You meet three people, the Liar (L), Truth-teller
(T), and Headache guy (HG). You don't know who is who. The three speak English but in their dialect "nes"
and "yo" mean "yes" and "no" and you're not sure which is which. The T always tells the
truth. The HG will respond "yo" to any question posed to him if he has a headache; if he doesn't have a headache
he will respond "nes." The L always lies, but sometimes he will lie and respond "I don't know"
to a question. The L and T are omniscient. You may ask only three separate yes-or-no questions, each to be answered by only
one of the three, and each question can be directed to any of the three. How can you identify the Liar with only
three questions?
Solution
We call the men A, B, and C. You ask A this rather long question:
1)"Of the two statements a) You are the Liar,
and b) The answer to the question of whether C is the HG is 'yo', is one and only one of them true?"
If the response is "yo" then either A or C is the HG. If it is "nes" then either A or B is the HG. You still don't know
what "yo" or "nes" means but the identity of the HG is narrowed to two possibilities.
If A responds, "yo" to question
1 then you ask B:
2) "Of the two statements, a) You are the Liar, and b) The answer to the question of whether A is the
HG is 'yo', is one and only one of them true?"
If the response is "yo" then A is the HG. If it is "nes" then C is the HG.
If A responds, "nes" to question 1 then you ask C question 2. If the response is "yo" then A is the HG. If it is "nes"
then B is the HG.
Once you have determined the identity of the HG, you then direct this question to one of the remaining
two:
3) "Is 'yo' the answer to the question of whether you are alive?"
If the response is "nes" then that person is
the Liar. If the response is "yo" then he is the T and the third person is the Liar.
Obviously, if you ever receive a
response of "I don't know" you have immediately identified the Liar. The following approach won't work. Suppose you pose this
question to either A or B: "If asked whether C is the HG, would you respond 'yo'?" If you posed that question to the Liar
he could respond with the word that means "yes." That reply wouldn't tell you whether C could be the HG. The reason is that
the Liar can always respond "I don't know" to a question and he could lie about that response with either "yes" or "no."
9.
The Red hat/Blue hat problem
A king tells 100 wise men that he will
place either a red or blue hat on each of their heads. All of the men are lined up facing the same direction. No man can see
the color of his own hat. However, each man can see the color of all the hats in front of him but none behind him. In other
words, the man in the back of the line can see the hat colors of the other 99 men. The man in front of him can see the hat
colors of the 98 men in front of him, etc. Each man will attempt to guess the color of the hat on his head. The man
in the back of the line is the first to guess. The king stands in front of all 100 men so every man can see him. If
the man guesses his hat is red he will silently raise his right hand. If he guesses blue he will raise his left hand. Once
he raises his hand the king shouts out the man's guess for all to hear. Once a man guesses the man in front of him will
guess next. Only the king can see any man raise his hand. No one can hear the man raise his hand. The men discuss
a strategy to enable them to correctly guess as many hats as possible. How many correct guesses are certain?
b. Assume a variation of this problem
with a point system. A correct guess is worth 1 point and incorrect guesses are worth 0 points. If a man wants to guess his
hat color before it is his turn it is worth 1½ points for a correct guess. However after he has made his guess every
man who was supposed to guess before him must then make his guess. What strategy should the wise men adopt to maximize their
score?
Solution
The number of correct guesses will be either 99 or 100. The wise men agree that the man in the back of the line
will raise his right hand and guess red if he counts an even number of red hats in front of him (0 is an even number). He
will guess blue and raise his left hand if the number is odd. If the first man guesses red and raises his right hand the man
in front of him will guess red only if the number of red hats in front of him is an even number (since he can count all of
the red hats in front of him). Otherwise he must guess blue. Each time a person guesses red, the number of red hats in front
of that person alternates between even and odd and at least 99 people will correctly guess the color of the hat on their heads.
b. The variation with additional points for guessing out of turn requires a different strategy. The wise men can adopt a strategy
in which the first seven men signal the number of the minority color. The first man calls the color with the fewest hats in
the last 93 men. The second man will call red if 12 is to be added to 23½ or blue if it is to be subtracted. The third man
will call red if 6 is to be added to the total or blue if subtracted. The fourth will call red if 3 is to be added to the
total or blue if subtracted. The fifth man will call red if ½ is to be added to the total or blue if subtracted. The last
two men will call either red or blue to add or subtract 1 from the total. This will yield the number of minority hats. Many
will correctly guess their hat color in advance using this scheme if the men take their time and reason through it.
10. The Dog
and the railroad car
A railroad car is 50 feet long and traveling on a railroad track at a
constant speed. A dog on the ground next to the track trots from the back of the car to the front, immediately trots from
the front of the car to the back, and finally trots to the front of the car again. At that point the car has traveled 75 feet.
Assuming the dog travels at a constant speed and loses no time in turning around, how far does the dog trot?
Solution
Let t equal the amount of time it takes for the dog to travel from the front of the car to the back. Let T equal
the amount of time it takes for the dog to travel from the back of the car to the front. Assume the car travels at the speed
of one foot per second (1). Therefore, the total time is 75 seconds (or 75). Let s equal the speed of the dog. Equation 1
must describe the dog's trip from the front of the car to the back (the back of the car and the dog are heading directly towards
one another from 50' apart):
1) t (s + 1) = 50
Equation 2 must describe the dog's trip from the back of the car to the front (the distance the dog travels in time t less
the distance the car travels in that time must equal 50):
2) T (s - 1) = 50
Equation 3 also must be true since the total time required for the trip is 75 seconds:
3) t + 2T = 75
Substituting the values for t and T from equations 1 and 2 into equation 3 yields:
3ss - 6s - 5 = 0
Using the quadratic equation we come up with one solution for s, namely s = 2.63. When multiplying that by the total time
of 75 seconds we arrive at the solution of 197.5 feet.
Here's my variation. Five people
play a game. Four of them stand forming a square and each faces in one of the four directions (N, S, E, or W). But they
don't all face the same direction. The fifth man (Al) turns his back so he can never see the positions of the four. Al
calls out one of six possible instructions:
A) "Any two people on a diagonal turn 180º." B)
"Any two people on a side of the square turn 180º." C) "Any one person turn 180º."
D) "Any two people on a diagonal turn 90º in any direction." E) "Any two people on a side
of the square turn 90º in any direction." F) "Any one person turn 90º in any direction."
The four decide which of them comply with the instruction. Al can call any instruction more than once. After
each instruction the men remain in the new position. Al wins the game if he causes all four to face the same direction at
any time calling 63 instructions. Can he always win?
Solution
Al can always win. Let S be these seven instructions: 1) A 2) B 3) A 4) C 5) A 6) B 7) A. The 63-instruction
solution is: S, D, S, E, S, D, S, F, S, D, S, E, S, D, S.
12. Roll the
dice
a) A and B are playing a game with two dice. A bets that
two 6s (totaling 12) will be rolled before two consecutive 7s are rolled. They continue rolling the dice until a player wins.
What is the probability A wins?
b) Assume only one die is rolled at a time.
A bets that a 6 will be rolled before two consecutive rolls totaling a two, three, or four will be rolled (i.e., 1 and 1;
1 and 2; 2 and 2; or 1 and 3 in any order). What is the probability A wins? An approximation to a whole percent is sufficient.
Solution
a) Three scenarios are possible:
1. The probability a 12 is rolled is (1/36).
2. The probability a 7 is rolled followed by a 12 is (1/6) (1/36).
3. The probability two 7s are rolled consecutively is (1/36).
In all other cases neither player will win; and we are back to square one. A wins with scenarios 1 and 2 above. B wins with
scenario 3 above. So, the probability A wins is [1 + 2]/ [1 + 2 + 3] = [(1/36) + 1/6(1/36)] / [(2/36) + 1/6(1/36)] = 7/13
Problem b) Whenever a 4 or 5 is rolled essentially we are back to square one in this problem and the following procedure
is repeated. So the problem begins once a 1, 2, 3 or 6 is rolled. For any one of those particular numbers to appear first
the probability is 1/4. For purposes of notation a roll of a 2 followed by a 1 will be denoted as follows: 21. A notation
of a number or numbers in parenthesis followed by * (e.g., 2* or (23)*) means a potentially infinite sequence whereby the
number(s) is repeatedly rolled (e.g., consecutive 2s are rolled or 23232323).
The solution basically involves determining all of the possible sequential rolls for each player whereby that player wins
on a particular roll and neither player has previously won.
For A here are all of the possible sequential rolls: 6,16, 26, 36, 33*6, 3*26, 23*6, 323*6, 2(32)*6, 23(23)*6 and 32(32)*6.
The probability A wins is the sum of all of A's possible sequential rolls divided by the sum of A's and B's possible sequential
rolls.
1) The probability A's rolls will occur is:
a) For the lone 6, (1/4).
b) For the three two-number rolls, (3)(1/4)(1/6) or 1/8.
c) For the three three-number rolls, (3)(1/4)(1/6)(1/6) or 1/48.
d) For the five four-number rolls, (5)(1/4)(1/6)(1/6)(1/6) or 5/684.
e) For the six five-number rolls, (6) (1/4)(1/6) (1/6)(1/6)(1/6) or 1/864.
The remaining terms are small enough that we can simply add a), b), c), d) and e) to approximate A's probability. The probability
of any of A's rolls occurring is (1/4) + (1/8) + (1/48) + (5/684) + (1/864) =.40430.
For B here are all of the relevant possible sequential rolls: 11, 12, 21, 22, 31, 13, 322, (32)*1, (23)*1, (23)*21, (23)*22,
33*1, 33*21, 33*22, 3(23)*1, 3(23)*21, 3(23)*22, 233*1, 233*22, and 233*21.
The probability B's rolls will occur is:
a) For the six two-number rolls, (1/4).
b) For the four three-number rolls, (4)(1/4)(1/6)(1/6) or 1/36.
c) For the seven four-number rolls, (7)(1/4)(1/6)(1/6)(1/6) or 7/864.
d) For the ten five-number rolls, (9)(1/4)(1/6)(1/6)(1/6)(1/6) or 10/5,184.
The remaining terms are small enough that we can simply add a), b), c) and d) to approximate B's probability which is .28780.
The probability A wins is [1]/[1 +2] or .40430/.69210 = 58.4%.
13. Poem Puzzle
"Design"
I
found a dimpled spider, fat and white, On a white heal-all, holding up a moth Like a white piece of rigid satin
cloth -- Assorted characters of death and blight Mixed ready to begin the morning right, Like the ingredients
of a witches' broth -- A snow-drop spider, a flower like a froth, And dead wings carried like a paper kite.
What had that flower to do with being white, The wayside blue and innocent heal-all? What brought the kindred
spider to that height, Then steered the white moth thither in the night? What but design of darkness to appall?-- If design govern in a thing so small.
Robert Frost wrote
this popular poem in 1922. Assume Frost described something he actually saw (and not a painting or drawing, etc.). What
is the most logical explanation for the scene? Hint: think of a relationship between the fifth line ("mixed
ready...") and the words "satin" and "froth."
Solution
The most logical explanation is that some workers, while painting a lamppost, spilled white paint onto a spider's web that
was attached to a blue plant. There is no other rational explanation for why a blue plant would be white and a spider would
appear white. Moreover, froth is foam that forms when mixing paint and satin is a type of paint.
14. An elaborate
hoax?
Here is a philosophical problem with no true solution.
It is designed to promote conjecture more than anything else.
Suppose
a man named Bob claims he will be reincarnated. To prove this he and another person both decide on a secret number-only
the two of them know the number. Bob does this with other people. He claims after he dies and is reincarnated he will
return and disclose the secret number to each person. The problem is that it could be an elaborate hoax. Bob will
simply reveal the numbers to another conspirator before he dies and the conspirator will find an accomplice to fool those
people. To guard against this possibility the people could always ask the "reincarnated" Bob details about the location
where they chose the number, and the time of day, etc. Only the real Bob would know this information.
But here is yet another factor. The more people who know of Bob's claim, the
more likely a hoax could be arranged. Information about Bob's contacts with those people can be more readily available
when a large group of people are aware of his claim. So his claim can only be disclosed to a relatively small group
of people.
Ultimately you should ask yourself this: Can Bob even
make an airtight case to support his claim?
