fortran - yuck!)

    For fun, calculated how many managers it would take to run a company, based on each manager having a maximum of 4 people under them (still working on those formulas - I leave it as an excerise for the reader to determine why I decided on 4, or you could go join the Discordians and find out for yourself).

    Calculated my wife's reading rate, to determine when she would be done with the Harry Potter book

    Cut down a recipe for 10 to a recipe for 2 1/2.

    Determined whether it would be a better idea to make an extra car payment, house payment, get a CD, or invest in a mutual fund

    Tried to figure out your age, based on how little math you have had (19? 20?)

    It's been a slow math week, too. In my job (systems programmer), I've used logic, binary arithmetic, calculus, trig, geometry, statistics, and other flavors of math.

    Understanding the materials is a long way from realizing practical applications. Even these folks who do math for a living don't comprehend possible applications yet, but it's been a safe bet that today's math theory is tommorrow's application.

  • Also, because the boat becomes lighter, it displaces less, so the level goes down.
  • One person is set as the check bit. He is the first to answer. He answers pass in under 10 seconds if there is an even numnber of red hats over 10 if there is an odd number of red hats. Then each person from then on out counts the red hats and and if it is suposed to be even, and he counts odd then he is a red hat, if it is suposed to be even and he counts even then he is a blue hat.
  • 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? =)

  • by typical geek (261980) on Tuesday April 10, 2001 @05:32AM (#303083) Homepage
    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
  • So, is it possible to increase your odds or not? I wondered this, so I created a Perl script that chooses three hat colors at random. It then applies the technique mentioned in the article. The result? And I quote:

    Out of 100000 tests:
    75093 (75.093%) resulted in success.
    24907 (24.907%) resulted in failures.

    Seems pretty obvious to me.

  • This made me curious for the 7 hats case. According to the article, since the number of hats is a power of two minus one (2**3 - 1), they claim there is a strategy that is succesful for 7 out of 8 times played. Ok, let's have a look.

    We have 7 hats. The following combinations can occur.

    rrrrrrr
    rrrrrrb
    rrrrrbb
    rrrrbbb
    rrrbbbb
    rrbbbbb
    rbbbbbb
    bbbbbbb

    Yep, I consider the order of the hats irrelevant. The article does not reveal if this assumption is correct; from what I found for this case I guess order is relevant. See my remarks at the end.

    Let X be a hat wearer.

    Case 1: X sees only 1 color of hats (6r or 6b)
    Case 2: X sees a distribution of 5/1 (5r/1b or 5b/1r)
    Case 3: X sees a distribution of 4/2 (4r/2b or 4b/2r)
    Case 4: X sees a distribution of 3/3 (3r/3b)

    Now, there are not many strategies to choose from, really. For either case, one can choose guess I'm red, guess I'm blue or pass. You could easily check them one by one and see which strategy works best. For instance, I tried (and I strongly believe this is optimal):

    Case 1: if 6r, choose b, if 6b, choose r
    Case 2: pass
    Case 3: if 4r, choose b, if 4b, choose r
    Case 4: pass

    This strategy makes sure everyone guesses wrong for the two cases that have only one permutation (ie. 7b/0r and 7r/0b, Case 1 applies), which is the best achievable result. Because consider: we did 14 wrong guesses, but lost in only 2 out of 128 cases (there are 2**7 = 128 permutations for 7 hats). Knowing that on the average, 50% of our guesses will be wrong, ditching 14/128 of them while loosing only twice is a good start.

    At the same time, this strategy makes sure that for 4r/3b and 3r/4b, we win. Since 4r/3b renders the most permutations (35 possible permutations for each), we guess right for 70 out of 128 permutations.

    A closer look on this: (Cases repeated from above for convenience)

    Case 1: X sees 6r/0b => b; or 6b/0r => r
    Case 2: X sees 5r/1b or 5b/1r; pass
    Case 3: X sees 4r/2b => b; or 4b/2r => r
    Case 4: X sees 3r/3b; pass

    rrrrrrr (7r/0b)
    => Case 1; everyone guessses b, all lose
    ====> (7 0) = 1 combination exists for this case

    rrrrrrb (6r/1b)
    => for the r's, Case 2 applies (5r/1b) so they pass
    => for the b, Case 1 applies, so it guesses b (and is right)
    ====> (7 1) = 7 combinations exist for this case

    rrrrrbb (5r/2b)
    => for the r's, Case 3 applies (4r seen), they choose r (wrong)
    => for the b's, Case 2 applies, they pass (irrelevant)
    ====> (7 2) = 21 combinations exist for this case (ouch)

    rrrrbbb (4r/3b)
    => for the r's, Case 4 applies, they pass
    => for the b's, Case 3 applies, they choose r (right)
    ====> (7 3) = 35 combinations exist for this case (wheee!)

    repeat above, but exchange the colors. Now for the totals:

    rrrrrrr / bbbbbbb: 2 wrong (2*(7 0))
    rrrrrrb / bbbbbbr: 14 right (2*(7 1))
    rrrrrbb / bbbbbrr: 42 wrong (2*(7 2))
    rrrrbbb / bbbbrrr: 70 right (2*(7 3))

    Right: 84
    Wrong: 44
    Total percentage: 84/128 = 66% or about 2/3.

    Conclusion: in about 2/3 of the cases the above strategy wins. Now the problem is, I don't really see how to improve on this. But the article claims that a strategy exists that wins 7 out of 8 times or 88%. That would would match with a strategy that is losing for the 7r, 7b 6r/1b and 1r/6b cases, and winning for the 5r/2b, 2b/5r, 4r/3b and 4b/3r cases. I don't think there is such a strategy.

    The only thing that I can imagine is that hat wearer X gets a piece of paper which shows information like this:

    rrXbrbbr

    Not only can he see how many red/blue hats there are, but also their relative locations, and his own. This would probably open up a host of other strategies.

    If not, please someone explain how to improve on the above strategy. I'm curious. In any case, a nice problem. It surprised me that the Slashdot crowd did not seem willing to take on this problem for other than the 3 hats case, which is why I did it myself.

  • by BillyGoatThree (324006) on Tuesday April 10, 2001 @06:13AM (#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.
    --
  • I once had a life or death situation where I could have used Pythagorus's Theorum if only I could remember it. I was anchored in a bad storm close to some rocks. I knew my depth and my anchor rode length. So I knew height and hypotenuse. I needed to figure out the A side to figure the radius of the circle my boat would spin around (provided it didn't drag the anchor.) Since I couldn't remember it I was forced to stay up all night to make sure I didn't end up on the rocks. I was exhausted the next day trying to beat back to port in the tale end of the storm. When I got back to my slip I went straight home and learned a^2 + b^2 = c^2 once and for good! Then I slept for 17 hours.

    The lesson of my story is that I once thought math wasn't worth my time to learn and found out the hard way how wrong I was.

    Now you may argue that that's 8th grade trig and more practical then calculus. That's fine but if you want to do anything in a science field then you will need the calculus. If you don't think you'll ever do anything in a science field then you'll probably be stunned to find out that when you graduate they don't need quite so many IT professionals and do need chemists and physicists and EE's (EE's who actually know how to build devices not just program in JAVA.) and if you don't know the advanced math you'll be someones secretary or pool cleaner.

  • Actually the limit of my orgasm approaches 1 long before the limit of her orgasm approaches infinity.
  • With the idea of guessing independantly, you have a .5 chance. The other solutions brings in conditional probability, where the probability of guessing colour x correctly is conditional on situation y, or: P(x|y) which can be broken down into the following situations:
    • Guess blue, saw 2 reds (which works .5 of the time)
    • Guess red, saw 2 blues (also works .5 of the time)
    • pass, saw one of each.
    You only fail when you are red, and see 2 reds or are blue and see 2 blues. The probability of you failing is equal to the probability of everyone being the same colour (.25). The probability of succeding equals 1 - probability of failure (1 - .25 = .75), this method is better than random guessing.
  • Assuming typical lakes, boats, and blocks, any difference in lake level caused by an action on the boat or block will be unmeasurably small, so we can assume it will remain effectively unchanged.
    --
  • I won't flame you because the vast majority of people, even with your level of math education, would agree with you. You're wrong about math, of course, but I think most of the blame lies in the way mathematics is taught. (I don't mean to suggest that I have a solution to this problem, either.)

    I like to think of math as a language. For one, it's a way of communicating ideas both abstract and concrete. Also like other languages, it provides us a conceptual framework which allows us to understand things better. And finally it's a powerful tool when you learn how to manipulate it.

    Much of our math education is just the manipulation part of math, which is useful but never by itself (except on tests, as you said). Learning to describe and understand things mathematically is at least as important. With these three aspects mastered, mathematics becomes indispensable to whatever you do. Well, except for sex -- but other than that, it's pretty useful.

  • 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?
    --
  • Forgive me for reordering your matrix:

    r r r r W W W W
    r r r b - - - C
    r r b r - - C -
    r r b b - C - -
    r b r r - C - -
    r b r b - - C -
    r b b r - - - C
    r b b b C W W W
    b r r r C W W W
    b r r b - - - C
    b r b r - - C -
    b r b b - C - -
    b b r r - C - -
    b b r b - - C -
    b b b r - - - C
    b b b b W W W W

    Ah! Now I understand. Players 2 - 4 are essentially playing their own game, with player 1 simply trying not to mess everything up. In this scenario, player 1 could pass every time and not affect the outcome.
    --

The use of anthropomorphic terminology when dealing with computing systems is a symptom of professional immaturity. -- Edsger Dijkstra

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