During a total solar eclipse we see the moon as a dark round disc silhouetted against the sun.
But which side of the moon do we see? Standardly, we do not see an object unless the object causes our perception. But in this case it is the far side of the moon which absorbs and blocks the sunlight from us and causes us to see the moon as a dark disc. So during an eclipse it is the far side of the moon, not its near side, that we see. But that seems quite contrary to the way we think of seeing.
8 Problems
Problem 1. Let us suppose that Akitans always tell the truth and Tokyoites always lie and that all people mentioned in this example are from either one of these two cities.
(a) We meet a personAwho says of herself and of another personB: “Either I am from Tokyo orB is from Akita.” Wherefrom areAandB?
(b) We meet three people,A,B andC, two of whom make the following assertions:
A: “All the three of us are from Tokyo.”.
B: “There is exactly one Akitan among us.”.
Where are they from?
(c) Another trio,A, B andC. A says thatB andC come from the same city. We then askC:
“DoAandB come from the same city?” What will she answer?
(d) Suppose thatAsays: “Either I am from Tokyo or 2+2=5.” What can we deduce from this?
Problem 2. A logic student meets someone very interesting. She asks him: “Will you do me a favour? I will make a statement. All I ask you is that if this statement is true, you will give me your phone number.” – He agrees. – “And there is another favour I wanted to ask you: if my statement is false, you will not give me your phone number. ” – And he agrees to this too.
The logic student then makes her statement and her interlocutor discovers, much to his surprise, that to keep his promise, he must give her a kiss!
What could be the statement that has such magical powers?
Problem 3. Let us suppose that Akitans always tell the truth and Tokyoites always lie and that all members of a family come from the same city.
(a) Two brothers from one of the two cities were asked whether they were married. Their answers were:
Kentaro “We are both married.”
Ryo “I am not married.”
Is Kentaro married? What about Ryo?
(b) The same question was asked to two sisters, who answered:
Akiko “Either both of us are married or none is.”
Misako “I am not married.”
Who is married?
(c) The same question was asked to a third duo. The older of two brothers has only said that at least one of them was married. Unfortunately, all records of what the younger has said were lost, but we know at least that either he said that he was married or that he said that he was not married. We do know, however, that based on this information, their interlocutor was able to figure out who of the two was married. Do you know which one?
Problem 4. Let us suppose that Akitans always tell the truth and Tokyoites always lie.
(a) What assertion can be made by an Akitan woman or an Tokyoite man, but cannot be made by a Akitan man or a Tokyoite woman?
(b) What assertion can be made by any woman, Akitan or Tokyoite, but cannot be made by any man, Akitan or Tokyoite?
(c) What assertion can only be made by a Akitan woman, but not by an Akitan man nor by any Tokyoite?
(d) What assertion can only be made by a Tokyoite woman?
Problem 5. Ski camp: there is a nocturnal control of sleeping arrangements in the ski camp.
Three sleeping bags are found, with two people inside each of them. The instructors know that in one of them, there are two men, in another one there are two women and in a third one, there are a man and a woman. They randomly choose one sleeping bag, open it, and find one man (the other person still being hidden inside). What is the probability that the other person in
that bag is also a man?
The chief instructor, Haiko, reasons as follows:
The probability that the other person in this bag is a man is 50%. Because we have already found a man inside, it cannot be the all-female sleeping bag. So it must be one of the two others. In these two other cases, either a man or a woman is the second person.
Is he right?
Problem 6. There are three boxes, one with a prize in it, the other two empty. Here is a piece of reasoning that seems to show that the probability of finding the prize in a given box is1/2:
Let us call the three boxesA,B andC. Suppose we have chosenAand think about the (certain!) fact that one of the other two boxes is empty:
P1 IfB is empty, then the probability that the prize is inAis1/2.
(Because it’s either inAor inC.)
P2 IfC is empty, then the probability that the prize is inAis1/2.
(Because it’s either inAor inB.)
We note that the disjunction is exhaustive: either B or C must be empty! So we have:
P3 EitherB is empty orC is empty.
By eliminating the disjunction, we conclude that the probability that the prize is in Ais 1/2.
Where lies the fault?
9 Solutions
Problem 1. The Tokyoites who always lie and the Akitans who always tell the truth (after Smullyan(1978, 21–25)).
(a) IfAis a Tokyoite, thenAhas lied: thenneitherAis a TokyoitenorisB Akitan. Under this supposition, in particular, A is both Tokyoite and not Tokyoite. Hence A is not Tokyoite and must be Akitan. Hence he has told the truth andB is Akitan too.
(b) Acannot be right: if he were Akitan, he would have told the truth, so he would be Tokyoite.
Hence A is Tokyoite and there must be at least one Akitan among them. If B is also a Tokyoite, then C must be Akitan, but then what B says is true andB is not a Tokyoite.
HenceB must be Akitan and tells the truth. HenceC must be a Tokyoite.
(c) IfAis Akitan, he tells the truth andB andC are from the same city. IfCis Tokyoite (and Bis too),Cwill answer “yes”, becauseAandB are not from the same city andC is always lying. IfC is Akitan (andB is too), he will answer “yes”, becauseAandB are of the same city andCnever lies.
If, on the other hand,A is Tokyoite,Ahas lied andB andC are not from the same city. If Cis Tokyoite (and henceBAkitan),Cwill answer “yes”, becauseAandBare not from the same city andC is always lying. IfCis Akitan (and henceB Tokyoite), thenC will answer
“yes” becauseAand B are from the same city andC never lies. In all cases,C will answer
“yes”.
In all four possible cases, therefore,C will answer “yes”.
(d) The only possible conclusion is that the supposition is false: A cannot have said this. If he had said it and were Akitan, then2 + 2 = 5, which is impossible. If he had said it and were Tokyoite, he would not have lied, which is also impossible.
Problem 2. How logic helps to flirt (fromSmullyan(1997)). A sentence that will do the job is (there are many others):
(D) You will neither give me your number nor a kiss.
If (D) is true, then he will have to give his phone number (that was the first promise). But then (D) would be false. Hence (D) must be false, i.e. the logic student will either get his phone number or a kiss. But he cannot give her his number (that was the second promise). So he must kiss her to keep his word.
Problem 3. Lying Tokyoites and honest Akitans again.
(a) Because Kentaro and Ryo are brothers, either they both tell the truth or they both lie.
Because the two statements are in conflict with each other, they cannot both be true. Hence the brothers are both lying: Ryo is married but Kentaro is not.
(b) If both statements are true, both sisters are unmarried. If both statements are false, then Misako is married but Akiko is not. In both cases, Akiko is unmarried, but we cannot say anything about the marital status of Misako.
(c) If the younger brother said that he was married, then it is both possible that both statements are false (and hence that both brothers are unmarried) and it is possible that both statements are true (and hence the younger brother married and the older either married or unmarried).
So the interlocutor would not be able to draw any conclusion.
If, on the other hand, the younger brother said that he was unmarried, it is not possible that both assertions are false: if the younger brother’s statement is false, then the older brother’s statement must be true. Hence in this second case, both statements must be true, which means that the younger brother is unmarried and the older brother married.
Problem 4. Here are some sentences that do the job (there are many others):
(a) “I am a woman” – made by an Akitan woman, the assertion is true; made by a Tokyoite man, it is false. The assertion cannot be made by an Akitan man (because it would be false) nor by a Tokyoite woman (because it would be true).
(b) “I am either an Akitan woman or a Tokyoite man” – made by an Akitan woman, the assertion is true; made by a Tokyoite woman, the assertion is false. The assertion cannot be made by any man from any of the cities.
(c) “I am not an Akitan man” – true if made by an Akitan woman, false if made by an Akitan man and true (and hence impossible to make) for any Tokyoite.
(d) “I am a Tokyoite man” – the Akitans cannot make this assertion (because it would be false), and Tokyoite men cannot make it because it would not be false.
Problem 5. Haiko makes a mistake which is quite common in probabilistic reasoning. In fact, the probability is2/3, i.e. 66.67 %. We can argue for this the following way:
1. Forget about the all-women sleeping bag. In the two others, there are three men, let us call them A,B andC. Suppose that Aand B share a sleeping bag and that C is with a woman. The man we have found may, with equal probability, be any one ofA,B orC. If he is A or B (which is the case with a probability of 2/3), then the other person in the sleeping bag is also male.
2. If we forget about the all-women sleeping bag, four persons are involved, three men and one woman. One of the men we have found. The unknown person sharing a sleeping bag with him may be any of the other three, two of which are men, which gives a probability of2/3.
3. Let us change the example and take four cards, three red ones and one black one. If I pick one randomly and its red, there are three cards left, two red and one black. If I now pick one of them at random, the probability of my picking a red one isobviously2/3.
Haiko makes the mistake of being misled by an apparently binary opposition (male vs. female), whereas really there are three options to consider.
Problem 6. Reasoning about the boxes:
P1 IfB is empty, then the probability that the prize is inAis1/2.
(Because it’s either inAor inC.)
P2 IfC is empty, then the probability that the prize is inAis1/2.
(Because it’s either inAor inB.) P3 EitherB is empty orC is empty.
P3 The probability that the prize is inA is1/2.
Their appearance notwithstanding,P1andP2are not best interpreted as material conditionals.
Rather, they are statements of conditional probability:
P1′ In half of the cases whereB is empty, the prize is inA.
P2′ In half of the cases whereC is empty, the prize is inA.
To have six equiprobable cases, we need to add:
P4 In both halves of the cases where Ais empty, the prize is not inA.
C′ In two of six equiprobable cases, the prize is inA(the probability is1/3).
The mistake may be due to the fact that the probability ‘operator’, which is in effect conditional, i.e. talks only about proportions among cases where the antecedent is true, is misleadingly placed
into the consequent. Less misleadingly, the premisses are put thus: P1′′ The probability is1/2that: ifB is empty, the prize is in A.
P2′′ The probability is1/2that: ifC is empty, the prize is inA.
This is a familiar phenomenon in natural language:
conditional bets “If you’re playing another game, I bet you’ll win” does not mean that I only make a bet if you’re playing, it means that I bet that: if you play you win.
conditional promises “If you trust me I promise that I won’t let you down” makes (uncondi-tionally) a promise, even if you do not trust me. The assertion gives you the choice between lack of trust and expectation of non-deception.
conditional insults “If he does not prepare, the idiot will fail the exam” is an (unconditional) insult. Whether or not he prepares, the poor guy is called an idiot for not succeeding in the exam unless he prepares (one is led to wonder how else he should be succeeding).
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