Poll: A little math problem

[Geoffrey42]

My sticking point is that I don't accept that we don't change the numbers once one of the events goes from a probability to a certainty.

As you said before, since this is math, isn't half the point knowing whether we got to the answer the right way? The only reason we care about that is for the sake of reproducibility. When you have two competing methods/models, that both accurately predict the outcome of a given scenario, the best way to find which one is right is to provide an example where one of them fails, and the other succeeds. Can you think of an example where our model fails? Up to that point, it is absolutely semantics; two ways of getting to the same place, reliably and speedily, and nothing but a pillow fight over "my way being better".

 

[Ancalagon]

In the case of coins, I'm not sure why you think a reflip is inappropriate, in the situation of two tails.

Because you can't change the sex of the puppies or pick a new set of puppies every time you get two females. So if we're using coins as a stand in for puppies, they have to obey the same rules as puppies.

The reflip wouldn't be modelling a hasty gender reassignment, it just models the fact that we know that two tails cannot have happened, or we cannot have had two female dogs to start with, so if we get it, we reflip. If there were a way of making it so that two flipped coins turning up tails simply couldn't happen, then we could do that instead.

All right, relating to you problem with Alex_P. To solve this problem, you simply place the number of pairs with 2 males over the number of pairs with at least 1 male.

Out of 100: 25/75

The sample space of the problem are the 75 pairs that contain at least 1 male, this fraction represents the number of pairs of dogs out of that where the other dog is a male. You made the fractions yourself.

Everything changes once we know one of the events--which are independent--is no longer in the realm of probability, but has become a certainty.

Absolutely. The only thing that changes the probabilities throughout the whole exercise is that we find out that at least one of the two dogs is definitely a male.

We start off with:
Dog 1/Dog 2
M/M
M/F
F/M
F/F
all being 25% likely. Where we go our different ways is when we are told that at least one of the two dogs is definitely a male. How does that affect the probability that each of these was the starting position? F/F cannot have been the starting position, since we know that at least one of the two dogs is definitely a male. So we change it to:

Dog 1/Dog 2
M/M 33% likely
M/F 33% likely
F/M 33% likely
F/F 0% likely

If on the other hand we were told that Dog 1 was a male, then F/M could not have happened either, so we would have:

Dog 1/Dog 2
M/M 50% likely
M/F 50% likely
F/M 0% likely
F/F 0% likely

But that isn't what we're told.

Okay, from first principles:

First, what can we agree on:

1.) There are two puppies, which we can call Dog A and Dog B.

2.) There are four equally likely combinations of sex that the two puppies can have: MM, MF, FM, FF, where the first letter indicates the sex of dog A, and the second letter indicates the sex of dog B.

3.) That if you remove one equally likely combination from consideration, the rest remain equally likely to one another.

Okay, I don't think anyone is claiming that any of these is false.

So when you remove FF from consideration, you still have three equally likely starting positions that are valid. Dog A could be male, and Dog B could be male. Dog A could be male, Dog B could be female. Dog A could be female, Dog B could be male.

Now what most of the 50%-ers have been saying is that you don't have three equally likely outcomes, you have two. The dog in your hand is male, the dog not in your hand is either male or female. So you have MM and MF, where the first letter denotes the dog in your hand, the second letter denotes the one that isn't. The problem is that these situations are not equally likely. Why are they not equally likely? Because the dog-bather chose a dog. He may have looked at both dogs, found a male and a female, and picked up the male. If, on the other hand, he had picked up a dog at random, not looked at the other, and the one he picked up was male, then there would be a 50% chance that the other is male.

Another way I've heard it put is that if you're saying that the starting situation FF must be removed, so must either MF or FM. Why? Which one would you remove? If you remove the first, what you're saying is 'Dog A can't be a male, while Dog B is a female'. If you say the other, then you're saying the opposite. You cannot make either of these assumptions based on what you know.

How do I know that my three combinations MM, FM, and MF are equally likely? because they always have been. We agreed that they were equally likely in the beginning. Nothing has happened to make any of these situations any more or less likely. So we have three equally likely starting situations, two of which lead to the other dog being female, one of which leads to the other dog being male. The problem comes from people thinking that given the set:

Dog A/Dog B
M/M
M/F
F/M
F/F

as equally likely initial starting points, it can be changed to:

Dog in Hand/Dog not in Hand
M/M
M/F
and that these two situations are equally likely. They aren't.

 

[Geoffrey42]

Thing is, we're not getting to the same place. Let's say this is a bet on picking a pair of male puppies. Even money when we don't know either puppy's sex is 4-1 odds, right? I'm saying that once we know that one puppy is male, that even money goes down to 2-1 odds. Everyone else is saying that the odds only go down to 3-1 for even money, if I'm doing the math right.

I misunderstood your earlier comment then. I thought we were all in agreement about the 33%, and the sticking point was how we get to that number (whether it be a recalculation of the odds, a reduction of the result set, whatever).

Let's try: puppies are paired at the puppy farm randomly. It turns out that for this particular breed of dog, pairs of females always instantly kill one another. Therefore, the universe of possible puppy pairs only consists of MM and MF pairs. Due to the nature of the universe (this is not an assumption of the problem, just a statement about the nature of the universe when sample sizes are sufficiently high), there are twice as many MF as MM pairs. What are your break-even betting odds of getting a MM pair?

How does the above differ from the original question?

 

[Death Magnetic]

1 in 3 baby.

-Ricky

 

[guyy]

SNIP
So why doesn't the 'set' change like I described in comment #175

from:

Puppy 1/ Puppy 2
M M
M F
F M
F F

to:

Referred To Puppy/ Other Puppy
M M
M F

The things you know are that there are two pups, that at least one pup is male, and that the chances of any pup being male are 50% - you can make no other statements of knowledge. That eliminates only one possible distribution out of four, that both are female. Ergo there are three remaining possible distributions. Of those three remaining possible distributions, only one results in two males; thus the percentage of two males is 1 in 3 or, expressed as an integer percentage, 33%.

This is where you go wrong.

Or more concisely: the question isn't "Is puppy #1 male?"; the question is "Is one of them male?", so possibility 3, "F M", is not eliminated. Only "M M" would make the other one also male, so the chance is 1/3.

 

[Ancalagon]

The reflip wouldn't be modelling a hasty gender reassignment, it just models the fact that we know that two tails cannot have happened, or we cannot have had two female dogs to start with, so if we get it, we reflip.

But that's not what we know. That's something we've deduced from knowing that one puppy is male. My problem is that you can't just incorporate half a fact--one puppy being male has more repercussions for our knowledge than just "we cannot have had two female dogs to start with"

There's a difference between being told that a particular dog is a male, and being told that out of two dogs, at least one of them is a male. You are being told that in this set of two dogs, at least one is a male, so M/M, F/M, and M/F are all valid. You aren't being told that a particular dog is male, which would reduce the likelihoods to either F/M and M/M, or M/F and M/M. Ah! But surely we do know one of them in particular is a male - the one in his hand is a male. Yes, that's true, but which dog is in his hand? Dog A or Dog B? He could've picked either up, and our initial supposition that MM, MF, FM and FF were equally likely was based on a Dog A/Dog B basis.

Now what most of the 50%-ers have been saying is that you don't have three equally likely outcomes, you have two. The dog in your hand is male, the dog not in your hand is either male or female. So you have MM and MF, where the first letter denotes the dog in your hand, the second letter denotes the one that isn't. The problem is that these situations are not equally likely. Why are they not equally likely? Because the dog-bather chose a dog. He may have looked at both dogs, found a male and a female, and picked up the male. If, on the other hand, he had picked up a dog at random, not looked at the other, and the one he picked up was male, then there would be a 50% chance that the other is male.

EDIT: Sorry, now I get you--there's a difference between 'dog in your hand' and 'dog referred to when saying 'yes'. That's where you're getting stuck. Whichever order he checked them, if he picked them up, etc. is irrelevant because we can't know about that. What we *do* know is that when we ask him the question, he's referring to one of the dogs and removing all probability that the dog is female.

It's subtle, but very meaningful, the difference between 'dog in your hand' 'dog checked first' 'dog checked second' and what I'm talking about, 'dog referred to when answering the question'.

You say that the how he checked the dogs is irrelevant, because we can't know how he did it. But we can suppose that he at least picked one of the dogs up. The chance that this first dog was a male is two-thirds, since we know it could be F/M, M/F, or M/M, and in two of those cases, the first dog he picks up will be a male. A third of the time, it'll be a female, so he will necessarily have to check the second dog. If this is the case, we know that he found it to be a male, since if he had not, that would give us an invalid state (two female dogs). Now, in the two thirds of the time the first dog is male, what sex is the other dog? It doesn't matter if he checks or not, the chance of that dog being male is 50%. Why? Because the first dog has already satisfied the assertion that one of the dogs is male, and that assertion no longer affects the probability.

So, the First dog is male two-thirds of the time. One of those thirds of a time, the other is male. The other third, the other is female. One third of the time the first dog is female. If the first dog is female, the second must be male. So only one third of the time are both dogs male.

 

[Lukeje]

Thing is, we're not getting to the same place. Let's say this is a bet on picking a pair of male puppies. Even money when we don't know either puppy's sex is 4-1 odds, right? I'm saying that once we know that one puppy is male, that even money goes down to 2-1 odds. Everyone else is saying that the odds only go down to 3-1 for even money, if I'm doing the math right.

I misunderstood your earlier comment then. I thought we were all in agreement about the 33%, and the sticking point was how we get to that number (whether it be a recalculation of the odds, a reduction of the result set, whatever).

Let's try: puppies are paired at the puppy farm randomly. It turns out that for this particular breed of dog, pairs of females always instantly kill one another. Therefore, the universe of possible puppy pairs only consists of MM and MF pairs. Due to the nature of the universe (this is not an assumption of the problem, just a statement about the nature of the universe when sample sizes are sufficiently high), there are twice as many MF as MM pairs. What are your break-even betting odds of getting a MM pair?

How does the above differ from the original question?

You're specifically looking for a male pair of dogs, not a single male dog after knowing that the first is male.

 

[Lukeje]

You're specifically looking for a male pair of dogs, not a single male dog after knowing that the first is male.

Maybe before you knew one of the puppies was male, but now that you know one is male, the inquiry has changed.

No; the main difference between the 1/2 crowd, and the 1/3 crowd, is their different interpretation of the question.
1. If you assume the question is asking what the probability of two dogs being male when there is no chance of both being female (as the above question states)
2. If you assume the question is referring specifically to the dog labelled 'Other'
(3. If you assume that the dogs are then going to be randomly selected, so are wondering what your odds of getting a male dog are (2/3))

BUT no-one has yet answered why the question itself makes no sense. If the person walks into the shop wanting a male puppy, why does he not just accept the first male one? Why would he ask if there were a second?

 

[bad rider]

what does it mean by pair?

 

[bad rider]

Aha, i have it, the other is a cleverly disguised squirrel, so the answer is squirrel.

 

[Jimmydanger]

Ok to everyone who thinks that the answer is 50% I can prove to you that it is in fact 33%. Forget all this complicated math stuff that has been posted so far its correct but forget it. Take two coins. No not theoretical coins real coins in your hands. These represent the puppies. Heads is male tails is female.

Flip the coins.

ask yourself "is one of these coins heads?" AKA "Is one of the dogs male"

If one is heads then ask "Is the second coin heads as well" AKA "is the other dog male"

Do this at least 20 times recording the results the more flips the better. If you have time to write a post you have time to flip 2 coins 20 times. The results will prove who is right.

my results were 14 no 7 yes exactly 35%

 

[bad rider]

Ok to everyone who thinks that the answer is 50% I can prove to you that it is in fact 33%. Forget all this complicated math stuff that has been posted so far its correct but forget it. Take two coins. No not theoretical coins real coins in your hands. These represent the puppies. Heads is male tails is female.

Flip the coins.

ask yourself "is one of these coins heads?" AKA "Is one of the dogs male"

If one is heads then ask "Is the second coin heads as well" AKA "is the other dog male"

Do this at least 20 times recording the results the more flips the better. If you have time to write a post you have time to flip 2 coins 20 times. The results will prove who is right.

my results were 14 no 7 yes exactly 35%

You are aware your test is based entirely on luck?

 

[Jimmydanger]

do the test yourself the more times you flip the more accurate your results if you do it 100 times luck stops mattering

 

[Lukeje]

do the test yourself the more times you flip the more accurate your results if you do it 100 times luck stops mattering

Technically as the number of tosses approaches infinity luck stops mattering.
It also doesn't solve the problem for anyone who interprets the fact that one dog can be labelled as 'other' suggests that we only care about the outcome of one dog. This would be the equivalent of tossing one coin till you get 'male' then tossing another coin.