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.
Poll: A little math problem
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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
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
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)
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， 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　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.
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)
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).Cheeze_Pavilion post=18.73797.846647 said:
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， 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　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.
i would say 50%.hemahemahema post=18.73797.847198 said:
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.
Wow, sorry. You understand it is a little bit difficult to keep up with some 900 posts.Alex_P post=18.73797.847423 said:
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.Cheeze_Pavilion post=18.73797.847945 said:
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)
