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The Dogs are not Particles; Discuss! Don Paolo's solution inside
Topic Started: Jul 8 2009, 05:47 PM (1,677 Views)
Don Paolo
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If you guys want to continue this debate here's the place to do it.

Don Paolo's Solution:

1/3, because there are four equal-chance possibilities at first (MM, MF, FM, FF), the shopkeeper narrows it to three (MM, MF, FM), and one of these three is MM.

When she shopkeeper says that one of them is male we do not know which is male the first or the second in the set. This is why HAVE to treat these as a pair. These can be described as MaleA(Ma) or MaleB (Mb). At this point we can establish all possibilities. There is only one possible female so we cannot account for two:

1. Ma + Mb 2. Mb + Ma 3. Ma + F 4. Mb + F 5. F + Ma 6. F + Mb

There only two out of these six which are all male, options 1&2. Hence 2/6 reduces to 1/3.

Some would argue 2/3. The order of the puppies is irrelevant, so the different options are MM, MF and FF. It is obviously not FF. We now have two possibilities: MM and MF. I will now label the two male puppies Ma and Mb. So our options are MaMb or MF. The person bathing the dogs checks to see if one of them is a male. He is either looking at:

* Ma, and so the other is Mb.
* Mb, and so the other is Ma.
* M, and so the other is F.

Thus the probability of it being MaMb (ie MM) is 2/3, right? Wrong.

Unlike particles, these dogs do not cease being distinct entities because the only thing you know about them is whether they are male or female. MF and FM are always different, though in permutations you count them as two instances of the same category. That doesn't mean you can ever ignore them.

Specifically, the three possibilities are MF, FM, and MM, which we know are all equally likely because we assume each dog initially had a 50% chance of being male or female. Which dog the checker looked at is irrelevant; in fact, why not assume he looked at both and just answered the question honestly (i.e., at least one was male, but he's not telling what the other was)? Regardless, all three possibilities are equally likely, so the solution is: 1/3
Edited by Don Paolo, Jul 8 2009, 05:55 PM.
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cjcurry

This is garbage. Permutations are not the way to solve this problem. Male-female is exactly the same as female-male. I will argue this to the grave if I have to, but applying permutations here is not the right approach.

One key point to my argument is that birth order is not important. If birth order was important, then I would say it is 1/2. But never have I seen a bigger logical and mathematical fallacy than this so-called "solution".
Edited by cjcurry, Jul 8 2009, 06:13 PM.
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Giant-Sandbag

Partical Dog would make a great superhero. Just saying.
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Don Paolo
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Here's a more intuitive way of thinking about it: what the bather has told us, by saying that at least one of them is male, is equivalent to just saying "they are not both female". And what the question is asking, by asking what the probability that the other is male as well, is equivalent to "what is the probability that they are both male". So read the question in full as "what is the probability that both dogs are male, given that they are not both female?" and it is much easier to see why 1/3 is the correct answer.

So, lets look at the case of a lazy checker. (you can apply this two the N dog case as well to make it more tedious, but obvious)

With a 50% chance, the first dog was the first male found with a 50% chance, the second dog was the first male found (1+ males found as premise) if the first dog was male, there is a 50% chance of a second male. if the second dog was the first male, there is a 0% chance of a second male. so, in 100 cases, 50 would have found the male first. 50 second. of these cases, 25 would have two males. this is 25 cases with two males, 75 cases without. or 1/3rd of the cases.

An interesting corollary is the situation where the two dogs are completely indistinguishable except for gender; that is, they behave like particles and Heisenberg applies (always a reasonable assumption for dogs, I know). In this case, the probability is indeed 50% because the FM and MF cases are treated as the same.
Edited by Don Paolo, Jul 8 2009, 06:12 PM.
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Tea Alchemist

Giant-Sandbag
Jul 8 2009, 06:08 PM
Partical Dog would make a great superhero. Just saying.
Particle Dog, Particle Dog, doing the things a particle can, what's he like? It's not important!

Hmm, doesn't quite have the same ring to it.

Really though, I think that in the context of the puzzle it could only have three possible options, not two. Although the way that the final question is worded does sort of indicate a "one or the other" kind of scenario.

Also, dogs are totally made of particles. Awhyeah!

... Maaaaybe we shouldn't let ourselves get too distracted by debating this.
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Pseudonym

cjcurry
Jul 8 2009, 06:04 PM
This is garbage. Permutations are not the way to solve this problem. Male-female is exactly the same as female-male. I will argue this to the grave if I have to, but applying permutations here is not the right approach.
Is it? I mean lets say unknown to us one dog was black and one white. A white female pup and a black male pup arn't really the same set as a white male pup and a black female pup. Am I thinking about this right? Or am I barking up the wrong tree?

(I gotta get my mind around all of this by the time I get home. My family thought it had to be 1/2, so now I have to explain 1/3!)
Edited by Pseudonym, Jul 8 2009, 06:19 PM.
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Don Paolo
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Tea Alchemist
Jul 8 2009, 06:14 PM
Giant-Sandbag
Jul 8 2009, 06:08 PM
Partical Dog would make a great superhero. Just saying.
Particle Dog, Particle Dog, doing the things a particle can, what's he like? It's not important!
It's okay I was thinking this too.


And I'd love you to put all your energy into this debate and not the real puzzles. Yes, that would be... ehhhcellent.
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cjcurry

The female-male pair is the same as the male-female pair. Yet, either dog from the male-male pair could easily be chosen as the "male" that the bather refers to. This means that, given one of those two dogs is chosen, in two cases out of two will the other dog be male. Adding on the zero cases out of one with the male-female pair gives two cases out of three.

A correct proof in only five lines.
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cjcurry

Pseudonym
Jul 8 2009, 06:16 PM
cjcurry
Jul 8 2009, 06:04 PM
This is garbage. Permutations are not the way to solve this problem. Male-female is exactly the same as female-male. I will argue this to the grave if I have to, but applying permutations here is not the right approach.
Is it? I mean lets say unknown to us one dog was black and one white. A white female pup and a black male pup arn't really the same set as a white male pup and a black female pup. Am I thinking about this right? Or am I barking up the wrong tree?
You are barking up the wrong tree, because colour has no significance. Neither does birth order.
Edited by cjcurry, Jul 8 2009, 06:20 PM.
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cjcurry

It's 4:30 in the morning where I am. I need sleep. But believe me, I will resume my post when I wake up.
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