Poll: A little math problem

[Samirat]

Ok its very simple.
it IS 50 % chance if you deny that you fail(exept when you count in the chance of a hermaphrodite but lets just ignore that for now because we will probably all get a headache from that).

The thing is i haven't seen it being explained properly yet (i haven't read the last page so i am sorry if its in there but its late and i am tired so i don't want to read the rest of all the silly 33.333endless theory's)i will explain it as best i can.

ok lets begin by calling one dog sparky and the other bill

so now comes the hard part

these are the things that can happen

male:bill male:sparky 25%
male:bill female:sparky 25%
male:sparky female:bill 25%
female:sparky female:bill 25%

ok now we count in the fact that one is male (see how im not only ruling out the fact that they can both be female because that is what all the 33% guys do wrong)

so now the chances are

male: bill male: sparky 25%
male: sparky male: bill 25%
female: sparky male: bill 25%
female: bill male: sparky 25%
see its that simple now the only option that remains is the the other one has a 50 % chance of being male.

Tell me, how are Case 1 and Case 2 different? You were right, until you created two identical situations. Take away one of these, and you are left with 33 percent, which is, by the way, the right answer. I'm not sure why that's in dispute. It's a premise of the original problem.

So. Your new solution set is
male: sparky male: bill 33%
female: sparky male: bill 33%
female: bill male: sparky 33%

I think by naming the dogs you made this a lot easier to understand. Thanks for that.

 

[Samirat]

All right, let me try this.

Say the first dog the dog washer checks is a female, and the second one is the male. This is one case where the other dog is female.

Now, suppose the first dog she picks up is male. Then there is a 50 percent chance either way that the second dog will be male or female. So this makes two more situations.

Out of these three cases, only 1 of them leads to both dogs being male.

 

[guyy]

I think the simplest way to explain it is this:

Before asking if one of the dogs is male, there are 4 possible scenarios: M/M, M/F, F/M, and F/F.
By learning that one is male, the fourth possibility gets eliminated, leaving only M/M, M/F, and F/M.
But we want to know about the other one, so removing the counted M from each one, you just get M, F, and F.
1 M out of 3, so it's a 1 in 3 chance.

Hopefully that's not gibberish. It's very counterintuitive, but 1 in 3 is definitely the right answer.

 

[werepossum]

SNIP
--the chances of one pup being male are 100% and the chances of the other pup being male are 50%

Why is that either incorrect, OR, why doesn't it require us to eliminate one of the pair of F/M or M/F?

We actually don't yet know anything about either pup; both still have a 50% chance of being male at this point. The only new piece of information we get from the groomer is about the set, not about individual pups. That is why we can only eliminate the possible distribution that both are female. Only when we get information about a specific pup - say, the brown pup is male, the older pup is male, the smaller pup is female - can we take away one possible distribution. At that point, we know something about one member of the set as well as something about the set, rather than only knowing something about the set itself.

 

[Ancalagon]

Okay, I'd promised myself I was quitting this thread, cold turkey, but I can't help myself. I've looked out my old university text book on probability, and looked for a similar problem. Here it is. I appreciate that the problem is different, but if you can follow the logic for this, you should be able to do the same for the beagle problem.

http://www.flickr.com/photos/26994968@N05/2935840081/sizes/l/

http://www.flickr.com/photos/26994968@N05/2935845645/sizes/l/

http://www.flickr.com/photos/26994968@N05/2935830717/sizes/l/

It's the three card problem, and the solution is Solution 3.2.

 

[TheGhostOfSin]

What do you mean? An African or European beagle?

I love you.

 

[Geoffrey42]

I'm not good enough at probabilities to figure this out, but I think the answer might be in the fact that you're not factoring in the probability that the first dog he picked up was male vs. the probability that the first dog he picked up was female.

In other words, it's not that we're eliminating the possibility that the dogs are in reality F/F, although that is true. It's that for purposes of probability, we have to also eliminate the chance that he might pick up an F puppy first that comes from the previously existing possibility that the puppies were F/F. I don't know how the numbers break down, but I do know that you have to eliminate more than just F/F.

You're giving too much import to the order. Out of the 4 starting options, F/F, M/F, F/M, M/M, the male being first, or the female being first, doesn't matter. It might as well be 1xF/F, 2xM/F, 1xM/M. When we learn that one of the dogs is male, we still have 2xM/F, 1xM/M. The order itself was never relevant (to this question).

If the question were: "given two dogs, what are the odds that the first one picked up turns out to be male?" then we suddenly care about order, because M/F and F/M are distinct options, making up 25% each.

 

[Easykill]

225 posts later... What?

 

[jim_doki]

I saw something like this on Numbers once...

it went if there were a choice of three when you made your choice, and then one was revealed to be wrong, you doubled your chances of being right if you changed your answer.

I think it's 66.6% based on the following (and i want to point out i suck at maths):
we have three scenarios here, either one dog is male, the other female, one dog is femail and the other male, and there are two males. if we knew which dog was male, then it would be 50 50. we don't, so we have to assume there is an equal chance that either dog is male.
so if we take that under consideration, it works out that there are two out of three scenarios that work out the "Other Puppy" is a male

ive gone cross-eyed

 

[Birras]

This exacerbates things a bit...

 

[werepossum]

SNIP
--the chances of one pup being male are 100% and the chances of the other pup being male are 50%

Why is that either incorrect, OR, why doesn't it require us to eliminate one of the pair of F/M or M/F?

We actually don't yet know anything about either pup; both still have a 50% chance of being male at this point. The only new piece of information we get from the groomer is about the set, not about individual pups.

That's not true, unless like I keep saying puppy gender is a quantum event. What we know about the set allows us to know something about the individual members of the set that we didn't know before. We can now be 100% certain that the set contains a male puppy, so how can saying they both still have a 50% chance of being male or female be correct?

We can now be 100% certain that THE SET contains at least A male puppy. But F/M and M/F are two different conditions - the dog that is male is different between the two, so there is still an equal chance of either being true. And there's an equal chance that M/M is true. Thus the 33% chance that the other dog is male, because in the two remaining options (F/M and M/F) the other dog must be female.

We still have no additional information about the individual pups, JUST ABOUT THE SET. Knowing something about the set does not necessarily tell us anything about the individual items in the set, although granted, in some special conditions it might. If we know that a set of ten pups has ten males, we know the sex of each individual pup. However, if we know that a set of ten pups has nine males, we do not know the sex of each (or any) individual pup; we have only the probability distribution. We can say that each pup has a 9 in 10 chance of being male, but we do not know the sex of any given member of the set; thus the sex of each pup in the set is still indeterminate. If we say the youngest pup is female, then granted, we now know the sex of all ten pups. But if we say the youngest pup is male, then we still do not know the sex of each individual pup except for the youngest. For the others, the chance of being male shifts from 9 in 10 to 8 in 9; thus we have added information, but our guessing ability for each individual pup has actually worsened. This should emphasize that added information about the set does not translate into information about individual items in the set unless with the added information the set becoems homogenous.

On the other hand, if in our two-pup set we learn that the older pup is male, now we know something about an item in the set AND thus about the set itself. If we assign the pups within the set to positions by age, eldest first, then we can eliminate F/M as well as F/F. Our remaining probability distributions are M/F and M/M, yielding a 50% chance the second dog is male. If we assign the pups within the set to positions by age, youngest first, then we can eliminate M/F as well as F/F, with the same net result. If we know ONLY that F/M OR M/F OR M/M is true, then we can eliminate ONLY F/F. That is the difference between knowing something ONLY ABOUT THE SET and knowing something ABOUT THE SET'S ITEMS.

Am I not saying this right? This shouldn't be this hard to understand. I figured it out in an eighth grade IQ test without ever having studied it; it shouldn't be this hard for obviously intelligent people to understand even given only Saskwach's explanation.

EDIT: I should add that no, the pups' sexes are not quantum occurrences. They are determined even before the pet store lady buys them, determined at birth - we just don't know what they are. What we are try to do is establish a probability for both pups being male without actually learning the sex of each pup.

 

[Apocalypse Tank]

Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

Never thought to use the Monty Hall problem for this. Nice.

Yeah good analysis I was thinking that too.
That or the 50-50 female-male ratio
Argh, either way, the question is simple and stupid.

 

[Fruhstuck]

maralin vos savant says it's 33% so i go with her lol
i'm not going to win any argument against her

 

[Samirat]

You're giving too much import to the order. Out of the 4 starting options, F/F, M/F, F/M, M/M, the male being first, or the female being first, doesn't matter. It might as well be 1xF/F, 2xM/F, 1xM/M. When we learn that one of the dogs is male, we still have 2xM/F, 1xM/M. The order itself was never relevant (to this question).

I say we only have 1xM/F and 1xM/M once we learn that one of the dogs is male:

http://www.escapistmagazine.com/forums/jump/18.73797.811307

I think the problem is that we've got Mendel's genetics in the back of our heads, and that is making us put in F/M and M/F the way we'd put Tt and tT if we were doing an offspring problem:

http://www.escapistmagazine.com/forums/jump/18.73797.811411

All right, let's do this slowly. Or slower.

I'm going to use a 2 coin analogy, because it's easier to work with. Let's say that you flip two coins, and it is guaranteed that at least one of the landed heads. The question is what the probability is that the other landed heads as well. Look okay?

All right, do you agree that the chances of getting one head and one tail are .5, while the chances of getting both of each is .25?

In this case, would this be an acceptable range of outcomes for you?

HT
TH
TT
HH

Do you agree that the chances for all of these are equal, at .25.

Now in how many of these are you guaranteed a heads? 3.

And only in one do you have another heads.

I'm not sure why you would eliminate one of the one tails one heads options. Because this outcome is twice as likely, it needs two spots on the sample space.

I think your error is assuming that the first coin examined is a heads. You mentioned earlier that it is like laying down one heads and only flipping the other one. That gives you two options, 1 heads and 1 tails, or 2 heads. But if the first coin examined is a tails, you have only 1 option, since one of them has to be heads. So, two options for the first, one for the second. Three situations, of equal probability. What you're doing is assuming a 100 percent probability that the first coin is heads. Having 2 tales isn't even an option in the problem, since it expressly states that there is at least one heads. So if you happen to roll two tails, you can very rightly flip again, since this situation doesn't even exist in the problem's premise. The two coins are flipped at the same time, there are 3 options, since one is not an option at all, and all of these will satisfy the problem's basic premise. It's not like having one heads

Okay, let's say that the two coins are differentiated somehow. Coin 1 and Coin 2. Knowing that one coin turned up heads doesn't mean that Coin 1 turned up heads. Either could be heads, or both. Again, three different situations. Coin 1 is heads, Coin 2 is tails. Coin 2 is heads, Coin 1 is tails. Or both are heads. Only one results in another heads.

All right, let me try another avenue of attack. Say you expand this problem to include 50 coins. If you are guaranteed at least 50 heads, is the probability that the other 50 are tails the same as the probability that all 100 are heads? Of course not. You don't lay down 50 heads and then flip 50 coins. The 50 heads and 50 tails are the most probable solution in a 100 coin toss, and the probability of that outcome, while still very low, is much much greater than the probability of tossing 100 heads in a row, since 50 heads and 50 tails can be arranged countless ways. The dog problem is no different, just on a smaller scale.

This is it, this is all I can do. If it's still not doing it for you, I'm sorry.