Poll: A little math problem

[Cheesus333]

How the hell did a dog-based probability question get 9 freaking pages!?

 

[Jimmydanger]

Well the labeling problem is just because some people can't get preconceived notions out of their head to actually read the problem for what it is not what they assumed it to be. I have worded the experiment exactly as is stated in the problem without any added constructs.

 

[The Admiral]

If you go with the perspective:

puppy 1/puppy 2
m/m
m/f
f/m

then you are left with a 33% chance of one of those combinations but one of the puppies has been identified leaving you with

puppy 1
m
m
f

or

puppy 2
m
f
m

in either case the probability of one being male is 66%.

 

[Jimmydanger]

If you go with the perspective:

puppy 1/puppy 2
m/m
m/f
f/m

then you are left with a 33% chance of one of those combinations but one of the puppies has been identified leaving you with

puppy 1
m
m
f

or

puppy 2
m
f
m

in either case the probability of one being male is 66%.

What you've done there is found the probability that either puppy is a male not whether the remaining puppy is a male.
Edit: The probability that 1 of them is male is 100% I'm not sure what you found there.

To find the probability that the remaining one is male you take away the M option from 1 of the sides because you know that 1 of the puppies has to be male.
Meaning
M/M = M
M/F = F
F/M = F

or 33%

 

[Lukeje]

If you go with the perspective:

puppy 1/puppy 2
m/m
m/f
f/m

then you are left with a 33% chance of one of those combinations but one of the puppies has been identified leaving you with

puppy 1
m
m
f

or

puppy 2
m
f
m

in either case the probability of one being male is 66%.

What you've done there is found the probability that either puppy is a male not whether the remaining puppy is a male.
Edit: The probability that 1 of them is male is 100% I'm not sure what you found there.

He's found the probability if you pick a dog at random after knowing that at least one is male.

 

[Shivari]

Didn't we figure this out 8 pages ago?

 

[The Admiral]

if puppy 1 has been identified then you can only have m/m or m/f
if puppy 2 has been identified then you can only have f/m or m/m
both cases make it 50% chance

If you make the square
m / m | m / f
f / m | f / f

since one is male f/f is omitted and f/m = m/f due to the commutative property. That leaves m/m or m/f(f/m) and thus 50%.

 

[Lukeje]

Didn't we figure this out 8 pages ago?

Technically yes, but most people refuse to look through 8 pages of probability matrices in order to find the correct answer among all the other answers proclaiming to be the correct answer.

 

[Jimmydanger]

if puppy 1 has been identified then you can only have m/m or m/f
if puppy 2 has been identified then you can only have f/m or m/m
both cases make it 50% chance

If you make the square
m / m | m / f
f / m | f / f

since one is male f/f is omitted and f/m = m/f due to the commutative property. That leaves m/m or m/f(f/m) and thus 50%.

if you are using the commutative property then m/f and f/m add together equaling 2M/F not 1m/f as in your example.

this leaves 1m/m and 2m/f or 33%

 

[The Admiral]

commutative property states a+b=b+a you are saying a+b=2(a+b)

it is like 5+8=8+5=13 not 26

sorry if this has been solved my brain was starting to hurt around page 3 so i kinda skipped

 

[Jimmydanger]

commutative property states a+b=b+a you are saying a+b=2(a+b)

it is like 5+8=8+5=13 not 26

sorry if this has been solved my brain was starting to hurt around page 3 so i kinda skipped

no that's not what I'm saying. We have 3 different conditions that we have to add together MM+MF+FM but since MF=FM then they can be added together. To make it simpler we can use a little algebra MM=A MF=B FM=B
then the equation would be A+B+B=A+2B.

As for us solving it pages back no one ever agreed on the answer because it is so hard to wrap your head around. If you actually do the experiment and flip coins to simulate dogs the result after enough time will always be 33%. If you don't believe me or my math just start flipping.

 

[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"

What if they both come up tails? What are a pair of tails AKA for?

I suppose the AKA for two tails would be if the shopkeeper had said "no there is no male puppy." In our problem however the shopkeeper did find a male puppy so that data is irrelevant. We are only concerning ourselves with situations where at least one coin is heads. You may record that data or not it is your choice but at the end you are still evaluating only the other possibilities.


Also for anyone feeling bad about difficulty with this problem I found this quote about the Monty hall Problem that is based on the same logic.

"no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer."

 

[Jimmydanger]

Umm, that's a really flawed approach to creating experiments that model reality. You can't set up an experiment and say it models reality, and then when you get results that can't happen in reality, you just, like, chuck them out the window. That's a totally flawed manner in which to evaluate an experiment.

Ok maybe I explained that poorly were not throwing away the data it is just not relevant. it's like if you were collecting data about the average snowfall in winter. Snow may fall in fall but it would not be relevant to what you are studying. In this case are studying "when at least one coin is heads. what is the probability of the other coin also being heads." So just like we would have to wait for winter to collect data for our first problem we would have to wait for one coin to be heads before we collected data for the experiment.

Yeah, this problem is not based on the same logic of the Monty Hall problem. It is similar to the Monty Hall problem, but there's an extra event in the Monty Hall problem that invalidates any direct comparison of the two problems to each other:

http://www.escapistmagazine.com/forums/read/18.73797?page=8#814324

I was more referring to the counter intuitiveness and controversy associated with the two problems I shouldn't have said same logic.

 

[Seann]

I want to know who the four people are who voted 0%.

 

[Jimmydanger]

your

Umm, that's a really flawed approach to creating experiments that model reality. You can't set up an experiment and say it models reality, and then when you get results that can't happen in reality, you just, like, chuck them out the window. That's a totally flawed manner in which to evaluate an experiment.

Ok maybe I explained that poorly were not throwing away the data it is just not relevant. it's like if you were collecting data about the average snowfall in winter. Snow may fall in fall but it would not be relevant to what you are studying.

You're confusing the gathering of data from events we have no control over with constructing a predictive model we can run whenever we want, how often we want.

Lol so you think we have control over the random flipping of coins? remind me not to bet with you. We are not making the experiment less random we are simply putting constraints on it. If you want another example that seems closer how about this one.

"When a coin is flipped and lands heads what is the chance when flipped again it will be heads a second time."
The answer is obviously 1 in 2 but in order to experimentally find the answer you would have to flip a coin then IF it were heads flip again and record the data. If the first flip resulted in tails then the next flip would have no bearing on your data.