Puzzle Corner

[Laird]

During a recent census, a man told the census taker that he had three children. When asked their ages, he replied, "The product of their ages is 72. The sum of their ages is the same as my house number." The census taker ran to the man's front door and looked at the house number. "I still can't tell," she complained. The man replied, "Oh that's right, I forgot to tell you that the oldest one likes chocolate pudding." The census taker then promptly wrote down the ages of the three children. How old are they?

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The prime factors of 72 are 3, 3, 2, 2, and 2. This gives us 3-factor combinations of:

3,6,4  (sum 13)
3,12,2 (sum 17)
9,4,2  (sum 15)
18,2,2 (sum 22)
6,6,2  (sum 14)
3,3,8  (sum 14)

Since the census taker couldn't tell from the house number, we assume that the sum was 14, which has two possible 3-factors, whereas all of the other 3-factor sums are unique.

The extra information that there is "an" oldest one implies that we can rule out the 6,6,2 3-factor given that for that one, there are two oldest ones, and thus we are left with the ages of 3, 3, and 8.

During a recent census, a man told the census taker that he had three children. When asked their ages, he replied, "The product of their ages is 72. The sum of their ages is the same as my house number." The census taker ran to the man's front door and looked at the house number. "I still can't tell," she complained. The man replied, "Oh that's right, I forgot to tell you that the oldest one likes chocolate pudding." The census taker then promptly wrote down the ages of the three children. How old are they?

I'm not sure, but

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I'm reckoning the two possibilities the enumerator had to choose from were 3, 3 and 8 and 2, 6 and 6 (both of which add up to 14). Then is the reference to the eldest one meant to rule out the second possibility, because the eldest two are the same age? (Supposing the baker is reckoning from conception, not birth???)

My take on the trick.

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So the question isn't "who is right?", but "how did the psychologist come up with one in six?" Because even with the given explanation, the mathematician should still be considered right...maybe? Yes, people do tend to stop on the last event included when they say "5 of the top 8". But not always, especially if you're talking about a "top 10". And there are puzzles out there where the "trick" is to not make the assumption the psychologist made. For example, "how many months have 28 days?"

Linda

It would seem that the optimal probability of guessing right would be a weighted average of the psychologist's probability and the mathematician's.

It would seem that the optimal probability of guessing right would be a weighted average of the psychologist's probability and the mathematician's.

Yes, I think it should be an average, because the question asked was "what are the odds of your correctly guessing which three they are?" [my emphasis].

(Edit: Or if "your" can apply to them individually, they can both be right about their own odds!)

The following isn't really a puzzle but a paradox. I have never seen a satisfying explanation of it. It has various formulations. Courtesy of Wikipedia, here is one by the novelist Andrew Crumey, from Mr Mee:

Tissot showed a similar misunderstanding of my teaching when, exasperated by his continuing moroseness and his near-permanent occupancy of my writing desk, I said to him, 'Next week I am going to bring your wife here so that you can speak to her in person and sort out your difficulties. I know you don't want to see her, and so I shall not tell you which day she will arrive; but you can be sure that you'll meet her before the week is out.'

Tissot knew his wife would not be brought to confront him next Friday, because in that case he could be certain by Thursday evening that she must be coming, and he could make himself absent. But equally, I would also have to avoid Thursday, since otherwise he would be forewarned when Wednesday passed without a scene. Dismissing every other day in a similar manner, Tissot concluded that his wife could never show up unexpectedly to harangue him; but on Thursday he answered the door to be greeted not only by her, but also by her mother, both of whom boxed him soundly about the ears while I made myself scarce, quietly judging that so poor a logician deserved everything he got.

https://en.wikipedia.org/wiki/Unexpected...ng_paradox

[Laird]

I have never seen a satisfying explanation of it.

My sense (not rigorously explored) is that the solution has to do with self-reference. On Wednesday evening, Tissot believes that he has logically deduced that a Thursday scene is impossible - therefore it will be a surprise to him if it does, in fact occur. The self-reference, then, is in that his elimination of Thursday is due to a deduction, but that the deduction depends on a premise which its being false would reverse (the premise based in surprise) - and so the deduction is in a sense invalid even though its invalidity seems to depend on its validity; admittedly, this is, as you write, a paradox, but I hope that I am providing something of a sufficient explanation.

[Edit: I have tinkered with this wording a bit, trying to get it right, but it seems it would take a lot more thought before I achieved that. Also, my use of "surprise" rather than the idea of "avoidance" is inspired by earlier conceptions of the problem (the day of hanging) on that Wikipedia page]

[Laird]

OK, here's a more rigorous analysis. Let's start with the original problem in terms of a prisoner's surprise hanging, and strip it back to a single day: Friday. It is currently Thursday evening. The prisoner reasons as follows:

1. I will only be hanged if the day of my hanging is a surprise to me.
   
2. If I am hanged tomorrow (Friday), it will not be a surprise, because that is the only day left.
   
3. Therefore, I will not be hanged tomorrow (Friday).
   

Premise two is false precisely because the conclusion invalidates it: by the conclusion, it will be a surprise if the prisoner is hanged on Friday (because s/he believes that s/he has deduced that s/he will not be, and thus is not expecting it).

Now, in terms of the framing of the problem in the terms you've given it, Chris, we can adapt the reasoning as follows:

1. My wife will only arrive if she knows that I do not expect her (because she knows that if I expect her, I will be absent).
   
2. My wife knows that I will expect her tomorrow (Friday) because that is the only day left.
   
3. Therefore, my wife will not arrive tomorrow.
   

Again, premise two is false precisely because the conclusion invalidates it: by the conclusion, assuming that his wife follows her husband's reasoning, her husband will not expect her (because he has deduced by an argument that she can follow that she will not arrive).

So, I hope that that substantiates my claim that the solution has to do with self-reference. In other words, the conclusion of the argument references (alters) one of its premises.

[Laird]

Expressing the argument a little more verbosely helps us to identify even better the faulty (due to self-reference) premise:

1. I will only be hanged if the day of my hanging is a surprise to me (premise).
   
2. My hanging would not be a surprise if there were only one day left on which it might occur (premise).
   
3. There is only one day left (Friday) on which I might be hung (premise).
   
4. My hanging (tomorrow; Friday) would not be a surprise if it were to occur (inference from 2 and 3).
   
5. Therefore, I will not be hanged tomorrow (Friday) (inference from 1 and 4).
   

Premise #2 is faulty. The prisoner's hanging tomorrow could still be a surprise - even if there were only one day left - if he had reasoned, per this argument and in particular its conclusion at #5, that he will not be hanged tomorrow.

To my simple mind, it seems that starting from the premise that he can be hanged without warning, one can conclude that he can't be hanged without warning. But once one concludes the premise is wrong, then it follows that he can be hanged without warning after all.

I think the condemned man is a close relation of the set of all sets that don't contan themselves, and the barber who only shaves all the men in the village who don't shave themselves.

Sorry - of course the first thing I should have done was thank Laird for his comments on this mystifying problem.

I agree that when one tries to write out the argument logically there are problems with one of the steps. But my difficulty is that what seems to invalidate that step is the conclusion of the argument, and if the step is invalid, so is the conclusion.