A father tells his children, a boy and a girl, to play in the backyard without getting dirty, but they play in the mud anyway. Both children get mud on their foreheads. The father says, “at least one of you has mud on his or her forehead, which I asked you not to do. I’m going to take away desert tonight, but I’ll give you a chance to get your desert back.” He tells them he will ask the following question twice: do you know if you have mud on your forehead? Each child must answer honestly and simultaneously each time the question is asked. Each child can see the other’s forehead but not his or her own. If both children can correctly tell whether or not they have mud on their foreheads, they can have desert that night. What should the children do?

Solution

When the father asks the question the first time, “do you know if you have mud on your forehead,” both children will answer “no.” Each child sees the mud on the forehead of the other one. Since they know at least one of them has mud on his forehead, neither child can conclude whether he also has mud on his own forehead. They must answer “no” the first time the question is asked.

However, after the question is asked and answered, the girl will immediately conclude she must have mud on her forehead since the boy answered “no.” The boy will also conclude that he must have mud on his forehead for the same reason. So, when the father asks the same question again, they both answer “yes” and enjoy desert that night.

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A nobleman is throwing a party in 24 hours, and he wants to serve wine from 1000 casks in his cellar. Unfortunately, an assassin has poisoned one of these casks. The poison takes 23 hours to take effect, and it shows no symptoms before causing death. The nobleman has servants who are “willing” to taste the bottles.

What is the least number of servants the nobleman needs to correctly identify the poisoned bottle, and what is his strategy?

Solution:
The nobleman needs 10 servants. He numbers each servant 0-9, and these represent the respective powers of 2. He numbers each bottle 1-1000. He then labels each bottle by converting the decimal number on each bottle to binary. For example, Bottle 7 becomes 0000000111 and bottle 762 becomes 1011111010. The servants drink from each bottle which has a 1 in the location of their number. So, servants 0, 1, and 2 all drink from Bottle 7 and servants 1,3,4,5,6,7, and 9 all drink from bottle number 762. The servants do this until every bottle has been tested by the appropriate servants.

The nobleman can then look at which servants died. If, for example, Servants 0, 3, 6, and 8 died, the nobleman knows that the poisoned bottle is Bottle 329 (1 + 8 + 64 + 256).

Puzzle 1

Suppose there are two trolls at this fork in the road. One always tells the truth and the other always lies but you don’t know which is which. You have only one gold coin to pay one of them to answer a question. What question will you ask one of them to be 100% sure of which fork to take?

Answer

You simply select one of the trolls randomly and ask, “If I asked the other troll which way I should take, what would he say?” If you selected the lying troll, he would know that the other troll would have told the truth, so the lying troll will lie about the answer. The correct answer, then, is the opposite of what the lying troll would say. On the other hand, if you selected the truthful troll, he would truthfully tell you what the lying troll would say, a lie, and you should therefore take the opposite of his answer. In either case, you should always choose the road that is the opposite of the answer you receive, guaranteeing with certainty that you select the correct road.

Puzzle 2

You are lost in a country in which one third of the inhabitants always lie. The other inhabitants tell the truth, but only three quarters of the time. Desperate to get out of the country, you come to a fork in the road guarded by a troll. One way leads out and the other to certain death, but you don’t know which is which. You can ask the troll which way is out if you pay him a gold coin. You have four gold coins.

1. You pay the troll a gold coin and he advises you to go right. Does that answer help you to decide?

2. You pay the troll another gold coin and he advises you to go right again. Does that answer help you to decide?

3. You pay the troll another gold coin and he advises you to go right yet again. Does that answer help you to decide?

4. You pay the troll your last gold coin and he advises you to go left this time. You must make a final decision. What should you do?

5. What is the chance you get out of this country alive after you’ve paid your last gold coin?

Answer

The variation is good bit harder, requiring some probability calculations. Let’s define some events first. S_n is the event that the troll gives the same answer n times. T is the event that the answer is true. D is the event that the troll is dishonest but not a total liar, i.e., he lies one fourth of the time. The probability we want is P(T|S_n), the conditional probability the response is true given that we have been given the same response n times. By the properties of conditional probability,

P(S_n )= [(3/4)ⁿ +(1/4)ⁿ]*(2/3)+1/3, since 2/3 of the time n truthful answers will be given with probability 3/4 or n false answers will be given with probability 1/4, or 1/3 of the time n false answers will be given with probability 1.

The troll’s first answer is “go right.” What is the probability that answer is truthful? P(S₁ )= [(3/4)¹ +(1/4)¹]*(2/3)+1/3 =1. Therefore, P(T|S₁) = (3/4) * (2/3)/1 =1/2. Unfortunately, the first answer doesn’t improve the odds from 50-50.

The troll’s second answer is “go right.” The probability it is truthful is P(T|S₂) = (3/4)² * (2/3)/ P(S₂ ). But P(S₂ )= [(3/4)² +(1/4)²]*(2/3)+1/3 = (10/16)*(2/3)+1/3 = 18/24 = 3/4. Therefore, P(T|S₂) = (3/4)² * (2/3)/(3/4) = (3/4)*(2/3) = 1/2. The second answer doesn’t help either—it’s still 50-50 odds.

The troll’s third answer is “go right.” The probability it is truthful is P(T|S₃) = (3/4)³ * (2/3)/ P(S₃ ). However, P(S₃ )= [(3/4)³ +(1/4)³]*(2/3)+1/3 = (7/16)*(2/3)+1/3 = 5/8. So, P(T|S₃) = (3/4)³ * (2/3)/(5/8) = 9/20. The probability that the third answer is true is slightly under a half. That helps a little, but not much.

However, the fact that the troll changed his answer on the fourth question helps tremendously, since it shows that he’s dishonest but not a total liar. We now know that he either gave three truthful answers and one false one, or he gave three false answers and one truthful one. Let R be the event that going right is a truthful answer and A be the event that he gave three truthful answers and one false one or three false answers and one truthful one. We want to calculate P(R|A).

When the troll changed his answer, we became 90% certain that going right is the correct road to take, even though he advised us to go left the fourth time we asked.

What would happen if the troll never changed his answer? You could never be certain the troll is a liar based on that, but you would become more and more convinced he’s a liar if he kept giving you the same answer. How many gold coins would you need to become at least 95% sure that his answer is a lie? You would need to find n such that P(T|S_n) <= 5%. If n = 13, P(T|S₁₃) = 4.5%, so you’d need 13 gold coins.

Crick’s Comma Free Genetic Code: Crick suggested the "adaptor hypothesis," which posits that amino acids do not interact directly with mRNA but are carried by small molecules that recognize specific codons. (Today, the adaptor molecules would be called tRNAs.) The codons are then thought to be non-overlapping triplets of bases. This code removes the restriction in the Gamow schemes that allowed only certain combinations of amino acids to be adjacent. However, care must be taken to ensure that each triplet uniquely identifies one amino acid. For example, suppose that somewhere in an mRNA there is the partial sequence ... UGUCGUAAG.... (Note that in RNA uracil replaces the thymine of DNA, and so the code is written with U rather than T.) The intended reading frame is ... UGU, CGU, AAG..., but the RNA molecule has no spaces or commas to indicate codon boundaries. The sequence could equally well be read as ... UG, UCG, UAA, G ... or ... U, GUC, GUA, AG.... Each of these alternatives would have a different meaning. Furthermore, adaptor molecules that attached to the messenger RNA in different reading frames might interfere with one another and prevent any protein from being produced at all. He suggested that adaptor molecules exist for only a subset of the 64 codons, with the result that only that subset would be meaningful; the rest of the triplets would be "nonsense codons." The trick is to construct a code in such a way that when any two meaningful codons are put next to each other, the frame-shifted overlap codons are always nonsense. For example, if CGU and AAG are sense codons, then GUA and UAA must be nonsense, because they appear inside the concatenated sequence CGUAAG. Similarly, AGC and GCG are ruled out by the sequence AAGCGU. The codons AAA, CCC, GGG and UUU cannot appear in any comma-free code, since they cannot combine with themselves without generating reading-frame ambiguity. If all the out-of-frame triplets are nonsense, then there is only one possible way to read the message. A code with this property is said to be comma-free, since messages remain unambiguous even when words are run togetherwithoutcommasorspaces. Show that such a comma free code also codes for 20 amino acids. One way to solve this problem is to write a Matlab code for it.

To solve this problem, start by assuming that a particular codon is allowed. Thus, if ACT is a possible codon then to eliminate ambiguity, ACTACT should be uniquely readable as two ACTs. This eliminates CTA and TAC as possible codons because they would result in an ambiguity in the reading frame. Now choose a different possible codon – say CGT and eliminate ambiguous codons.... and so on. Note that not all starting choices will lead to 20 codons. They often end up with fewer than 20. The point is that the maximum number of codons starting with some sequence of choices will result in 20 codons

Solution: The first thing to notice is that AAA, CCC, GGG and TTT cannot be codons in the comma free code because they would each cause reading frame ambiguity (i.e. in a sequence AAAAAAAA, the reading frame is ambiguous). It is also easy to see that if XYZ is an allowed codon, then its cyclic permutations YZX, ZXY cannot be codons. For example, if both XYZ and YZX were allowed, then the sequence XYZXYZ would have a reading frame ambiguity: XYZXYZ and XYZXYZ. Thus, for any allowed codon, its three cyclic permutations are disallowed. Hence the 60 possible codons that remain after discarding AAA, CCC, GGG, TTT should divide into 20 distinct classes. In principle, it should be possible to choose one codon from each class to avoid ambiguity of the reading frame.

Although this seems plausible, the actual exercise of devising such a code is not trivial.

Below is a Matlab program that iteratively finds the codons. The nucleotides have been replaced by numbers: A==1, C==2, G==3, T==4.

When you run the code, it asks you for an input codon. The number of codons it finds which have no reading frame ambiguity (and so can be used in a comma-free-code) depends on the input codon. Starting with some of the codons does not produce twenty codons. For example, starting with 112 ends up finding only 9 codons:

[112,311,411,321,421,312,412,322,422]. However, starting with codon 123 generates a complete set:

C: [123,211,311,411,122,322,422,213,313,413,323,423,433,214,314,414,124,324,424,434]

Other comma free codes can be obtained by choosing different start codons for example, 231 and 312.

The second code below will check that your list of codons has no reading frame ambiguity. The code writes the output array C into a local file called C.csv.

Code to find the codons:

______________________________

% Code to find the codons for Crick's comma free code. The number found

% depends on starting codon (line 16) - the initial value 123 gives all 20

clc;clear all;close all;

C00 = zeros(1,60);

ic = 0;

for i=1:4

for j=1:4

for k=1:4

if abs(i-j)+abs(i-k)+abs(k-j)>0

ic = ic+1;

C00(ic) = i+j*10+k*100;

end

end

end

end

C(1)=input(' give initial codon \n')

for ij=1:20

C0=C00;

n = length(C);

iw = zeros(n,1);

nc = 0;

for i=1:60

for j=1:n

if C0(i)==C(j)

nc = nc+1;

iw(nc) = i;

end

end

end

C0(iw(:))=[];

% now try to add codons one by one and check

m = length(C0);

n=n+1;

err = 1;

kk = 0;

while (err == 1)

kk = kk+1;

if kk>m

C(n)=[];

n=n-1;

for ij=1:n

fprintf('Codon %i = %i \n',ij,C(ij))

end

fprintf('*********\n')

fprintf(' %i codons found \n',n)

break

end

jerr = 0;

C(n) = C0(kk); % add one more codon from C0

% check that this still works

for i=1:n

c3 = mod(C(i),10);

c2 = mod((C(i)-c3)/10,10);

c1 = (C(i)-c2*10-c3)/100;

for j=1:n

d3 = mod(C(j),10);

d2 = mod((C(j)-d3)/10,10);

d1 = (C(j)-d2*10-d3)/100;

I1 = d2*100+d3*10+c1;

I2 = d3*100+c1*10+c2;

I3 = c2*100+c3*10+d1;

I4 = c3*100+d1*10+d2;

for k=1:n

if (I1==C(k))||(I2==C(k))||(I3==C(k))||(I4==C(k))

jerr = 1;

% fprintf(' error %i %i %i %i %i \n',I1,I2,I3,I4,C(k))

end

end

end

end

if jerr == 0

err = 0;

end

end

end

csvwrite('./C.csv',C)

________________________

Code to check the codons:

clc;clear all;close all;

C = csvread('./C.csv');

n=length(C);ierr=0;

for i=1:n

c3 = mod(C(i),10);

c2 = mod((C(i)-c3)/10,10);

c1 = (C(i)-c2*10-c3)/100;

for j=1:n

d3 = mod(C(j),10);

d2 = mod((C(j)-d3)/10,10);

d1 = (C(j)-d2*10-d3)/100;

I1 = d2*100+d3*10+c1;

I2 = d3*100+c1*10+c2;

I3 = c2*100+c3*10+d1;

I4 = c3*100+d1*10+d2;

for k=1:n

if (I1==C(k))||(I2==C(k))

fprintf(' error 1 %i %i %i %i %i \n',I1,I2,I3,I4,C(k))

ierr=1;

end

if (I3==C(k))||(I4==C(k))

fprintf(' error 2 %i %i %i %i %i \n',I1,I2,I3,I4,C(k))

ierr=1;

end

end

end

end

if ierr==0

for ij=1:n

fprintf('Codon %i = %i \n',ij,C(ij))

end

fprintf(' codons have no reading frame ambiguity \n')

end

Gamow’s order free overlapping code: Gamow suggested yet another overlapping-triplet code with an even simpler description: Each codon is defined entirely by its base composition, ignoring the order of the bases within the codon. In other words, any reordering of the three nucleotides does not change the codon. Thus ACT, ATC, CAT, CTA, TAC and TCA are equivalent and specify the same amino acid. Show that the number of codon families in this scheme again turns out to be exactly 20.

Solution: If the codon is defined only by its nucleotide base composition, then with 4 bases A, C, G, T, one possible combination are the ones in the set, where we eliminate all permutations for each codon.

S = {AAA, CCC, GGG, TTT, AAC, AAG, AAT, CCA, CCG, CCT, GGA,

GGC, GGT, TTA, TTC, TTG, ACG, ACT, AGT, CGT}.

We see that the set S has exactly 20 members. A permutation of nucleotides within each of these codons represents the same amino acid.

Gamow’s Diamond Code: Gamow’s proposed the so-called overlapping “diamond genetic code” (Gamow, G. Possible relation between deoxyribonucleic acid and protein structures. Nature, 1954. 173, 318). In his code, each amino acid corresponds to a configuration of 4 bases. Two of the bases are paired on complementary strands (so only one of them is free to vary, the other is the complement of the first A=T, G=C) and the other two are adjacent bases on both sides of the central paired base so they could be different. The amino acids fit into the groove formed by the diamond and are then stitched together into a linear chain to create a protein. Gamow assumed that that the diamond formed by the four bases could be flipped end-for-end or flopped side-to-side without changing its meaning (i.e. it would code for the same amino-acid). For example, the triplet CAG becomes GAC when it is flipped end-for-end, hence both of these codons must specify the same amino acid. Flopping CAG side-to-side changes the middle A into a complementary T, so that CTG and GTC are also members of the same family of equivalent codons. Show that if one accounts for these degeneracies, the number of possible configurations is exactly 20.

Hint: The diamond code is really a 3 nucleotide code because one of the nucleotides is always paired with its complement. Suppose we place the paired nucleotide in the middle. Then, the codon is represented by XYZ where Y is paired with its complement. If Y is the complement of Y, then when X≠Z, the equivalency operations (which do not change the amino acid) are represented by the transformations:

X Y Z = X Y Z = Z Y X=Z Y X.

On the other hand, when X=Z, the equivalency operation is

X Y X = X Y X

Solution: The above symmetry operations give the following 20 equivalence classes for the 4 possible values of X,Y,Z:

AAA=ATA

CCC=CGC

GGG=GCG

TTT=TAT

CAC=CTC

GAG=GTG

ACA=AGA

TCT=TGT

AAC=ATC=CAA=CTA

AAG=ATG=GAA=GTA

AAT=ATT=TAA=TTA

GGA=GCA=AGG=ACG

ACT=AGT=TCA=TGA

CAG=CTG=GAC=GTC

CAT=CTT=TAC=TTC

CCG=CGG=GCC=GGC

CCT=CGT=TCC=TGC

GAT=GTT=TAG=TTG

GCT=GGT=TCG=TGG

ACC=AGC=CCA=CGA

It is easy to check that each one of the 64 possible ways to combine the nucleotides three at a time is represented exactly once in the above set of equivalence classes. Gamow’s diamond coding would choose one member of each equivalence class to assign to a unique amino acid.

A man needs to go through a train tunnel to reach the other side. He starts running at 10 mph through the tunnel in an effort to reach his destination as soon as possible. When he is ¼ th of the way through the tunnel, he hears the whistle of the train behind him. Assuming the tunnel is not big enough for him and the train, he has to get out of the tunnel in order to survive. We know the following: Given 1. If he runs back, he will make it out of the tunnel by a whisker. Given 2. If he continues running forward, he will still make it out through the other end by a whisker. What is the speed of the train?

Solution: Let S = Speed of train, s = Speed of man = 10 mph, L = Length of the tunnel, D = Distance of train to start of tunnel. Distance of man from start of tunnel = L/4

By Given 1, if he runs back, he will make it out of the tunnel by a whisker. Hence, D/S = L/(4s)
By Given 2, if he continues running forward, he will still make out through the other end by a whisker. Hence,
(D + L)/S = 3L/(4s)

Combining the two equations:
S/2 = s, S = 2s = 20 mph !

More elegant solution: (submitted by Mr. Gerard Delaney of Goa, India)

We know he is 1/4 of the way into the tunnel when he hears the whistle. And he will just escape if he runs back towards the train (Given 1). This means that if he keeps running after he hears the train, he will be 1/2 way into the tunnel when the train enters the tunnel. And (Given 2) we know that the time he takes to run the remaining half of the tunnel will be the same as the time the train takes to travel the whole tunnel.

Hence the speed of the train is twice his speed or 20 mph.

Five mathematically minded pirates discover a chest containing 100 gold coins. They sit down and devise a distribution strategy. The pirates are ranked Pirate 1 to Pirate 5, where Pirate 1 is the highest ranked. The highest ranked pirate gets to propose a plan on how to distribute the gold and then all the pirates vote on it. If at least half of the pirates agree to his plan, the gold is split according to the proposal. If not, the proposing pirate is thrown off the ship and the process will continue with the remaining pirates until a proposal is accepted. The first priority of the pirates is to stay alive and second is to maximize the gold they get. Pirate 1 devises a plan which he knows will be accepted for sure and will maximize his gold. What was the plan he proposed?

To understand the Solution, we need to reduce this problem to only 2 pirates. If there are two pirates, Pirate 1 can easily propose any plan he wants, and it will be accepted. So, he proposes the plan: 100 to me and nothing for you. Since he constitutes 50% of the pirates, the proposal has to be accepted, leaving Pirate 2 with nothing. Now let us look at the situation with 3 pirates. Pirate 1 knows that if his proposal does not get accepted, then pirate 2 will get all the gold and pirate 3 will get nothing. So, he bribes pirate 3 with one gold coin. Pirate 3 knows that one gold coin is better than none, so he has to back pirate 1. So with three pirates, Pirate 1 proposes: {pirate 1, pirate 2, pirate 3} = {99, 0, 1}. Since pirate 1 and 3 will vote for it, it will be accepted. If there are 4 pirates, pirate 1 needs to get one more pirate to vote for his proposal. Pirate 1 realizes that if he fails, pirate 3 will get nothing (according to the proposal with 3 pirates) so he bribes pirate 3 with one gold coin to get his vote. So, the distribution proposed is {99, 0, 1, 0}. With 5 pirates, Pirate 1 needs 2 votes and he knows that if he fails, pirates 1 and 3 will get nothing. He bribes pirates 3 and 5 with one gold coin each to get their vote proposing {98, 0, 1, 0, 1}. This proposal will get accepted. All of this assumes that the pirates are perfectly rational. What would likely happen in reality is that they would fight and the one who wins keeps all the gold.

You walk up a mountain starting at 6 AM and get to the top at 6 PM. You spend the night at the top and on the next day, you start walking from the top at 6 AM and get to the bottom at 6 PM along the same path. You do not necessarily travel at the same speed on the two days. Show that there must be at least one clock time of day when, while going up and coming down, you were at the same height.

Solution: Let d be the height of the mountain. Suppose you plot a graph of time versus height when you go up and when you descend. The graph will start at 6 AM on the time axis on the first day and reach a height of d at 6 PM. The shape of the graph will depend on the speed with which you traveled. On the next day, at 6 AM the return graph will start at a height d and reach a height zero at time 6 PM. Regardless of the shapes of the outbound and return trips, there must be at least one point between the top and bottom where the graphs cross. At this time, you were at the same height at the same clock time on both days.

You go to the bank and take out some money (say X dollars and Y cents). By mistake, the teller gives you dollars instead of cents and cents instead of dollars, i.e. you got Y dollars and X cents. On the way home, you buy a newspaper for 20 cents. When you get home, you find you have twice as much money in your wallet as what you had originally intended to withdraw. What are X and Y?

The equation for X and Y is:

100 Y + X – 20 = 2 (100 X + Y) which can be simplified to

98Y = 199X+20

We have only one equation for two unknowns. However, X and Y are constrained in value. They are both nonzero, positive integers and less than 100: 0 < X < 100, 0 < Y < 100. The trick to find the solution is to express X and Y in terms of some other integer valued parameter. The steps are as follows:

Y = 2X + (3X+20)/98. But since X and Y are integers, so must (3X+20)/98 be.

Let t = (3X+20)/98, t = integer.

Then X = 33t – (t+20)/3

Thus, (t+20)/3 must also be an integer.

Let w = (t+20)/3. Then, t = 3w-20 and hence,

X = 33t – w = 98w - 660 and

Y = t + 2X = 199w - 1340

Substituting w = 1,2,3,.. we see that the only possible solution which satisfies 0 < X < 100 and 0 < Y < 100 is obtained for w = 7. This gives, X = 26, Y = 53. We can check that there are no other solutions by noting that for w = 6 or less, the solutions are negative and for w = 8 or greater they are greater than 100. Hence the unique solution is that you asked for $26.53 from the bank. Instead, the teller gave you $53.26. After buying the newspaper, you had $53.06 left, and this is twice what you had asked the teller for because $53.06 = 2 x ($26.53) !

Five pirates are marooned on an island and spend the day collecting coconuts, intending to divide them up the next morning. One of the pirates decides to get more than his share. So, he gets up at night and finding that the coconuts they had collected do not divide by five unless one coconut is removed. So he gives one coconut to a monkey, removes 1/5 of the remaining hides them. Each pirate does this in turn, giving one to the monkey and hiding 1/5 of the rest. When they wake up, everyone notices the diminished pile but since they all stole, no one says anything. So, they give one to the monkey and take 1/5 th of the rest each. What is the least number of coconuts for which this is possible?

Solution: The easiest way to solve this puzzle is to use a trick. We start with - 4 coconuts. Each pirate, when he wakes up at night and takes some of the coconuts give one to the monkey and then keeps 1/5. With - 4 coconut, giving one to the monkey leaves - 5 coconuts. Take away 1/5 and we are back to - 4. Thus - 4 is a fixed point of the process. Since there were 6 divisions by five after giving one to the monkey, the minimum required is 5⁶-4 = 15,621.

Let us see how this works:

· First pirate: Gives one to monkey leaving 15,620 coconuts. Keeps 1/5 th and leaves 4/5 th = 12,496 coconuts.

· Second pirate: Gives one to monkey leaving 12,495 coconuts. Keeps 1/5 th and leaves 4/5 th = 9,996 coconuts.

· Third pirate: Gives one to monkey leaving 9,995 coconuts. Keeps 1/5 th and leaves 4/5 th = 7,996 coconuts.

· Fourth pirate: Gives one to monkey leaving 7,995 coconuts. Keeps 1/5 th and leaves 4/5 th = 6,396 coconuts.

· Fifth pirate: Gives one to monkey leaving 6,395 coconuts. Keeps 1/5 th and leaves 4/5 th = 5,116 coconuts.

· Morning: Give one to monkey leaving 5,115 coconuts. Divide equally so each gets 1023 coconuts in the last division.

It is said that Paul Dirac, the physicist, on being told the puzzle, immediately came up with the number -4 as the fixed point of the problem, possibly because he invented the concept of “holes” or “negative particles” (also called anti-particles) in his formulation of the famous Dirac equation for electrons.

The general solution with n pirates is that the minimum number of coconuts required is nⁿ⁺¹ – n + 1.

This is a problem which is extensively discussed online with videos and text. Here we present the general solution.

Here is the Problem: N people are lined up facing in the same direction. With their eyes closed, hats are placed on their heads. The hats are of two colors, either white or black. Each person can see the hats on the heads of people ahead of him. Starting from the last person in the line, they are asked to predict the color of the hat on their heads. What is the maximum number of correct answers possible if they use some rule?

If there are N people in the line, then the optimal algorithm can predict N – ½ correctly. The key idea is a concept called “Parity”. The idea goes as follows : the last person says white if he sees an odd number of white hats in front of him (parity odd or 1) and black (parity even or 0) if he sees an even number of white hats in front of him. Each person in line counts the white hats in front of him and adds to it the number of white hat calls from behind him, including the call from the last person. If this number is odd, he says W and if it even he says B.

Let us see how this works for one case with N = 7 with the sequence: LASTFIRST: BBWBWWB

We start with the last person who sees 3 white hats in front of him. Three is odd so its parity is 1. The last person calls W. The next person also sees 3 white hats in front of him and one W call from behind which add up to an even number, so he calls B.

Now the partial call sequence is WB.

The next person sees 2 white hats in front of him and there was one W call from behind which when added to the hats he sees is 3, an odd number, so he calls W, and we get the sequence: WBW

And in this way, it proceeds up the line. The key is to find the parity of the sum of the number of white hats you see in front of you plus number of white calls made so far, including the one from the last person. If this sum is odd call W else call B. The final call sequence is:

WBWBWWB.

Except for the wrong call from the last person in line, everyone else got their hat color right. Since the first person to make the call had a 50:50 chance that his hat would be the color he calls out, on average this algorithm gets N – ½ predictions right. To check if you understood see if you can work out the result for the sequence: WBBBWWBWBW. You should find that the called sequence is BBBBWWBWBW, which is correct except for the last person’s call.

You and your friends want to play a game of craps, which requires two fair dice. The problem is that you have three dice, two of which are fair and one of which is loaded. You don’t know which of the dice are fair and which is loaded and you don’t know how the loaded die has been loaded. Is there a way to roll the dice to have a fair game of craps?

Solution: Choose two of the dice randomly and then discard the other. Roll the two selected dice once and take the sum mod 6. The value is the first fair dice roll. Then roll the two selected dice again, taking sum mod 6. The value is the second fair dice roll. For example, if you rolled a 1 and a 4 on the first trial, then the sum mod 6 is 5, so the first roll would be a 5. If on the second roll you got 3 and 6, then 3 + 6 mod 6 = 9 mod 6 = 3, so the second roll would be a 3. Therefore the roll is (5,3).

The reason this works is that either you chose the two fair dice randomly or you chose one fair die and one loaded die. If you chose the two fair dice, then rolling both and taking the sum mod 6 will be equivalent to one fair die roll. But if you chose one fair die and one loaded die, taking the sum mod 6 is also equivalent to one fair die roll.

To see this, let F be the value of the fair die roll and L be the value of the loaded die roll. Assume the loaded die has the following probabilities

where

Then

Another way to see this is to enumerate all possible ways to get a specific value. For example, the following table shows all the ways to get an outcome of five along with their associated probabilities.

Thus, the probability of rolling a five is

The other possible outcomes can be readily verified.

A wants to send B a secure message. He puts it in a locked box and sends it via the postal service. The postal service is reliable and will get the box to its destination without the lock being opened and the message read. However, B does not have the key to open the box. How will he get the message?

Solution: When B gets the box, he will lock the box with another lock and send it back to A. A will remove his lock and send it back to B, who can then use his own key for the lock he used to open it.

Why does this not work if instead of a lock, the message is encrypted?

Once upon a time, in a far away kingdom, there lived a king who was a mathematician. He wanted to find a suitable friend with whom to discuss mathematics. He invited the best mathematicians in his kingdom to come and solve some riddles that he posed to them. The first one of these was to determine which of the above two boxes has the gold.

Can you solve it?

Solution: You can argue as follows: The statement on box B can either be true or false. If it is true, then the statement on box A must be false. On the other hand, if the statement on box B is false, then either both statements are false or both are true. But since we have assumed that the statement on box B is false, then both statements must be false. In either case, the statement on box A is false. So you conclude that the gold must be in box A. Alas, when you open box A you do not find the gold. The gold is in fact in box B.

Why is the logic above incorrect? The reason is that the statement on box B is self referential and therefore cannot be used to make any conclusions. It is like the statement: “This statement is false”. Hence the only valid statement in the problem is the statement on box A, which says the gold is in the other box - box B. As indeed it is.

An Arab sheikh wants to will all his fortune to only the wisest among his two sons.He makes them a proposition. His two sons will ride their camels in a race, and the son whose camel crosses the finish line last will inherit it all. The two brothers wander aimlessly for days, neither willing to cross the finish line. In desperation, they ask a Wise Dervish for advice. He tells them something. The brothers leap onto the camels and charge toward the finish line. What did the Dervish tell them?

Solution: He told them to switch camels. Perhaps the sheikh should have left his fortune to the Dervish. But then again, the Dervish may not want it.

There is a wonderful book by Raymond D. Smullyan that everyone who likes puzzles should read.

Its name is: What Is the Name of This Book?: The Riddle of Dracula and Other Logical Puzzles.

Here are some puzzles from this book. All the puzzles below are about the “Island of Knights and Knaves.” Knights always tell the truth and Knaves always lie.

Puzzle 1: A stranger from out of town meets three inhabitants A, B, C. He asks A: “Are you a Knight or a Knave?” A mumbles something the stranger could not hear so he asks B: “What did he say?” B said: “He said he was a Knave.” C said: “Don’t believe B, he is lying !” What are B and C (Knights or Knaves)?

Solution: Note that a Knave can never say “I am a knave” because then he would be telling the truth. So B must be a Knave because A cannot have said: “I am a Knave.”It also follows that C must be a Knight since he made a true statement. But it is not possible to decide what A is.

Puzzle 2: Now the stranger meets two people A and B

A says: “At least one of us is a Knave.” What are A and B?

Solution: If A was a Knave then he would have told the truth when he said at least one of us is a Knave. That would be a contradiction. So, A cannot be a Knave. Hence A must be a Knight. Hence, his statement “at least one of us is a Knave” must be true. So, one of A and B is a Knight and one is a Knave. Since A is a Knight, B must be a Knave.

And here is one more that I will leave you to solve on your own:

Puzzle 3: The stranger from out of town meets three other inhabitants A, B, C. He asks A: “How many Knight are among you?” A mumbles something the stranger could not hear so he asks B: “What did he say?” B said: “He said there is one Knight among us.” C said: “Don’t believe B, he is lying !” What are B and C (Knights or Knaves)?

An old farmer had 19 horses that he willed to his three children as follows: The oldest child gets half the horses, the middle child gets one fourth of the horses and the youngest child gets one fifth of the horses. When the old man died, the children did not know how to divide the horses without killing them. They asked for help from a wise neighboring farmer, who came over the next day riding on his horse. How did he solve their problem?

Solution:

The wise farmer added his horse to the 19 to make 20 horses. He gave the oldest child half of 20 horses = 10 horses; the middle child one fourth of 20 horses = 5 horses; and the youngest child one fifth of 20 horses = 4 horses. This left the wise farmer his own horse to ride home on. Why does this problem need one more horse? Because 1/2+ 1/4 + 1/5 adds up to 19/20.

Ramsey theory is a branch of mathematics that looks for ordered subsets in random systems when they get large enough. The simplest problem in Ramsey theory is the following: What is the smallest number of people to guarantee a clique of size 3? A clique in this context is defined as a set of people who are either friends or strangers. To make the problem simpler, we will give the answer. A clique of size 3 is guaranteed among six people. The problem we will pose is to prove that this is true.

Prove that in a group of six people, there must be three people who are friends (know each other) or strangers (do not know each other).

Solution: We will use the images below to prove this statement. Suppose we consider the people as vertices and relationship (friends or strangers) as links colored blue and red respectively. Pick a vertex. Since there are only two colors available, and there are 5 links out of v, at least 3 of them must be the same color. Let us choose this color as red (shown in image). Let us say that these links connect to vertices a, b and c. Now consider the triangle a, b, c. Since there are only two colors available, either one of the edges, there are only a limited number of possibilities to color them.

1. All are blue. But this would make a blue clique of size 3 from the triangle abc.

2. Two are blue and one, say ab, is red. In this case, we have the red clique vab.

3. Any other choice of colors with more reds would give at least one red clique of size 3.

This proves that with six people, there must be a clique of three people, who either all know each other, or all do not know each other.

Two people, let us call them A and B, decide to have a duel. They agree to meet at a designated spot sometime within the hour between noon to 1 PM. Each would arrive at a random time and stay for 5 minutes. If the other arrived within these 5 minutes, they would fight. Otherwise, they would leave, honor served. What is the probability that they will fight?

Solution: The figure below shows the possible arrival times of A and B on the two axes. It is clear that they will fight only of their arrival times are within the diagonal band in the figure. Hence the probability that they will fight is P = (60² – 55²)/60² ~ 0.16. There is a 16 % chance of the duel taking place.