8.
The Alternating Liar, Truth-teller, and Normal guy
You meet three people, the Alternating Liar,
Truth-teller, and Normal guy. 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. To every other question
posed to him, the alternating liar will lie, otherwise he tells the truth. He will lie to the first question you pose to him.
Another guy always tells the truth. The third is a normal guy-- he can answer "yo" or "nes" to any question
posed to him regardless of whether it's the truth. The three also know which of them is the alternating liar, truth-teller,
and normal guy. 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. However, if you ask the same person two questions, both questions must
be identical. How can you determine the identity of all three 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, You are the alternating
liar and The answer to the question of whether C is the normal guy is 'yo', is one and only one of them true?"
If the
response is "yo" then either A or C is the normal guy. If it is "nes" then either A or B is the Normal guy. You still dont
know what "yo" or "nes" means but the identity of the Normal guy is narrowed to two possibilities.
If A responds, "yo" to question 1 then you ask B:
2) "Of the two statements, You are the alternating liar and The answer
to the question of whether A is the normal guy is 'yo', is one and only one of them true?"
If the response is "yo" then
A is the Normal guy. If it is "nes" then C is the Normal guy.
If A responds, "nes" to question 1 then you ask C:
"Of the two statements, You are the alternating liar and The answer
to the question of whether A is the normal guy is 'yo', is one and only one of them true?"
If the response is "yo" then
A is the Normal guy. If it is "nes" then B is the Normal guy.
Once you have determined the identity of the Normal guy, you then repeat the question you had previously asked the person
you know is not the normal guy. If he changes his response he is the alternating liar. If not, he is the truth-teller.
9.
The Red hat/Blue hat problem
Here's a problem that invokes considerable
conjecture. 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 a slight variation. Only
one die is rolled at a time. A bets two consecutive 6s will be rolled before three consecutive rolls totaling five will
be rolled (i.e., 1, 2, and 2 or 1, 1, and 3 in any order). They continue rolling the die until a player wins. What is
the probability A wins? A calculator is needed. Note: this problem is hard.
Hint:
it is advisable that you solve problem a) first.
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
b) The scenario of rolling three die that total five is much more complicated. Only when a player rolls a 4 or 5 can neither
player win and we are back to square one. In all other cases either player can still win. Because 1, 2, and 2 and 1, 3, and
1 each can be arranged in three distinct orders, the probability a five appears in any three consecutive rolls is 1/36. This
is the same probability as two consecutive 6s being rolled. Therefore we know A still has a slight advantage but it is less
than in problem a) because B requires three individual rolls as opposed to four in problem a).
Another method exists for calculating any similar dice probability question. One way is by adding up all of the individual
probabilities that any single roll will be the winning roll for that player. Since an infinite number of terms exist, we dont
need to add up every single term. We only need to compute enough terms so we can project the ratio of the remaining terms
for both A and B.
Since we already know the answer to part a) here's how this approach can be used to solve part a). The approach involves a
tedious process in which you have to compute the number of possible dice combinations that will win for either A or B. We
figure A probability (which we call P). A has a 1/36 chance of winning before the first roll of dice. Actually before any
dice roll A has a 1/36 probability multiplied by the probability that neither A nor B won on the previous dice roll(s). Since
only A could win on the first dice roll, P for the 2nd roll was 35/36 (1/36). For the 3rd dice roll the probability is [17/18
(35/1,296)] (1/36) or .917438 since we have to factor in the probability that either A or B won on the first two rolls. We
always compute the number of possible winning combinations for A and B and subtract that number from 1/36 to determine the
probability. For the 4th roll P is [17/18 (29/36) (1 pp) - 30/7,776)] (1/36). The last term, 30/7,776, is always 180 times
the number of excluded combinations in the second term preceding that term divided by the number of possible combinations;
PP is the probability of the previous roll times 36.
So, A's P looks like this for part a):
We also apply the same process in computing B's probability (P) of winning in part a). B has a 1/36 chance of winning before
the second roll of the dice (assuming the first roll was a 7). Before any subsequent dice roll B has a 1/36 probability multiplied
by the probability that neither A nor B won on the previous dice roll(s). Before the 3rd roll the probability is 29/36 (1/36)
because the first roll cannot be a 12 or a 7 (otherwise A or B would have won on the first or second roll). Before the 4th
roll P is 1,015/1,296 or .783179 because the first roll cannot be a 12 and the second roll cannot be a 12 or a 7. Before the
5th roll P would be [17/18 (29/36) (1 pp) - (210/7,776)] (1/36) or .742777 where pp is the probability of the previous roll
multiplied by 36. Before the 6th roll P would be [17/18 (29/36) (1 pp) - (8,430/279,936)] (1/36) or .707122.
The sum of A's and B's probability must equal one. Moreover we also note that the terms diminish proportionally. In other
words, .874078/.742777 roughly equals .83154/.707122. Therefore, the following equations are true:
When solving for R we find the probability A wins is 51.275%.
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. Assuming Frost was describing something he actually saw in nature, 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?
