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Rachel MargoleseMath 7May 23, 2013

Hats on Hats

Gentlemen in hats. Prisoners in hats. Dwarves in hats. White hats. Black hats. Red hats.

Blue hats. Every hat on every head proposes a new puzzle and a new way to consider puzzles.

Just as the characters in the multitude of hat puzzles all have a different perspective of all of the

hats and what different people are wearing, each puzzler who encounters a hat puzzle will have a

slightly different insight into the puzzle. For the seasoned mathematician, hat puzzles have

provided a gateway into the study of the mathematical branch of coding theory—an absolutely

fascinating discipline for some but for the layperson perhaps not the most engaging topic. What

is exciting however is a convoluted story of a group of prisoners forced to wear strange hats and

who can only escape from their dire situation if you, the puzzler, can use your wits and clever

deduction to set them free. These are two ways in which to consider a hat puzzle, as a captivating

story and diverting game of wit, or as an intriguing analogue to coding theory giving way to new

insights in the discipline of mathematics. Hat puzzles accomplish this duel purpose through their

intrinsic property of being easily expanded or varied. In practice, any slight variation of a hat

puzzle—either varying the number of hats or changing the information that each character has

access to—produces a remarkably distinctive puzzle with an equally distinctive solution. This is,

perhaps, not unique to hat puzzles in particular but it is particularly striking within the hat puzzle

genre.

The puzzle precursor to the modern hat puzzle genre appeared in a puzzle book by

George Gamov and Marvin Stern in 1958 and depicted three commuters with sooty faces.i

During the following decades, the puzzle was re-imagined into the three hat puzzle that appeared

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in Todd Ebert’s 1998 Ph.D dissertation and became the classic hat puzzle that puzzlers use

today.ii The puzzle begins simply with three prisoners. As each prisoner enters a room, a coin is

flipped, with the outcome determining what color hat that prisoner will wear: white if heads,

brown if tails. Once all three prisoners are wearing their hats, each prisoner will be taken aside

and given an opportunity to guess the color of her own hat, at which point she may either guess a

color or pass. The deal is that, if all of the prisoners who decide to venture a guess, guess

correctly, all of the prisoners will be set free. If none of the prisoners guess, or if any prisoner

guesses incorrectly, then all of the prisoners will be sentenced immediately to death row. The

prisoners are given time to strategize before they are given their hats, but once they are wearing

their hats, they may have no further communication with one another, nor can they see their own

hat. What strategy gives the prisoners the best odds of survival?

The immediate solution that many people come up with is to decide to have one prisoner

guess and one prisoner only. Since the combination of colored hats is chosen completely at

random, she’ll have exactly a 50% chance of guessing correctly. But let’s see if we can get better

odds than that. Consider the situation from the viewpoint of each prisoner, starting with prisoner

A. There are three distinct possibilities of combinations of hats that she might see, which are

outlined in Table 1. As the color of the hats that she sees have no baring on the color of her own

hat, this information is insufficient for the prisoner to make an educated guess with regards to her

personal hat. Thus, we must conclude that there is no solution which can guarantee her safety

and, as such, we are only searching for a solution which gives the highest possible chance of

survival. Now, move from the single prisoner’s viewpoint to the omniscient viewpoint. From

here, we see that there are eight possibilities of combinations of hats, outlined in Table 2. From a

probabilistic overview of these combinations, we see that, out of the eight combinations, there

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are only two in which all three hats are the same color. Phrased another way, we see that six of

the eight hat combinations consist of two hats of the same color and one hat of the opposite

color. Thus, if you are a prisoner and you see two hats of the same color, chances are six out of

eight (or 75%) that you are wearing the opposite color hat to those that you see. On the other

hand, if you see two hats of opposite colors, your hat is, with equal probability, either one of

those colors, and thus you have no way of increasing your chances of a correct guess to above

50%.

This analysis leads us to a possible solution to maximize the chance of survival. Since

you have a 75% chance of correctly guessing your hat color if you see two hats of the same

color, you should guess your hat color in these cases. If you see two hats of opposite colors, you

should pass your guess because you only have a 50% chance of guessing correctly. As we can

see from Table 2, this strategy will give the prisoners a 75% chance of survival. To the

layperson, this is a clever and satisfying solution, puzzle complete. But the curious few may

wonder how does that work? How can a person have a greater than 50% of guessing a randomly

chosen color using information that has no effect on her chosen color? Phrased that way, yes, the

solution appears to be a strange anomaly, a trick of logic. However, that ‘trick of logic’ is

grounded in coding theory and has spawned exploration into an area of coding theory known as

error-correcting codes.

Before getting into the mathematical theory, let’s examine three alternative variations of

the original three-hat puzzle. The first puzzle brings us to four men captured by Nazis. The Nazi

general wants to have a little fun with these prisoners and so he engages them in a wager of wits

with their freedom as the prize. The general takes three of the soldiers into one room and

immobilizes each man in a chair, one behind the other. Soldier A can see soldiers B and C,

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soldier B can see soldier C, and soldier C can only see the wall in front of him. The general then

brings soldier D to another room where the soldier can only see the four cement walls

surrounding him. The general then shows the soldiers four hats: two white hats and two black

hats and tells the soldiers that he will put one hat on each soldier. If any one of the four soldiers

can guess the color of his hat without communicating or in any way signaling to the others, all

four men will be set free. The general then blindfolds the soldiers, places a hat on each soldier’s

head, and takes the blindfolds off. Will the soldiers earn their freedom? If so, who will set them

free?

As the omniscient puzzle solver, you have the power to set up the puzzle as you see fit;

you decide which soldier has on which hat with each hat placement depending on the locations

of all the other hats’ placements. Knowing all of the hat locations doesn’t help you understand

how each soldier will guess. Thus, in order to solve the puzzle, you must disregard part of the

information that you know as the omniscient puzzle solver and put yourself into the puzzle and

into the place of each soldier one-by-one. Start with the isolated soldier. Obviously, he has no

information about any of the hats, so he has no chance of definitively guessing his hat correctly.

Next, sit in the seat of soldier A, who can see the two men and their hats in front of him. From

this vantage point, there are three different views you can have: two white hats, two black hats,

or a white and a black hat. If you see two hats of the same color, you know you must have the

other color hat because there are only two hats of each color. Therefore, only in these two cases,

soldier A will set the group free. But, in the situation where you see one of each color, you could

be wearing either a black or a white hat; there is not enough information for you to know what

color hat you are wearing. Now, sit in soldier B’s seat. From here, you can see only the hat in

front of you, which is either black or white. In this seat, however, you have more information

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than just what you see. If the soldier behind you saw two black hats or two white hats, he would

immediately say the opposite color that he saw. However, in the case where the soldier behind

you says nothing, you know that he must see one of each hat color. His silence tells you that you

must be wearing the opposite color of the hat that you see in front of you.

Just as in all other hat puzzles, the solution to this four-hat variation lies not through

information supplied to the omniscient viewer, but through the eyes of the characters in the

puzzle. In this analysis, we can begin to see the foundation of the error correcting code theory

imbedded in the solution to the problem. In computer science, error-correcting codes are codes

which are essentially able to check a string of bits and detect any single error that is contained

within that string—an error being a flipped bit, either a 0 has become a 1, or a 1 has become a

zero. For these codes to work, there must be some factor which is dependent on the status of each

of the bits and would thus change if any of the bits were to change. The four-hat puzzle offers a

very basic view of how error codes work; in this puzzle, each element within the puzzle can look

around, add up what he sees, and deduce whether he has enough information to decide the status

of his own color. In much the same way, error-correcting codes use the status’ of the

information-containing bits to determine the status of a checking bit, that checking bit being

analogous to a soldier who is looking at the other hats. In both the code and the puzzle there is a

constraint which is integral for the checking to occur—in the puzzle it’s the fact that of the four

hats, two are white and two are black, and in the code it’s the fact that the checking bit holds

information which relates to the orientation of all of the other bits.

The four hat puzzle gives us a rudimentary look at how error-correcting code relates to

hat puzzles. Now let’s consider a hat puzzle which involves a greater internal dependence from

one hat to the next. For entertainment, a sadistic wizard captures ten dwarves. He proposes a

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deal. The ten dwarves will line up in a straight line, and each will be wearing either a blue or an

orange hat. They could all be wearing blue hats or all orange hats, or any combination of blue

and orange hats. The dwarf in the back will be able to see all nine of dwarves in front of him, the

second to last dwarf will be able to see all eight dwarves in front of him, and so on down the line

until the first dwarf in line who will not be able to see any dwarves. The deal the wizard proposes

is this: each dwarf will guess the color of his own hat (which he cannot see), starting with the

dwarf in the back of the line, who can see all nine dwarves in front of him. If he guesses right,

that dwarf is set free. If he guesses incorrectly, he is immediately, and silently, killed. The wizard

allows the dwarfs to conceive a plan before they are lined up. All of the dwarves hear each guess

that is made, but they do not know if the dwarf who is guessing is killed or not. They are not

allowed to turn around once they are in line, so each dwarf can only see the hats on the dwarves

in front of him. Additionally, they are not allowed to communicate in any way once they are

lined up. However, all dwarves do hear the single hat guess that each dwarf makes. How many

dwarves can be saved and how?

As before, let us first try to insert ourselves into the puzzle. Start by imagining yourself as

the last dwarf in line, the dwarf who can see all nine dwarves in front of him. From this position,

you can see nine hats, some of which are blue and some of which are orange. Seeing these nine

hats gives you no information pertaining to your own hat. So you must guess randomly. Now put

yourself in the place of the ninth dwarf in line, the one who can see eight dwarves in front of

him. You just heard the dwarf behind you guess a color, say blue. You also see the eight colors

in front of you. In total, you have eight hats to look at, some blue, some orange, and one guess

from behind. Without any preconceived plan, this information tells you nothing about your own

hat and is insufficient information to produce a reasonable guess about your hat. You need more

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information. The only way you can glean more information while in line is through the guesses

that each dwarf behind you makes. Now consider possible plans that you could have come up

with before getting into line which could be used to pass along additional information about hats.

Consider a simple plan where every other dwarf says the color of the hat in front of him. In this

manner, every other dwarf would know the color of his own hat and be safe, while the other five

dwarves only have a 50% chance of being saved. Of course, assuming that this is an evil wizard,

he could arrange it so that the hats alternate colors such that the sacrificial dwarves will all be

killed. Thus, this plan can save 5 dwarves. Is there a better stragety?

Through their planning, the dwarves must come up with a scheme by which to convey

information about all of the hats in front of them, as we have seen that only conveying

information about a single hat in front of you cannot save a majority of the dwarves. Considering

the possible characteristics of the hats in front of you, one characteristic is the number of each

color and more specifically the oddness or evenness of that number. Say you are the second

dwarf in line; you might see five orange hats and three blue hats. How can you convey, in one

word, the number of each hat color that you see? You only need to convey information about one

of the colors, because if you know how many orange hats there are, you can then easily deduce

the number of blue hats. As we saw in the beginning, the dwarf in back has no information

besides the nine hats he sees and he cannot access any additional information before he makes

his guess. From this, we can see that there is no way to guarantee his survival. But his death need

not be in vain. He sees in front of him all of the dwarves’ hats, say six orange hats and three blue

hats. Let’s focus on just the orange hats. One characteristic of the number of orange hats is

whether there is an odd or even number. So, since the last dwarf in line cannot guarantee his own

safety, he can guess a color that will tell the rest of the dwarves whether he sees an odd number

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of orange hats or an even number of hats. Say he conveys that he sees an even number of orange

hats. Then, imagine yourself as the ninth dwarf in line. You look ahead and see five orange hats,

which is an odd number. Ergo, since the dwarf behind you saw an even number of hats, you must

be wearing an orange hat. You announce your guess of orange. Then, the eight dwarf in line

looks ahead and sees five orange hats. Since the first dwarf said there were an even number of

orange hats, and the second dwarf guessed orange, that second dwarf must have seen an odd

number of hats. Therefore, if this third dwarf to guess also sees an odd number of orange hats, he

must be wearing a blue hat. Using this logic, the dwarves continue down the line and all of them

are set free—except for the unfortunate dwarf in the back of the line, may he rest in peace.

The only detail left is to figure out how that first dwarf conveys the oddness or evenness

of the number of orange hats. Since the dwarves have time to plan before they line up, they could

decide that the first dwarf should guess orange if there are an even number of orange hats, and

blue if there are an odd number of orange hats. In this manner, 9 out of 10 dwarves can be saved,

which is an excellent survival rate.

At this point, we have reached an excellent and logical solution—even without

continuing into the coding theory, a puzzler could leave this puzzle content. While solving this

puzzle, we saw that the method of inserting yourself into the puzzle will only get you so far into

the solution. The leap from discovering that the dwarves need more information besides the hats

that they can see and deciding that the backmost dwarf must convey the oddness or evenness of

the number of orange hats requires some familiarity with working with numbers. Though a

difficult leap to make when solving this puzzle, the use of the oddness or evenness characteristic

of the number of orange hats is key to error-correcting codes in this example. Slightly different

from the previous example, with the dwarves, each hat guess depends on all of the previous

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guesses with the first guess being a pre-determined bit of information. All of the guesses depend

upon the first guess and in this way the first guess also contains information about all of the other

nine guesses, which is the crux of error-correcting codes.

Finally, lets consider one last variation on the original hat puzzle. In this version, given

one more piece of information, and a slight variation of rules, we can guarantee survival for all

prisoners. As in the first puzzle, there are three prisoners each wearing a randomly selected

colored hat, either brown or white. Once each prisoner is wearing a hat, the prison guard asks the

prisoners to raise their hands if they see a brown hat. All prisoners raise their hands. The

prisoners are then asked keep their hands up if they can guess their hat color. All hands go down.

The guard tells them that at least one prisoner must guess correctly for them to be set free. How

do the prisoners survive?

Once again, consider what prisoner A sees. She looks out and sees either one brown hat

and one white hat, or two brown hats. Let’s say she sees a white and a brown hat. Now she

considers what the prisoner wearing the brown hat must see. Since that prisoner raised her hand

to say that she had seen a brown hat, one of the two hats that she is looking at must be brown.

Therefore, since the third prisoner’s hat is white, prisoner A’s hat must be brown.

What if prisoner A looks out and sees two brown hats? Then she has no way of knowing

if her hat is brown or white, so she will remain silent. Now, move on to prisoner B, who looks

out and sees either a white and a brown hat or two brown hats. If she sees one of each color, she

will correctly guess her own hat using the same logic as we went through with A, but, if she sees

two brown hats, then she too will remain silent, hoping that someone else will have enough

information to deduce their hat color. Finally, prisoner C, hearing the silence of the first two

prisoners, can deduce that neither has seen two different colored hats because, if they had, they

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would have guessed their own hat color. Therefore, prisoner C can correctly deduce that she is

wearing a brown hat. Using this logic, the prisoners have a 100% chance of survival.

Much like in the four-prisoners variation, this puzzle requires logical interdependence

deduction where each prisoner must deduce what all of the other prisoners must be thinking in

order to make her own guess. Walking through this puzzle logically while considering the

situation from each of the prisoner’s viewpoint produces a very elegant pathway to find the

solution. In this solution, we can again see how the error-correcting codes are ingrained in the

dependence each prisoner has on the other two prisoners; each prisoner depends upon the logical

conclusions that the other two prisoners reach, i.e. that if prisoner A and prisoner B both remain

silent, they must both have deduced that they need prisoner C’s information in order to guess

their own hats. Thus, all color guesses are linked together and therefore interdependent.

Error-correcting codes, specifically Hamming codes which are a family of error-

correcting codes, make all of these hat problems very interesting for mathematicians. These last

three puzzles lay out different aspects of the error-correcting codes which helps to understand

just what is meant by an error-correcting code and how they relate to puzzles, but the original hat

puzzle is the real link to understanding the fascination mathematicians have with hat puzzles and

error codes. The three-hat original puzzle can be expanded to include seven, sixteen, even

hundreds of hats and by using coding theory, mathematicians have proven that the prisoners have

a better than 50% chance of survival. (To see the math behind error codes for large numbers of

prisoners read "The Hat Problem and Hamming Codes" in Focus (2001))iii This ability to expand

the puzzle to infinite many prisoners is what is so captivating for mathematicians looking to

expand modern mathematics, yet at the same time the puzzles have no need to be expanded to

provide entertainment and an excellent exercise in logical problem solving for other puzzle

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solvers. Hat puzzle have reached such a wide audience by both appealing to the mathematically

inclined crowd and to the recreational puzzler looking for a diversion. As each hat puzzle

revolves around the fact that the puzzle is solved through the eyes of the wearer and not the

beholder, hat puzzles are also unique from other puzzles; in these puzzles you get to be a part of

the puzzle and become a part of the puzzle story which is fun in its own right. Be a part of the

problem to find the solution.

Prisoner B Prisoner C

Brown Brown

Brown White

White White

Table 1: Possible combinations of hats which prisoner A could see.

Actually wearing Guess Survived?

Prisoner A Prisoner B Prisoner C Prisoner A Prisoner B Prisoner C

White White White Brown Brown Brown N

White White Brown -- -- Brown Y

Brown White White Brown -- -- Y

White Brown White -- Brown -- Y

Brown Brown Brown White White White N

Brown Brown White -- -- White Y

White Brown Brown White -- -- Y

Brown White Brown -- White -- Y

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Table 2: Possible combinations of hats that each prisoner could be wearing and each prisoner’s guess given the condition that they only guess if they see two of the same hat color.

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References

1. Bernstein, Mira. "The Hat Problem and Hamming Codes." Focus (2001): 4-6. Mathematical Association of America. Web. 21 May 2013

2. Guo, Wenge, Subramanyam Kasala, Bhaskara Rao, and Brian Tucker. The Hat Problem And Some Variations. Rep. N.p.: n.p., n.d. University of Cincinnati. Web. 15 May 2013. <http://web.njit.edu/~wguo/Hat%20Probelm.pdf>.

3. Hardin, Christopher S., and Alan D. Taylor. "An Introduction to Infinite Hat Problems." The Mathematical Intelligencer 30.4 (2008): 20-25. Spnnger Science + Business Medta, Inc. Web. 13 May 2013.

4. Robinson, Sara. "Why Mathematicians Now Care About Their Hat Color." The New York Times. The New York Times, 10 Apr. 2001. Web. 16 May 2013.

5. Stern, Marvin. "Three Soot Smeared Faces." Puzzle-math. By George Gamow. New York: Viking, 1958. 77-78. Print.

Endnotes

i Stern, Marvin. "Three Soot Smeared Faces." Puzzle-math. By George Gamow. New York: Viking, 1958. 77-78. Print.ii Hardin, Christopher S., and Alan D. Taylor. "An Introduction to Infinite Hat Problems." The Mathematical Intelligencer 30.4 (2008): 20-25. Spnnger Science + Business Medta, Inc. Web. 13 May 20 13.iii Bernstein, Mira. "The Hat Problem and Hamming Codes." Focus (2001): 4-6. Mathematical Association of America. Web. 21 May 2013.