Poll: A little math problem

[Lukeje]

NEW PROBLEM　FOR THIS THREAD:

I have 3 cards, card A has 2 black sides, card B has 1 white side and 1 black side, card C
has 2 white sides. Out of these cards, I randomly show you one side which is black, then turn it over to show you the other side. What is the probability that the other side is white?

We've already mentioned the Three-Card Problem several times now.

-- Alex

Wow, sorry. You understand it is a little bit difficult to keep up with some 900 posts.

Try keeping up with it when about 600 of those are directed at you, personally! :-D

I'd say nearer 500 (only 70 posts left...).

 

[PMetal]

The chance that both dogs are male is 25%. Flip two coins, whats the chance of both being heads or both being tales? 25% with a 50% that one will be heads one will be tails.

The same thing applies here. I think. I might have read the question wrong.

Yes but we already know that the first one is male so the single probability is isolated.

 

[Lukeje]

Wow, sorry. You understand it is a little bit difficult to keep up with some 900 posts.

Try keeping up with it when about 600 of those are directed at you, personally! :-D

I'd say nearer 500 (only 70 posts left...).

Yeah, probably--you know I've been re-playing God of War during this, and so 'puppies' are in my head as those devil dogs this whole time?

Hmm... this does seem to be getting a little off-topic now... I wonder if the OP-er actually cares about this thread anymore?

 

[hemahemahema]

No, because those are three categories that people fall into. The shopkeeper woman is not making a statement about the three categories that puppy pairs fall into, she's making a statement about which of three possible pairs this pair is. My point was that if we take the statements about at least one being male as being equivalent to a certain kind of knowledge that depends on taking something so literally it becomes an abstraction, we have to treat the whole problem that same way. That if you start talking about biology here, you have to talk about puppy selection later, and for the same reasons in the Three Card Problem you were talking about mean that seeing a red side makes the all-red card more than 50% likely, seeing a male puppy makes the all-male pair more than 33% likely, namely, 50%.

Wow, you are actually a good logician. A slightly confused one though. Let's just start with the card example, for which your analogy had two errors.

A), you equated p(1)'showing a black side' to p(2)'at least one is male' where 'black' is equivalent to 'male'. And your arguement is that since p(1) makes the possibility of the other side being black higher than 33.33%, p(2) should also make the possibility of the other dog being male higher than 33.33%, and you assert it is 50%. Now the first sign the analogy is wrong is that the possibility of the other side is black is actually 66.66% (2/3), so how can the possibility of the other dog being male be 50%?

The reason the analogy is wrong is that p(1) require us to have seen one side of the card, where as p(2) requires us to have seen both puppies (if the first puppy seen is female, then the statement is neither proved true or false, so the 2nd one needs to be seen as well). So to make your analogy more correct, we change p(1) to p(3)'at least one side of the card is black' and we equate this to p(2).

This interestingly, makes the possibility of the all black card 50%, which I presume is why you have been claiming the answer to the dogs question is 50%. Which brings us to the 2nd error

B) In the Three Card Problem, you have been told there are 3 cards, two of them have black sides and one of them is all black. There is no mention of such information in the puppy question. Now this is where you disagree with most people on this forum so my first advice is still to TRY WITH TWO COINS (50-60 tims should make it fairly obvious head-tail combination accounts for 1/2 the outcomes).

And to be honest, I don't think anyone on this Forum even begins to understand your obsession with this conspiracy theory that some person or organisation is going around screening pairs of dogs to keep down the number of straight pairs, despite the fact there is no mention whatsoever of this grand project in the question. I mean christ, this is the kind of thing a shopkeeper would mention when selling pets to a customer "oh by the way, you've caught us in the middle of our 'Homo Month'! each pair of dog you buy is now 16.66% more likely to be in a gay or lesbian relationship!" I mean why!? What breeder would do that? They are not even selling them in pairs! When the customer said they are only interested in male she just called to ask if at least ONE is male, ONE, why would they care about the possibility of a combination within a pair when they are not even sold in pairs!? What motivation could they possibily, possibly have to reduce the number of dogs they have on the market by 1/4? Why cant you assume they are just a normal pet service ran by normal pet selling people selling normal bets bred by normal means with good old 1/2 chance to be male or female, that the pets people have no political, social or humanity (bestiality) agenda which they somehow hope to premote by manipulating the composition of pairs of beagles they have on the market?

And if you are still somehow insisting the dogs are just born that way, do reply If and only if you can explain how male male or female female pairs are 33.33% each, when calcualted from the possibility of a single dog being male or female.

(that is, if you agree with if probability of a is A and probability of b is B then the probability of a then b is given by AxB)

 

[whoops1995]

this doesn't deserve 1,000 posts "slams head on desk out of jealousy of the op's lucky idea"

 

[Samirat]

Wow, you are actually a good logician. A slightly confused one though. Let's just start with the card example, for which your analogy had two errors.

A), you equated p(1)'showing a black side' to p(2)'at least one is male' where 'black' is equivalent to 'male'. And your arguement is that since p(1) makes the possibility of the other side being black higher than 33.33%, p(2) should also make the possibility of the other dog being male higher than 33.33%, and you assert it is 50%. Now the first sign the analogy is wrong is that the possibility of the other side is black is actually 66.66% (2/3), so how can the possibility of the other dog being male be 50%?

Well, first off, thanks! As for the Three Card Problem, I wasn't saying it was a direct analog, just that the same reasoning applies, only in this case, if we're bringing biology into this, picking a pair with a male makes it equally likely that you've gotten the one all male pair than either of the two mixed pairs, for the same reason that picking a black side makes it more than likely that you've gotten the all black card.

Just in this case, the Three Card Problem is a Four Card Problem, with one all black card, two mixed cards, and one all-white card, representing the all male/mixed/all female puppy combinations.

B) In the Three Card Problem, you have been told there are 3 cards, two of them have black sides and one of them is all black. There is no mention of such information in the puppy question.

My point was that if we're going to take the 'at least one dog is male' statement so literally that we don't think about how it was arrived at, then we should take the shopkeeper woman's statement that they are either a male pair, a female pair, or a mixed pair as describing the actual pool of possible pairs, not just the possible combinations of male and female you find in them, so she's really describing a Three Card Problem.

Can't use one kind of logic to read one half of a problem, and a different kind of logic to read the other half, you know? This isn't the Second Amendment, with some kinda distinction between a prefatory and an operative clause!

And to be honest, I don't think anyone on this Forum even begins to understand your obsession with this conspiracy theory that some person or organisation is going around screening pairs of dogs to keep down the number of straight pairs, despite the fact there is no mention whatsoever of this grand project in the question.

My obsession is explained by the fact that people are thinking this describes a paradox that holds true in the real world every time you hear "at least one is male." Look at the survey that Marilyn Vos Savant took to try and prove she was 'right': she asked for parents with two children and at least one male to write in--sounds a whole lot like someone 'screening pairs of kids to keep out any FF pairs' now, doesn't it!

My point was that seeing this, people will start to think this is the way the whole world works. Now think of another scenario where the statement "there is at least one male" is every bit as true: tell every parent with two children to pick one of their children at random, and if that child is male, write in with the sex of the pair. Half the responses will be from parents with two male children, even though they only make up 25% of the population by average, just like in a Four Card Problem.

In short, brain teasers are cute, but really understanding why the math works the way it does is more important. That a few of the people arguing with me really don't get the deep workings of the math is proven by the fact that they kept trying to argue that I read a word problem wrong by throwing more math at me, or complaining about the fact that a word problem had come down to "semantics": that's like complaining about a problem about the sex of puppy pairs coming down to biology!

Basically, it seems a lot of people felt very smart for 'knowing' it was 33%, and didn't like that I came along and pointed out that you could read the problem a different way from the one they did without falling into the 'trap' that the problem throws at you.

One could say that I harshed on the squee of some people who know one math trick, and they got all emo about it!

Not to say that's true of everyone who disagreed with me, but, it happened. In the end I won, though, because I now know a lot more about probability than I did before. Even if I still don't know what werepossum was talking about with set vs. sequential probability.

Interesting, the child problem. All right, so children have a clearly defined order, the order in which they're born, correct?

So if you ask the mother "is at least one male?" and she says "yes," there are again the three situations:

1rst child male, 2nd child male
1rst child male, 2nd child female
1rst child female, 2nd child male

All of these are outcomes where at least one child is male, therefore in all of these situations, the women would answer "yes." The dog problem is no different. It may be confusing to have the two dogs, with no clearly defined order, but the situation remains the same. The only important part is that the order remains consistent.

Your interpretation of the word problem as presenting three equal possibilities of MM, one male one female, and FF is obviously incorrect, and your understanding of probability is flawed. Though I admire you for holding fast to your belief, even in the face of all contradictory logic and evidence. Oh wait. No I don't.

 

[Samirat]

Wow, we are determined to reach 1000 posts on this problem eh?

NEW PROBLEM　FOR THIS THREAD:

I have 3 cards, card A has 2 black sides, card B has 1 white side and 1 black side, card C
has 2 white sides. Out of these cards, I randomly show you one side which is black, then turn it over to show you the other side. What is the probability that the other side is white?

i would say 50%.

there is equal chance that it will be black or white. it must be either card A or B, as it has a black side.

But how many black sides are there? Three, correct? All with an equal chance of being face up. And two of them are on the black and black card, whereas only 1 of them is on the white and black card. Therefore, the chances that the other side will be white are actually only 1/3.

 

[Samirat]

So if you ask the mother "is at least one male?" and she says "yes," there are again the three situations:

1rst child male, 2nd child male
1rst child male, 2nd child female
1rst child female, 2nd child male

All of these are outcomes where at least one child is male, therefore in all of these situations, the women would answer "yes."

What if you asked a mother-to-be what sex her children will be, and she said:

"oh, I don't know, but I know I was fertilized with one of three sets of fraternal twins, one is two boys, one is two girls, and the other is mixed."

And then you asked the person doing the sonogram if at least one of them was male, and she answered "Yes!"

Then what would the odds be?

But that's not the question, is it? Do you really think it's all right to alter the premise this way? If you actually took dogs, as they said in the problem, and asked what the probability is that one's male and one's female, what would it be? 33 percent?

Let me alter your problem, and I want you to see that nothing I'm doing here is more illegal in probability than what you did. That is to say, it's very illegal. What if you asked a mother-to-be what sex her children will be, and she said:

"oh, I don't know, but I know I was fertilized with one of 9 sets of fraternal twins, 2 are two boys, 6 are two girls, and the other is mixed."

And then you asked the person doing the sonogram if at least one of them was male, and she answered "Yes!"

Then what would the odds be?

Well, I suppose they would be 2/3's that the other dog is male, and 1/3 that it is female. This problem is no different than the one you described, except in the numbers.

And none of it is relevant to the OP's problem. Because you're creating different problems, in your mathematically unreal fantasy world. And it's not good for you.

Arbitrarily changing probabilities, altering the problem, doctoring the outcomes. You aren't doing math, anymore, you're using the kind of debating tactics I would expect from a self-righteous 13 year old.

Two separate dogs = 2 separate events. Are you even trying anymore?

 

[Xanadu84]

Red Herring Answer:
Possible puppy pairings:

Male/Male
Male/Female
Female/Male
Female/Female.

Since its random, each of these circumstances have an equal chance of being what happened. We find out that one is male, leaving possible pairings to be m/m, m/f, and f/m, appearing to be a 33% chance of the other puppy being male.

Actual answer:
We know going into it that one puppy is male, either puppy 1, or puppy 2. First, assume puppy 1 is male being identified. Possible pairing are m/m and m/f. Chances are 50/50 of other puppy being male. The other option is that puppy 2 is the one being identified as male. Possible pairings are M/M and F/M, 50/50 of other puppy being male. Put them both together, and you have a 50/50 chance of the other puppy being male, and no gamblers fallacy. I'm fairly confident that is the answer. Deceptive.

 

[Alex_P]

Actual answer:
We know going into it that one puppy is male, either puppy 1, or puppy 2. First, assume puppy 1 is male being identified. Possible pairing are m/m and m/f. Chances are 50/50 of other puppy being male. The other option is that puppy 2 is the one being identified as male. Possible pairings are M/M and F/M, 50/50 of other puppy being male. Put them both together, and you have a 50/50 chance of the other puppy being male, and no gamblers fallacy. I'm fairly confident that is the answer. Deceptive.

Not quite.

The most straightforward way to resolve conditional probabilities is to start with the initial distribution and then apply the limiting conditions. That means both of your scenarios (I'll call them "Jesus is the known male" vs. "Satan is the known male") draw from a common initial distribution. You're basically double-counting the male/male sets.

In Bayes' Theorem terms, P ( Jesus is the first male found | Jesus is male and Satan is male ) = 1/2. (Not 1!)

-- Alex

 

[Narthlotep]

In another 54 posts, I say that those who argue that the answer is a 50/50 chance elect one representative and the 33% chance people also elect a representative to go down to a bar and settle it once and for all with a giant drinking contest, whoever passes out first has the losing argument. Either that, or settle it with drinks and darts. Just make sure that the solution comes from a friendly bar-like establishment.

 

[hemahemahema]

Red Herring Answer:
Possible puppy pairings:

Male/Male
Male/Female
Female/Male
Female/Female.

Since its random, each of these circumstances have an equal chance of being what happened. We find out that one is male, leaving possible pairings to be m/m, m/f, and f/m, appearing to be a 33% chance of the other puppy being male.

Actual answer:
We know going into it that one puppy is male, either puppy 1, or puppy 2. First, assume puppy 1 is male being identified. Possible pairing are m/m and m/f. Chances are 50/50 of other puppy being male. The other option is that puppy 2 is the one being identified as male. Possible pairings are M/M and F/M, 50/50 of other puppy being male. Put them both together, and you have a 50/50 chance of the other puppy being male, and no gamblers fallacy. I'm fairly confident that is the answer. Deceptive.

Eat you heart out, Cheese_Pavilion!
You thought you were the king of faulty logic but then this dude showed up and you got your paradoxical arse handed to you.

Way to go Xanadu84!! Take his bottom!

Oh and as for settling the solution with a drinking game, how about a Speed Pub Crawl? The participants choose a town, go into each bar and down a pint as fast as possible, the person who gets to the end of the town first without having 'relieved' himself at all wins.

 

[hemahemahema]

Oh and just ONE LAST LAST LAST EFFORT to explain the Three Card Problem.

It's all in the wording. You are shown on SIDE which is black.

One side, not a card with a black side but one side. So if the cards are A, B and C, and they have sides A1 A2 B1 B2 C1 and C2, you have not seen the card A, B or C. You have seen A1, A2, B1, B2, C1 or C2.

Note: at this point, you should not be able to answer the question 'has the card got at least one black side?' since you are not allowed to see the other side. Luckily a) someone is screening the side you are seeing so you only see a black side, which allows you to solve the problem but totally messes up the probabilities involved in 'has the card got at least one black side?', but then again b)THAT'S NOT WHAT THE FRIGGING PROBLEM IS ABOUT.

So it follows that there are 3 black sides out of the 3 cards thus 2 out of 3 cases when you turn over you see a black side and. All the information you have are deduce from the side you are seeing, and the probability you calculate applies to the other side

side - side

Now this is DIFFERENT from seeing a CARD with a black side, for the abysmally trivial reason you need to have seen BOTH sides of the card.

i.e. when you see a white side, you still have to turn it over before you know if it has a black side or not.

The wording that applies to the faulty interpretation would be: You are shown a CARD with at least one black side, what is the probability the other side is also black.

In other words, you have seen the cards A, B or C. NOT the sides A1, B1 etc.

In this case, there are two cards with at least one black side. So the probability is 1/2. Or for the 4 card question Cheese_Pavilion mentioned the probability is 1/3

And remember the information you are given is about the cards, and the probability you calculate is about the cards

cards - cards

To sum it up, 'At least one black side' does not equate to 'seeing a black side', for the simple reason that the former can still be true even if you see a white side, forcing you to look at both sides. You see whole cards, not just sides. Whereas the latter just involves sides.

Note: in the 'At least one black side' question you might get someone else to look at both sides of a card for you, otherwise there would be no point in using probabilities if you already know everything. Although REMEMBER if this is the case you would not be shown the cards at all, even one side. The information is given to you through that person and that person only, and it is about the card as a whole, or, if you like, 'pairs' of sides. Not individual sides.

 

[whoops1995]

this thread has gotten to a point where everyone is repeating like this post.

 

[Saskwach]

With only 50 posts to go, I figure it's almost cruel to not give the OP his shiny new badge. So here are my latest results for 100 coin flips: 35 heads, 65 tails.

 

[Lukeje]

With only 50 posts to go, I figure it's almost cruel to not give the OP his shiny new badge. So here are my latest results for 100 coin flips: 35 heads, 65 tails.

Only 47 now... How come all the arguing's stopped... did we all reach an agreement?

 

[Saskwach]

With only 50 posts to go, I figure it's almost cruel to not give the OP his shiny new badge. So here are my latest results for 100 coin flips: 35 heads, 65 tails.

Only 47 now... How come all the arguings stopped... did we all reach an agreement?

I think we all agreed that no one would change their minds because they were wrong but stubborn while we were right and...stubborn.

 

[geizr]

With only 50 posts to go, I figure it's almost cruel to not give the OP his shiny new badge. So here are my latest results for 100 coin flips: 35 heads, 65 tails.

Only 47 now... How come all the arguings stopped... did we all reach an agreement?

I think we all agreed that no one would change their minds because they were wrong but stubborn while we were right and...stubborn.

I stopped because Cheeze demonstrated to me that he doesn't understand things as well as he pretends, nor is he experienced enough to know how to use the math. He knows just enough to confuse himself, and all he has to attack with is hairsplitting on words. I started spectating just to see how long he's willing to drag this farce on.

 

[Lukeje]

With only 50 posts to go, I figure it's almost cruel to not give the OP his shiny new badge. So here are my latest results for 100 coin flips: 35 heads, 65 tails.

Only 47 now... How come all the arguings stopped... did we all reach an agreement?

I think we all agreed that no one would change their minds because they were wrong but stubborn while we were right and...stubborn.

I stopped because Cheeze demonstrated to me that he doesn't understand things as well as he pretends, nor is he experienced enough to know how to use the math. He knows just enough to confuse himself, and all he has to attack with is hairsplitting on words. I started spectating just to see how long he's willing to drag this farce on.

Apparently not long enough to reach 1000 posts...

 

[Fraught]

With only 50 posts to go, I figure it's almost cruel to not give the OP his shiny new badge. So here are my latest results for 100 coin flips: 35 heads, 65 tails.

Only 47 now... How come all the arguings stopped... did we all reach an agreement?

I think we all agreed that no one would change their minds because they were wrong but stubborn while we were right and...stubborn.

I stopped because Cheeze demonstrated to me that he doesn't understand things as well as he pretends, nor is he experienced enough to know how to use the math. He knows just enough to confuse himself, and all he has to attack with is hairsplitting on words. I started spectating just to see how long he's willing to drag this farce on.

Apparently not long enough to reach 1000 posts...

I'm pretty sure it will.