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?

Cheeze_Pavilion:
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.

Cheeze_Pavilion:

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.

Alex_P:
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?

Cheeze_Pavilion:

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.

Cheeze_Pavilion:

Alex_P:
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.

Cheeze_Pavilion:

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".

Here's a random coin flipper for you. 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.

This is the answer to this question. It is well-documented, and has been around a long time. 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:

Cheeze_Pavilion:

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.

50%, it is either a boy or a girl, it cant be 25% because it does not ask whether both of them are boys or not, it asks what the chances are that the second dog is a boy. It cant be anything else because that is just stupid.

Cheeze_Pavilion:

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.

Cheeze_Pavilion:

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

Geoffrey42:

This is the answer to this question. It is well-documented, and has been around a long time. 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?

Samirat:

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

I understand it almost as well as I understand the Monty Hall problem ;-D

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?

What beginning are you talking about? The beginning of the *question* or the beginning of the *experiment*? Of course I would manipulate my experiment at the start and call it realistic--as long as I was manipulating my experiment to model the reality as described by the question, the WHOLE question, not just the beginning of the question. So help me Monty Hall.

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.

Umm, the probability of any event that is actually known to have occurred is 1. Solution sets are for figuring out the probability of events that are not yet known either because they have not yet occurred or we do not have full knowledge of them. So yes--if you lay Coin 1 down Heads up, the probability that it is Heads 100%

If that changes later on in the problem, you're equivocated along the way about one of the terms you used, which is where your mistake is.

Samirat:

Geoffrey42:

This is the answer to this question. It is well-documented, and has been around a long time. 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 wouldn't consider a Wikipedia page that lists its reference as The Second Scientific American Book of Mathematical Puzzles and Diversions to be a "mathematical proof or incontrovertible evidence."

What's next--a 4chan picture of a nekkid Carol Vorderman proving Fermat's Last Theorem with Sudoku?

In all seriousness and with the hyperbole aside, I saw that page a long time ago, when I first started reading this thread. There is nothing there that actually proves anything--the whole problem is that it doesn't show why First Question is different from Second Question.

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?

Both.

Samirat:
[
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.

Or s/he used the Puppy Sex Detector 3000. We don't know how s/he checked the sex of the puppies, just that s/he did in a way that makes his/her response of "Yes!" reliable.

Alex_P:

Cheeze_Pavilion:

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.

I most definitely do--the puppy the Puppy Washing Man was using for warrant when asserting "Yes!" in response to the question.

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.

Jimmydanger:

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

When someone tells me I've won one of my bets, all I know is that I know I didn't lose both of them? Really? When someone tells me I've won one of my bets, I don't know that I've one at least one of my bets?

Think about it...

Geoffrey42:

Cheeze_Pavilion:

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.

This is math--nitpicking is as important as Jurassic Park huge flying insect the size of a car picking when it comes to math.

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,

Then it's not the sample set in the problem--the number of "Yes"s from the Puppy Washing Man will come out to 100%. That's given to us as fact in the problem itself.

This is the answer to this question. It is well-documented, and has been around a long time. I bid this thread adieu, and stand by my earlier statement that I thank-his-noodly-appendages public opinion doesn't dictate school curriculum.

Ehh, not really: http://www.escapistmagazine.com/forums/jump/18.73797.815492

One final attempt to explain before I give up. You believe that the problem states that one of the puppies is being examined and therefore you know that they are male. Then you state that since you know this puppies gender it is only a 50-50 chance of the second one being male.

Where you are making your mistake is we do not know which of the two puppies is male all we know is at least one of the two is.

I ask you again if this man had both puppies in two different boxes and merely stated one of the puppies is male how would you know which of the two boxes contained a male dog? You still have three options
dog on left - dog on right
M M
M F
F M

Please explain to me in this example how you could remove one of those options or tell me why this example is not analagous.

Cheeze_Pavilion:

Alex_P:
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?

Okay, but what I believe that the proper parallel with the original problem is if you were told that there was one out of the two dice was definitely even, so:

Even Even
Even Odd
Odd Even

are the possible dice.

What I think that you think the proper parallel is: you've been told that the first dice is even. Or the second, doesn't matter, so you've got either:

Even Even
Even Odd

or

Even Even
Odd Even

But you're never told which dog is male, only that at least one of the two is.

Cheeze_Pavilion:

What beginning are you talking about? The beginning of the *question* or the beginning of the *experiment*? Of course I would manipulate my experiment at the start and call it realistic--as long as I was manipulating my experiment to model the reality as described by the question, the WHOLE question, not just the beginning of the question. So help me Monty Hall.

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.

Umm, the probability of any event that is actually known to have occurred is 1. Solution sets are for figuring out the probability of events that are not yet known either because they have not yet occurred or we do not have full knowledge of them. So yes--if you lay Coin 1 down Heads up, the probability that it is Heads 100%

If that changes later on in the problem, you're equivocated along the way about one of the terms you used, which is where your mistake is.

All right, first, your experiment isn't random.

But relating to your comment on my proposed problem. I wasn't saying that laying down coins is the way you're actually supposed to do this. I'm saying this is the way you would do it, based on your response to the original problem. Essentially, you say that if you know one is heads, you lay down Coin 1 as heads. So if they then say that one is tails, you've got to lay down Coin 2 as tails. This is the way you're thinking about the original problem.

But what are the chances that Coin 1 is heads? Not 100 percent, as it is in this problem. Coin 2 could be heads, and Coin 1 could be tails, and the premise of the problem would still be satisfied. But the way you're going about it, the results are incorrect here, with the chances for each coin being absolutely determined.

So where we differ in our logic relates to how we set up the problem. You would like for the washer woman to say "yes" 100 percent of the time. But don't you see that this is not realistic. There is a certain chance that she would say "yes" and a certain chance that she would say "no," if you actually did this experiment. Manipulating it so she definitely says "yes" leads you to an answer which is, ultimately, incorrect. Are you saying that the washer woman made sure there was a male dog before the question was asked? That's what this amounts to.

The fact that she *happens* to say yes doesn't mean she says yes 100 percent of the time. It only means that in this case, there *happened* to be at least 1 male dog. This is why it's a random experiment. Assuring that she will always say yes invalidates the whole question. The point is that these are 2 random dogs, and that either or both of them could be female or male, and that this case just happened to fall within the 75 percent where there was at least one dog.

I'm not sure if you are deliberately misunderstanding this, or what. You've seen proof, you've had relatively solid explananations. But I've tried to explain problems like this before (such as Monty Hall) to people who didn't believe the answer, and I know how difficult it can be. But I mean, even the OP said the answer was 33 percent, from where he got the problem, and more than half of people still think that the answer is 50 percent.

Cheeze_Pavilion:

Alex_P:
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?

Think of it this way -- same dice game as above. Two dice have been rolled and are under separate cups.

Compare these two scenarios:
A. Your friend looks under one of the cups and says you have won at least one of your bets.
B. Your friend looks under both cups and says you have won at least one of your bets.

Do you see why A actually provides you with more information than B?

-- Alex

This is silly.

The is how The Ultra Joe would do it.

*Picks Up Dog*

Hmmmm...

*Checks Sex*

Problem solved, all this maths is silly, you can do it all you want but i'm the one with the puppy.

I have no actual answer, this is my way of covering my shame.

This thread is irrefutable proof that the Chinese will take over the world. Soon.

And, hopefully, institute a curriculum of mathematics and logic.

Jimmydanger:

I ask you again if this man had both puppies in two different boxes and merely stated one of the puppies is male how would you know which of the two boxes contained a male dog?

I would know it was whichever box contained the puppy that he was using to back up his claim that at least one of the puppies is male.

I don't need to know the *specific* box, I just need to know that whichever puppy provided the warrant for his claim that there was at least one male puppy, it's in one of the boxes in under discussion.

kailsar:

Cheeze_Pavilion:

Now, someone tells me I've won one of my bets.

Okay, but what I believe that the proper parallel with the original problem is if you were told that there was one out of the two dice was definitely even, so

Wait, how is telling me that "one out of the two dice is definitely even" different from telling me that "I've won one of my bets" when we know that I've bet on even?

Samirat:

So where we differ in our logic relates to how we set up the problem. You would like for the washer woman to say "yes" 100 percent of the time. But don't you see that this is not realistic. There is a certain chance that she would say "yes" and a certain chance that she would say "no," if you actually did this experiment.

I see that it conforms to the conditions as set by the word problem, and under the conditions set by the word problem, there is ZERO percent chance she will ever say no. Everytime someone asks this question the washer woman WILL say yes. If she doesn't say yes, then it's a different word problem.

I think the reason you're not getting what I'm saying is that you're failing to distinguish between figuring out the probabilities of any pair of random puppies we encounter in the world, and figuring out the probability of one specific pair of puppies in the context of the facts given to us in the word problem.

Are you saying that the washer woman made sure there was a male dog before the question was asked?

I'm saying the person who made up the word problem made sure the washer woman would find a male dog every time that person repeated the word problem.

I'm not sure if you are deliberately misunderstanding this, or what. You've seen proof, you've had relatively solid explananations. But I've tried to explain problems like this before (such as Monty Hall) to people who didn't believe the answer, and I know how difficult it can be. But I mean, even the OP said the answer was 33 percent, from where he got the problem, and more than half of people still think that the answer is 50 percent.

Like I've said many times before, if I agreed with you that your formulas captured the information in the word problem, I would be totally convinced it's 33%. My disagreement is that you start setting up your probability formulas before you read the whole word problem, and that is leading you to make mistakes when you try and adjust those formulas to capture the additional information that comes at the end of the word problem about the washer woman.

p.s. when people didn't believe you about Monty Hall, are you *sure* you explained it correctly? Because I could never get my head around it until someone made me aware of all of the conditions, i.e., Monty never opens the winning door, so that while your choice is one of pure random guessing, his choice is radically different from yours because his is made with perfect knowledge of what's behind all three doors.