Poll: A little math problem

[Samirat]

I'm afraid that you're now starting to convince me that you know even less about probability than I would have guessed. A combination, like that of a male female pair, has a probability defined by the number of permutations it contains.

Not if you are picking from among a set of three baskets, one of which has two male puppies, one of which has two female puppies, and one of which has a male puppy and a female puppy.

No. The number of outcomes is equal to 2 * 2, the male female chance of the first dog multiplied by the male female chance of the second dog.

 

[hemahemahema]

so to start with there are mf, fm, mm and ff of equal probability

at least 1 m brings us to fm, mf or mm, again of equal probability

out of which for mm the other 1 is male, so 1/3

the counter arguement seems to be that fm=mf, so really there are just 3 equal possibilities to start with

but this is like saying, when you flip 2 coins, getting a head and a tail is just as likely as getting both heads, or both tails, which is wrong.

a comparable example is flipping 3 coins, where getting hhh (all heads) and ttt (all tails) is clearly less likely compared to the total probability of getting a mixture of heads and tails (hht and tth)

if you still dont believe it, just get two coins and flip them

 

[Samirat]

You've confused there being two different ways to *arrange* something with there being two different ways to *achieve* something.

This is...absurd. I give up, this argument is hopeless.

It's absurd to think there's an equal chance of picking a red lego stuck to a white lego as pulling two red or two white legos stuck together from a pool where they are the three and only three combinations, even though there are two permutation of the red and white lego, where the red might be stuck into the white or vice versa?

Why is this absurd?

Here, I'll show you why you are twice as likely as likely to find a red lego white lego pair.

Say you are the one sticking legos together.

Say you pick up a white lego. You will then stick the next lego you pick up to it. There is a 50 50 chance it will be white or red.
WW
WR
Say you pick up a red lego. There is an equal chance of picking up a red or white lego next. So you get the two equally possible outcomes:
RW
RR

And there are your four permutations:
WW
WR
RW
RR

 

[Samirat]

or maybe even read my response to someone else with the same question, that they miss something as basic as mistaking the number of was to arrange a permutation for necessarily the number of ways to achieve the combination that the permutation is an arrangement of.

There is only one way to arrange one permutation. A permutation specifies order, so there are no different "was to arrange a permutation," only one. It is part of the definition.

 

[Samirat]

I'm afraid that you're now starting to convince me that you know even less about probability than I would have guessed. A combination, like that of a male female pair, has a probability defined by the number of permutations it contains.

Not if you are picking from among a set of three baskets, one of which has two male puppies, one of which has two female puppies, and one of which has a male puppy and a female puppy.

So if you doctor your possible outcomes, what you say is true? Nice. Whatever happened to the random experiment? Then again, I suppose you always insisted on doctoring your problem somehow, at first by placing a male dog in your problem, and now by doing... I don't even know.

 

[geizr]

Guys, you are wasting your time to continue arguing with Cheeze. He is going to constantly use senseless hairsplitting and redirection to debate. Just let him mentally masturbate in a corner by himself. He knows just enough to confuse himself.

 

[Samirat]

You've confused there being two different ways to *arrange* something with there being two different ways to *achieve* something.

This is...absurd. I give up, this argument is hopeless.

It's absurd to think there's an equal chance of picking a red lego stuck to a white lego as pulling two red or two white legos stuck together from a pool where they are the three and only three combinations, even though there are two permutation of the red and white lego, where the red might be stuck into the white or vice versa?

Why is this absurd?

Here, I'll show you why you are twice as likely as likely to find a red lego white lego pair.

Say you are the one sticking legos together.

The legos are already stuck together and in the pool, in the all-red, white-and-red, or all-white combinations.

So it is a non-random pool? Created specifically to be 1/3 all red, 1/3 white and red, and 1/3 all white? Whatever happened to the random bit? How is this pool created? How did the lego blocks get stuck together? Whatever, it has no relation to the two dogs problem. I still want your answers though.

or maybe even read my response to someone else with the same question, that they miss something as basic as mistaking the number of was to arrange a permutation for necessarily the number of ways to achieve the combination that the permutation is an arrangement of.

There is only one way to arrange one permutation. A permutation specifies order, so there are no different "was to arrange a permutation," only one. It is part of the definition.

I was unclear, I should have said "alternate ways to arrange a permutation."

All right, so there are two different ways to achieve the male female combination, and only one way to achieve the male male combination. How is this supporting your point?

 

[Samirat]

I was unclear, I should have said "alternate ways to arrange a permutation."

All right, so there are two different ways to achieve the male female combination, and only one way to achieve the male male combination. How is this supporting your point?

No, there is only one way to achieve--though two ways to arrange--the male female combination if you take the shopkeeper woman as talking about not just the possible compositions (arrangement) of a random pair (achievement) , but about the possible compositions (arrangement) of the possible pairs (achievement) when we read:

"A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair."

So are you saying that the chances here are

33 percent male male
33 percent male female
33 percent female female?

Cause that's not true. These are all the possibilities, but they're not equally probable.

If someone says, after flipping ten coins, that they're not sure if they got 1,2,3,4,5,6,7,8,9, or 10 heads, you better believe the chances of her having gotten 5 heads are greater than the chances of her having gotten all ten.

 

[Narthlotep]

I didn't want to jump on the "lol I no maths" bandwagon, but there are some who are still confused, so I'll give an in-depth answer - the one I used to explain this to myself.

Firstly, forget this business about "more than one male". What are the possible configurations before this?
Two males, two females, one of both.
The probability of either dog being a male or a female is 50% either way.
Therefore:
Bearing in mind that the total probability of something happening is 1,
Prob of two males = 0.5 x 0.5 = 0.25
Prob of two females = 0.5 x 0.5 = 0.25
Prob of one each = 1 - (sum of other possibilities) = 1 - 0.25 x 2 = 1 - 0.5 = 0.5
OR = 0.5 (using the reasoning that, no matter whether the first dog is male or female, we now just look at the probability of the other dog being the other sex. There are many ways to skin a cat.)
This part is simple, and I don't expect anyone was confused; it just had to be stated.
Now let's take our next piece of information: there is at least one male. What does this mean?
It means that one of our three different possibilities is no longer a possibility: we cannot have two females any more. So we strike that probability out:
Prob of two males = 0.25
Prob of one each = 0.5
Prob of two females = 0 (no longer possible)
BUT WAIT. The probability of getting our only two remaining possibilities is 0.75. What is this other 0.25 probability? That we have Schrodinger's Dog? This is the step that has thrown everyone. The probability of something happening must be 1 (unless "nothing happens" is given as an option in a particular problem, in which case that would also technically be 'something' - but this isn't happening here).
Clearly we have to go back to our probabilities.
Prob of two males (before two females was discounted) = 0.25
Prob of one each (before two females was discounted) = 0.5
Sum of both possibilities = 0.75
So we need to make this 0.75 become 1, as it is the new probability that "something happens".
Prob of two males = 0.25/0.73 = 0.333333333333333...
Prob of one of each = 0.5/0.75 = 0.66666666666666...
Now what was the question? Assuming that at least one dog was male, what is the probability that the other dog is male?
So we take one male from both of these probabilities and we are left with two cases: for one of each we are left with a female; for the two males we now have a second male. The second answer is the outcome we were asked to calculate and its probability, as we have shown, is 1/3.
*pumps fist in triumph. Now rests easy knowing he hasn't lost all his maths mojo*

Well said. What they don't teach you in school is that the equations are easy - framing the problem is the hard part.

Obscenely easy. I loved high school Probs and Stats more than any other maths subject because once you figured out what the problem was, you just had to put the numbers you were given into the equations you were taught. No proofs or complex integrations here, nosiree.

Sorry to intrude (I stopped reading after page three) but these types of questions are the reason I avoided statistics courses in school, and instead went for the "harder" mathematics of trig and calculus (plus they seemed more likely to help me with pool, trajectory and general machining mojo,) and this is most likely why those of you who understand these forms of arcane numerology will be my lords, ladies, and masters in a few years time.

 

[Saskwach]

Sorry to intrude (I stopped reading after page three) but these types of questions are the reason I avoided statistics courses in school, and instead went for the "harder" mathematics of trig and calculus (plus they seemed more likely to help me with pool, trajectory and general machining mojo,) and this is most likely why those of you who understand these forms of arcane numerology will be my lords, ladies, and masters in a few years time.

Ha! Our situations are practically reversed: I know my probability fairly well but wish my calculus weren't so shabby. How about we be each others' lord and master?

 

[Alex_P]

How about we be each others' lord and master?

Kinky!

-- Alex

 

[positrark]

From what I can tell I really don't think any of you are really wrong. Cheeze_Pavilion just makes different and dare I say, less logical, assumptions about the problem. If we as he says look at the two puppies as a single unit with 3 equally likely configurations, two male, two female or a pair, then by assumption he's correct. But again the logical thing to assume is that the gender of the two puppies are independent events, which gives two permutations of a pair. We always assume something is fair unless spesified otherwise. If someone says we can get values from 1-12 by tossing two dice, we don't assume it's equally likely to get 12 as 7. It all comes down to interpretation, but the large number of people siding with the latter aproach is evidence that this interpretation makes more sense. I will repeat that the biggest problem is that the problem is to some extent ambigously written.

 

[Lukeje]

How about we be each others' lord and master?

Kinky!

-- Alex

Sounds like a Depeche Mode song!

Damn--we're only 83 posts away!

83 Posts away from what?

 

[The Blue Mongoose]

oooooooooh.... kaaaaaaaaaaaaaaaaay....

if we consider the beagle on its own... it is 50%.

if we look at a pair of beagles the probability that they will both be male is 25%. but that is assuming that they are taken as at random from a population of more beagles.

basically... the shop could have purchased 2 male beagles, in which case the probability that they are both male is 100%.

there are flaws in the question. however if we are randomly selecting beagles from a larger population, the chance of picking one male beagle is 50%. picking out 2 male beagles, the chance is 25%. 3 male beagles? 12.5% and so on and so on.

basically... from an infinite population of beagles there is a .5^n probability of picking out n beagles.

hope that helps

Edit: the above assumes that there is an equal percentage of male and female beagles

 

[hemahemahema]

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

anyway, it's ironic how lego been used as an example. By Cheeze_Pavilion.
because, you know, a red lego stuck to a white one is actually different from a white one stuck to a red one

R
W

and

W
R

because of the specific shapes of Lego blocks, the two above do actually look different, which demonstrates the point they are different permutations, so the permutations of equal probabilities are

W
R

R
W

R
R

W
W

and each of these lumps of Lego is unique, thus the possibility for each is 1/4 (providing the pool of lego is indeed randomly arranged)

and Cheese_Pavilion, do try with a pair of coins, things become quite clear after tossing them a million times ( or 30)

"A shopkeeper says she has two new baby beagles to show you, but she doesn't know whether they're male, female, or a pair."

So are you saying that the chances here are

33 percent male male
33 percent male female
33 percent female female?

Cause that's not true. These are all the possibilities, but they're not equally probable.

If those are the all and only possibilities--and not just the possible combinations of the elements that the possibilities are made from--then yes, they are equally probable.

And that's the way to read what she said without bringing in extra information--she didn't say it's a fair pair of puppies, she said the pair is one of those three possibilities, with no other qualifications as to their probability.

I would say to bring in extra information, would be to assume it is a fair pair of puppies. Assuming they are not a fair pair is the same as assuming a pair of coins are biased when you are asked to toss them, in which case extra information is needed since there are so many ways two coins can be biased (6/4, 3/7, 1/9 you name it). Therefore she has to outrightly state the possibility of a pair is 1/3 before your arguement can be reasonably supported. But she didn't, she merely put it as an item in a list of 3. If this make the 3 items equally probable then I can claim homosextuality, heterosextuality and bisexuality are equally probable, which is false (but you can dispute me on that of course, Cheeze_Pavilion).

Another problem is that how do you make the possibility of a pair 1/3? Actually forget that, let's concentrate on making the possibility of two males 1/3 like you claim. Denote the possibility of 1 male p(male).

So we have p(male)^2 (i.e. p(male) squared) = 1/3

so p(male) = |1/root3| where root3 = square root of 3

looks fine, so we repeat the process for p(female), and similarly we obtain p(female) = 1/root3

here's the problem
root3<root4 so root3<2
so (1/root3)>(1/2)
so [P(male) + P(female)]=(1/root3 +1/root3) > (1/2 + 1/2) =1
so [P(male) + P(female)] > 1
which is obviously a contradiction.

So not only did the shopkeeper NOT claim that p(male/male)=p(female/female)=p(male/female)= 1/3&#65292; There is also NO way to biase the possiblities to make the claim true even if she wanted to. So it is best to assume the puppies are a fair pair.

NEW PROBLEM&#12288;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?

Essentially the same problem as the begining of the thread. You can solve this by making a million copies of each of A, B and C, hire a helicopter and scatter the cards from 100 metres in the air on to a large, flat, and windless square(which you may have to construct). Then take a clip board and go down to turn over every card with a black side up and make a tally chart of the colour of the other side. But if you had that sort of money and time, you surely have better ways of spending them, maybe.

 

[The Blue Mongoose]

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

NEW PROBLEM&#12288;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.

 

[Alex_P]

NEW PROBLEM&#12288;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

 

[TheDean]

i think there is a 1/6 chance

 

[Lt. Sera]

How the hell did this get to 27 pages

 

[hemahemahema]

NEW PROBLEM&#12288;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.