Poll: A little math problem

[Alex_P]

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"

So, other than "the set must contain at least one male," what are the repercussions of "at least one puppy is male"?

-- Alex

 

[Jimmydanger]

So, other than "the set must contain at least one male," what are the repercussions of "at least one puppy is male"?

When we adjust our probability matrix, we can't just pull out the F/F, we also have to pull out either one (exclusive) or the other of M/F or F/M. It doesn't matter which, but one of them has to go. One dog's sex is no longer a matter of probability but of certainty, and leaving both in reflects a situation where both dogs' sex is still a matter of probability.

Ok so that is where you are making your mistake. At least one puppy is male does not take out one of the M/F. Neither of the dog's sex is a certainty.

Say you could see both dogs were in boxes and you could not see them and someone told you that at least one was male. Then he pointed at one box and asked whether it was male or female. You would not know. Same situation if he pointed at the other box. You still have all three probabilities M/M M/F F/M. The only thing that "at least one dog is male" tells you is that BOTH are not female.

 

[Samirat]

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.

*Even if you don't want to read the first part, I request that you read the second. It might help you understand a little bit*

Whoa, you don't understand this at all, eh?

You would manipulate your experiment at the start, and call it realistic?

Do you deny that at the beginning, there is a random chance that either dog could be male or female? Putting one coin down as heads in the beginning is assigning one dog a gender, at the start. No, you have to find a random situation which is analogous to this. If you flip two coins, and ask someone whether one is heads, there is a chance he could say yes or no. It is not a 100 percent chance that he'll say yes, which is what you're assuming. You're buying one male dog, and then one of random gender. Essentially, not the same as the problem, at all. No. Here, there's a random chance that the washer could say "yes," or "no."

Let me propose an extension to this problem, it might help.

So, say you have two coins, Coin 1, and Coin 2. Just like this problem, when you ask if one of them is heads, the answer is "yes." So you lay down Coin 1 as heads. Then, in the problem, you ask if one is tails, and the answer is "yes" so you lay down Coin 2 as tails. This is how you would approach this problem, correct? Yes, based on what you've said so far.

So, what are the odds that Coin 1 is heads? 100 percent? Really? No. The odds are 50 percent. Because in the solution set:

MM MF FM FF, you have ruled out option MM, and option FF, but it could be either MF or FM.

 

[Jimmydanger]

"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.

Okay, at first glance that looks fine to me, but, unless you've now going to show me data from that experiment...what's your point?

My point is that in this experiment we are waiting until the first flip is heads before we collect data and in the other one we are waiting for "at least on coin to be heads" before we collect data.

the data you are calling "impossible" is not impossible just irrelevant.

 

[Jimmydanger]

So, other than "the set must contain at least one male," what are the repercussions of "at least one puppy is male"?

Think of it this way--two dice have been rolled, and they are under separate cups. I've bet an equal amount of money on each to come up even. The matrix would look like:

Even Even
Even Odd
Odd Even
Odd Odd

right? I've a 25% chance to make money, a 50% chance to break even, and a 25% chance to lose money.

Now, someone tells me I've won one of my bets. What are my chances of breaking even vs. winning now? How would you draw the new matrix?

All you know when the man says you won one bet is that you didn't lose them both, removing odd odd.

Even Even
Even Odd
Odd Even

you would win at least one bet 75% of the time. 50% of the time you only win one bet 25% two and 25% none.

 

[Geoffrey42]

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.

You're nitpicking with his presentation of a perfectly valid way to model the events in question. Flip two coins. Ask two questions "Is at least one of them heads?", then "Are both of them heads?". Record both answers. Rinse, repeat, as many times as you like. The distribution of HH, TH, HT, and TT will be analogous to the distribution of MM, FM, MF, and FF beagle pairs, especially at higher numbers of repetition. The number of "Yes"s for the first question will come out to approx. 75% of your sample set, and the number of "Yes"s for the second question will come out to approx. 25% of your sample set. 33% of the time that the answer is "Yes" for the first question, the answer will be "Yes" to the second question.

1 out of every 3 instances != 50%. No matter how fancy your argument for deducing that one of the split TH/MF should be excluded from the recalculation of the odds, you cannot alter the probabilistic outcome such that 50% of the time the first answer is "Yes", the second answer is "Yes".

[a href=http://www.random.org/coins/?num=2&cur=60-usd.0025c-pa]Here's a random coin flipper for you[/a]. I did 100 goes, with HH-25, HT-17, TH-27, and TT-31. 69% answer "Yes" to the first question, because the sex-checker confirms that one them is a boy. Of those 69, 36% are "Yes" to both being boys, and 64% are "No". Does 36/64 seem closer to 50/50, or 33/66? I dunno. That's tough.

[a href=http://en.wikipedia.org/wiki/Boy_or_Girl_paradox]This is the answer to this question. It is well-documented, and has been around a long time.[/a] I bid this thread adieu, and stand by my earlier statement that I thank-his-noodly-appendages public opinion doesn't dictate school curriculum.

EDIT:

Now, someone tells me I've won one of my bets. What are my chances of breaking even vs. winning now? How would you draw the new matrix?

The only thing you have learned is that you have not lost money. 2 times out of 3, you will break even at this point, and 1 time out of 3 you will make money. 66% break-even. 33% make money.

 

[Samirat]

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.

Why couldn't she have picked up a Female puppy first? As long as the puppy she checked second is male, she would still respond "yes." So there are the three situations, where the first dog she picked was female, and the second dog was male, or the first dog she picked was male, and the second dog was either male or female. All of these would result in a response of "yes" from the washer woman, therefore they're all valid.

So your logic breaks down early, at your assumption that the first dog she checks is male.

 

[Alex_P]

So, other than "the set must contain at least one male," what are the repercussions of "at least one puppy is male"?

When we adjust our probability matrix, we can't just pull out the F/F, we also have to pull out either one (exclusive) or the other of M/F or F/M. It doesn't matter which, but one of them has to go. One dog's sex is no longer a matter of probability but of certainty, and leaving both in reflects a situation where both dogs' sex is still a matter of probability.

Untrue. You most definitely don't know which dog is male.

-- Alex

 

[Samirat]

[a href=http://en.wikipedia.org/wiki/Boy_or_Girl_paradox]This is the answer to this question. It is well-documented, and has been around a long time.[/a] I bid this thread adieu, and stand by my earlier statement that I thank-his-noodly-appendages public opinion doesn't dictate school curriculum.

Thanks. I looked for some mathematical proof or incontrovertible evidence earlier, but couldn't find any. I think people will be much more open to persuasion, and find it easier to understand the solution, if they first accept the solution.

And you're right, it's kind of depressing how many people answered 50 percent, and how some of them even mocked the 33 percenters. I mean, they sounded stupid doing it, but what's the point if they don't actually know that they sound stupid. In fact, Cheese here is the only one actually pursuing it. I'm not sure if he actually thinks he's right, or if he's actually seeking some sort of understanding. Which is it, Cheese?

 

[Jimmydanger]

but we don't know WHICH puppy of the two he said yes about!

I am now convinced that cheeze is just screwing with us we have explained this too him every possible way. Either he does not understand probabilities, logic, and experimental design or he's f'ing with us. I hope for his sake it's the latter.