The probability of the other puppy being a boy is one half, 0.5, 1/2, 50%. Nothing else.

With two puppies, there are four possible gender combinations.

G,G
G,B
B,G
B,B

Because the question states that _at least one_ is a male (as opposed to 'the first one' or 'the second one') then the order doesn't make a difference. BG and GB are the same thing. If _at least one_ is a boy, then the options are like this:

B,B
B,G/G,B

That's only 2 options. It's 50:50.

I disagree with the geniuses and the schoolteachers and the mathemeticians because I know I'm right. I know I'm right about this, the Kennedy Assassination, the Moon Landing, 9/11, any moral or ethical decision and everything else in the universe.

Goodnight.

Bakery:
The probability of the other puppy being a boy is one half, 0.5, 1/2, 50%. Nothing else.

With two puppies, there are four possible gender combinations.

G,G
G,B
B,G
B,B

Because the question states that _at least one_ is a male (as opposed to 'the first one' or 'the second one') then the order doesn't make a difference. BG and GB are the same thing. If _at least one_ is a boy, then the options are like this:

B,B
B,G/G,B

That's only 2 options. It's 50:50.

I disagree with the geniuses and the schoolteachers and the mathemeticians because I know I'm right. I know I'm right about this, the Kennedy Assassination, the Moon Landing, 9/11, any moral or ethical decision and everything else in the universe.

Goodnight.

There is a way to prove its 1/2; but that's not it. You've made the same assumptions the 1/3ers have, but then forgotten that 'two options' does not mean that each has the same probability. If there are twice as many microstates for a particular outcome, then it will have twice as much probability, hence your answer should read '1/3'.

Lukeje:

Bakery:
The probability of the other puppy being a boy is one half, 0.5, 1/2, 50%. Nothing else.

With two puppies, there are four possible gender combinations.

G,G
G,B
B,G
B,B

Because the question states that _at least one_ is a male (as opposed to 'the first one' or 'the second one') then the order doesn't make a difference. BG and GB are the same thing. If _at least one_ is a boy, then the options are like this:

B,B
B,G/G,B

That's only 2 options. It's 50:50.

I disagree with the geniuses and the schoolteachers and the mathemeticians because I know I'm right. I know I'm right about this, the Kennedy Assassination, the Moon Landing, 9/11, any moral or ethical decision and everything else in the universe.

Goodnight.

There is a way to prove its 1/2; but that's not it. You've made the same assumptions the 1/3ers have, but then forgotten that 'two options' does not mean that each has the same probability. If there are twice as many microstates for a particular outcome, then it will have twice as much probability, hence your answer should read '1/3'.

I think he was being funny...I think

Lukeje:
'two options' does not mean that each has the same probability.

Order makes no difference in this question. If there's a boy _and_ a girl, then theres a boy and a girl, not a boy _then_ a girl/girl _then_ a boy.

To look at it any other way is to look at it too deeply and miss the point of the question.

EDIT:

Doug:

I think he was being funny...I think

I was in the ending sentance but alas, no, my answer of 50% is dead serious.

Bakery:

Lukeje:
'two options' does not mean that each has the same probability.

Order makes no difference in this question. If there's a boy _and_ a girl, then theres a boy and a girl, not a boy _then_ a girl/girl _then_ a boy.

To look at it any other way is to look at it too deeply and miss the point of the question.

EDIT:

Doug:

I think he was being funny...I think

I was in the ending sentance but alas, no, my answer of 50% is dead serious.

I was talking about you! Look at my previous answers.

EDIT:

Misread, my bad. You are reading the scenario incorrectly though.

Bakery:
Order makes no difference in this question. If there's a boy _and_ a girl, then theres a boy and a girl, not a boy _then_ a girl/girl _then_ a boy.

To look at it any other way is to look at it too deeply and miss the point of the question.

To look at it your way is to misread the situation and the maths of the scenario all together, I'm afraid.

So, the initial statement was:

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. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

So, breaking it down, we have

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, assuming P(male)=0.5, P(female)=0.5, we get:

Hence:
P(Male and Male) = 0.25
P(Male and Female) = 0.25
P(Female and Male) = 0.25
P(Female and Female) = 0.25

But if we assume the order isn't important, then we have:
P(2 Males) = 0.25
P(1 Male and 1 Female) = (0.25 + 0.25) = 0.5
P(2 Female) = 0.25

Next:

You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile.

Thefore, we can exclude P(2 Females) as that case has been removed, so we can multiply:
P(2 Female) = 0.25
P(2 Males) = 0.25/(0.25+0.5) = 0.25/0.75 =(approx) 0.3333 (or 1/3 exactly)
P(1 Male and 1 Female) = 0.5/(0.25+0.5) =(approx) 0.6667 (or 2/3 exactly)

What is the probability that the other one is a male?

P(2 Males) = 0.33333 or 33.33% or 1/3

EDIT:
For example, what are the odds of rolling 2 dice, one after the other, or at the same time, and getting a total of 5

36 (i.e. 6 * 6) possible combinations, of which the following match:
1+4=5
2+3=5
3+2=5
4+1=5

P(rolling a total of 5)=4/36

if you assumed order wasn't important, you'd get:
1+4=5
2+3=5
P(rolling a total of 5)=2/36

Which would be wrong

werepossum:
Scenario #2: Toss a fair coin ten times in a row without looking at the results. Now begin looking at the results for the first nine. They are all heads. What is the chance of the tenth being heads? About one in a thousand. This is set probability.

Just back in this quagmire for a second this was explained wrong and I think it can help with the main problem.

you cannot look at the first nine coins in a set and still treat them as a set. In this example you have reduced yourself to only two options coin 10 is heads or tails leaving a 50-50 chance.

For this to remain a theoretical set no knowledge must be known about individual coins. Say if someone told you at least nine of those coins were heads. Then you would have 11 options. First all coins are heads OR coin 1 is tails OR coin is tails 2 OR coin 3 etc. Then the chance of all the coins being heads would be 1 in 11.

This is equivalent to both puppies male OR first puppy female OR second puppy female. Then the chance of both male is 1 in 3

Cheeze_Pavilion:

fedpayne:

Unless, when people have been saying that the female male combination can also be written as male female, can male male also be written as male male.

I *thought* you were on to something, and you were. You've got the solution right there.

We start with this, which everyone agrees on:

M M
M F
F M
F F

Then we get the information that at least one dog is male. Let's call this the Known Dog, or K for short. Now we have to go back to our table, and work on the line in bold, applying our new knowledge.

M M
M F
M F
F F

Well, we know Known Dog can't be female, so we have to get rid of that line completely, so we get (understanding that the "-" is just a placeholder that indicates no possibility):

M M
M F
F M
- -

Now we go to the next line, the one in bold, and once again apply the knowledge that Known Dog is Male.

M M
M F
F M

Well, Known Dog is male, there's only one M, so by process of elimination, we must have to replace the lone M with a K, so we get:

M M
M F
F K

Now we go to the middle line, the one in bold, and apply the knowledge that Known Dog is Male a third time.

M M
M F
F K

Well, same as before--one M, that must be Known Dog

M M
K F
F K

Now we move to the top line, and apply our knowledge.

M M
K F
F K

Oh Noes! How do we apply our knowledge to this line? We have to apply that knowledge, we can't just pretend we don't have it. The solution? We have to create a new line, a line that reflects the fact that in pairs where both puppies are male, the Known Dog could be either dog. So we get:

K M
M K
K F
F K

And as all are equally likely because they are all (1x.5) or (.5x1), and K=M for purposes of the question 'is this a pair of male dogs,' two out of four equally likely possibilities means a probability of 50%.

Ok this got glossed over because someone started talking about 4 coins. By doubling the MM chances you have brought order into the problem when is isn't really needed but we can go through using it anyway. This much is true you cannot ever create new options for results halfway through, all the results must be there from the beginning so by your logic the original options are.

MM
MM
MF
FM
FF
FF

which as we have pointed out is obviously wrong because then the chance of getting 2 heads or 2 tails is 1/3 instead of 1/4. To fix this you must also double FM and MF. To make this more clear I will attach numbers to the dogs 1 and 2 this looks very repetitive at this point because it is it only becomes important once you start calling one of the results K.

1M 2M
2M 1M
1F 2M
2F 1M
1M 2F
2M 1F
1F 2F
2F 1M

Your final equation removing all FF now looks like this

1K 2M
2K 1M
1K 2F
2K 1F
1F 2K
2F 1K

1F 2F
2F 1F
Returning us to our 1/3 33% probability

edit: I noticed that the MF FM looks like I'm doubling because I still have repeated pairs like 1K2F and 2F1K. It only looks more repetitive than the MM's because the K is obscuring the doubling there. I added in the FF pairs show that it is the same as the grouping above and if it didn't have those doubles then we would have the 1/3 of the time you get 2 heads problem again.

Okay, here's the problem. There's a difference between the sexes of a pair of puppies being unknown because we know nothing about the puppies, and the sexes of a pair of puppies being unknown because we know nothing about which pair was pulled from a pool of four known pairs, composed of one all male pair, one all female pair, and two mixed pairs.

What people are doing is they are confusing the permutation of possible results for two unknown puppies with a listing of the pool of puppy pairs from which a pair will be drawn because they both look the same.

In order for set probability to play a role, we need to know the composition of the pool from which the pair were drawn. WE DO NOT KNOW THAT! In order to know that the question would have to look like this:

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 they are one of four pairs that the breeder had, and the breeder picked the pair from a pool that was made up of one all male pair, two mixed pairs, and one all female pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

See the kind of additional information you need to start using set probability? You can't just pretend because of a table of the possible permutations that you're picking from a pool of 4 pairs or 100 pairs like werepossum was talking about or 574543985770 pairs--until you get information that the pair of puppies was picked from a certain pool of pairs and you know something about the distribution of pairs in that pool, you can't just start using set probability.

Like I've been saying for a while now, you can't just pretend every symbol in a math problem refers to something in the real world. People have been reifying the possible permutations of two unknown events into actual pairs of puppies, and that is why they think they can just eliminate the F/F and rebalance the probability.

In short, the people who think it is 33% are mistaken because they think the information that at least one puppy is male means that the unknown pair was drawn from a pool of known pairs consisting of two mixed pairs and one all male pair, when what it really means is that a pair with at least one male was drawn from a pool of unknown pairs.

dirtface:
Ok this is like the monty hall problem.
If atleast one is male, you have three possible situations
Beagle1 Beagle 2
Male Male
Male Female
Female Male

Of these three possible outcomes, only one of them results in the second beagle being male.
Thats where the 33% comes from...
That's the logic behind the problem, however there's also the idea of independent events... the mind boggles.

This guy is right:

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. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

As long as you havn't seen them yet "The least one" could be any of them.

Makes perfect sense.

Jimmydanger:

Your final equation removing all FF now looks like this

1K 2M
2K 1M
1K 2F
2K 1F
1F 2K
2F 1K

1F 2F
2F 1F
Returning us to our 1/3 33% probability

Look again at that table--I've bolded the one pair that is identical and underlined the other. You can't leave both of an identical pair in. And see how once you pull one entry for each of the identical pairs out, it goes to 50%?

Or you can always add these in: 2M 1K & 1M 2K, if you prefer four out of eight to two out of four ;-D

See what you did there? You treated the order in which the puppies are picked as important in the case of the One of Each pairs, but irrelvant in the case of the All Male pairs. Like I keep saying, the people who get 33% are treating some lines of the matrix like they are a permutation, and others like they are a combination, and that's how they wind up double counting the One of Each pairs.

Jimmydanger:

werepossum:
Scenario #2: Toss a fair coin ten times in a row without looking at the results. Now begin looking at the results for the first nine. They are all heads. What is the chance of the tenth being heads? About one in a thousand. This is set probability.

Just back in this quagmire for a second this was explained wrong and I think it can help with the main problem.

you cannot look at the first nine coins in a set and still treat them as a set. In this example you have reduced yourself to only two options coin 10 is heads or tails leaving a 50-50 chance.

For this to remain a theoretical set no knowledge must be known about individual coins. Say if someone told you at least nine of those coins were heads. Then you would have 11 options. First all coins are heads OR coin 1 is tails OR coin is tails 2 OR coin 3 etc. Then the chance of all the coins being heads would be 1 in 11.

This is equivalent to both puppies male OR first puppy female OR second puppy female. Then the chance of both male is 1 in 3

Patently untrue. Mathematicians, scientists, and engineers constantly work with sets whose values are completely or partially known. Even if you have identified one or all puppies, you can still count them as a set - it's just that set probabilities no longer apply once you've identified a particular set item's value. While there is nothing wrong with your math, it represents a third set formation, not either of the two I discussed. My sets were a random set of ten coin tosses, of which the first nine (not any nine as in your example) came up heads; and a set of ten formed with the knowledge that the first nine (not any nine as in your example) are heads. In the latter case, the unlikely event of tossing nine heads in a row happens before you form your set. In the former case, each heads becomes more and more unlikely. I probably phrased it badly, but the first set was meant to represent the possibility that you have randomly thrown ten heads in a row before you know you have thrown nine heads in a row. The second set shows the change in the relative odds as you gain more information either by forming the set with more known information OR by discovering information about the set and/or its items.

My point with the two coin sets was that to show that when you form a set affects what possible probability distributions that set can take as much so as does what you subsequently learn about the set or its individual items. Each possible probability distribution has an unchanging absolute probability determined solely by the set construction - by the number of items and the possible values each can take. The relative value of each possible probability distribution is its absolute probability divided by the sum total of all possible probability distributions. When nothing is known about a particular set, the relative probability of each possible probability distribution is its absolute probability because the sum total of all possible probability distributions is one. As we learn information about the set OR about its individual items, we can eliminate some probability distributions that are possible for the set construction BUT which have been proved false for our particular set. The absolute probability of each possible probability distribution for the set construction remains unchanged because it is solely a function of the set construction, but the relative probability of each possible probability distribution for our particular set changes because by eliminating some possible probability distributions we reduce our divisor. This is true with both combinations and permutations.

To go back to our pup example, by learning there is at least one male we now have a 25% chance of a male/male combination and a 50% chance of a mixed pair combination. However our divisor has changed to 75% because we have eliminated the possibility (25% probability distribution) of a female/female combination. Thus the relative probability distributions remaining for our particular set are 25%/75% (or 1 in 3) for an all-male combination and 50%/75% (or 2 in 3) for a mixed pair combination. You cannot change the dividend without reforming the set; you can only change the divisor as you eliminate possible probability distributions. The problem is much the same for permutations, except that the 50% chance of a mixed pair combination becomes two separate 25% chances for male/female and female/male.

To see again how odds shift, let's look at the permutations. When I have two random pups, I know I have a 25% chance of having two male pups. When I learn that I have at least one male pup, my odds of having two increase to 33% because the 25% of having two females drops out of my divisor. It's still applicable to the set as a function and if I repeat this test a large number of times I'll get female/female pairs 25% of the time, but it's been proved false for my particular set. Now my chance of having a pair of males is 25%/(25%+25%+25%) = 33%. When I learn that one identifiable member of the set is male, my chance of having two males increases to 50% because one 25% permutation (that the identifiable member of the set is female whilst the other is male) drops out. Now my chance of having a pair of males is 25%/(25%+25%) = 50%. When I learn the other pup is male, my chance of having two males increases to 100% because I have all the knowledge I need to eliminate every other possible set probability distribution from this particular set.

Anyone who hasn't should read the Marilyn vos Savant link. http://en.wikipedia.org/wiki/Marilyn_vos_Savant
Besides shedding light on this problem, she's a fascinating person in her own right.

Those who remain absolutely convinced that 50% is the correct answer should definitely go into mathematics as a career, as you are smarter than every actual mathematician. Think of the opportunities!

Cheeze_Pavilion:
Okay, here's the problem. There's a difference between the sexes of a pair of puppies being unknown because we know nothing about the puppies, and the sexes of a pair of puppies being unknown because we know nothing about which pair was pulled from a pool of four known pairs, composed of one all male pair, one all female pair, and two mixed pairs.

What people are doing is they are confusing the permutation of possible results for two unknown puppies with a listing of the pool of puppy pairs from which a pair will be drawn because they both look the same.

In order for set probability to play a role, we need to know the composition of the pool from which the pair were drawn. WE DO NOT KNOW THAT! In order to know that the question would have to look like this:

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 they are one of four pairs that the breeder had, and the breeder picked the pair from a pool that was made up of one all male pair, two mixed pairs, and one all female pair. You tell her that you want only a male, and she telephones the fellow who's giving them a bath. "Is at least one a male?" she asks him. "Yes!" she informs you with a smile. What is the probability that the other one is a male?

See the kind of additional information you need to start using set probability? You can't just pretend because of a table of the possible permutations that you're picking from a pool of 4 pairs or 100 pairs like werepossum was talking about or 574543985770 pairs--until you get information that the pair of puppies was picked from a certain pool of pairs and you know something about the distribution of pairs in that pool, you can't just start using set probability.

Like I've been saying for a while now, you can't just pretend every symbol in a math problem refers to something in the real world. People have been reifying the possible permutations of two unknown events into actual pairs of puppies, and that is why they think they can just eliminate the F/F and rebalance the probability.

In short, the people who think it is 33% are mistaken because they think the information that at least one puppy is male means that the unknown pair was drawn from a pool of known pairs consisting of two mixed pairs and one all male pair, when what it really means is that a pair with at least one male was drawn from a pool of unknown pairs.

But you don't even need set mathematics to solve this. By your own example there are three out of four pup pairs that fit the description of "at least one male", but only one of those three sets will give you two males. What is the chance you'll draw the ONE of the THREE pairs of pups that gives you two males? The chance is ONE in THREE or 33%. You can't get any simpler than that.

The correct solution to this kind of question is quite simple.

If one is male and you wish to know if the other is male as well, you grab the damn phone and ASK.

werepossum:

But you don't even need set mathematics to solve this. By your own example there are three out of four pup pairs that fit the description of "at least one male", but only one of those three sets will give you two males. What is the chance you'll draw the ONE of the THREE pairs of pups that gives you two males? The chance is ONE in THREE or 33%. You can't get any simpler than that.

No, there are actually only three pairs. EDIT:

"she doesn't know whether they're male, female, or a pair"

that's three things in italics, which represent the possibilities which therefore number, um, three. What is screwing everyone up is the language--it would be clearer if the problem stated that she didn't know if they were:

a male pair, a female pair, or a mixed pair

Let's say RED represents a male puppy, and WHITE represents a female puppy. So a card with two RED sides represents a male pair, a card with one RED side and one WHITE side represents a mixed pair, and a card with two WHITE sides represents a female pair.

When we learn there is at least one male, that information allows us to remove the card with two WHITE sides, right? So our 'pool' is one card that is RED on both sides, and one card that is RED on one side, and WHITE on the other. What are the chances of picking either card? 50/50, right? So what are the chances of a male pair vs. a mixed pair? 50/50, just like the cards that represent those two possibilities.

Where we were all getting screwed up is that we thought the matrix of permutations represented the matrix of possibilities. Which it would if we had a known set of two male puppies and two female puppies, and we were picking two puppies of the four (see the @ note below). However, that's just the thing: we don't *know* our set contains two male puppies and two female puppies. We only know it contains two *unknown* puppies, which means our set could contain two male puppies, one male and one female puppy, or two female puppies, all of which are equally likely.

So while the permutation matrix is:

M M
M F
F M
F F

the probability matrix is

Two male .25 (.5 x .5)
One of Each .25 (.5 x .5)
Two female .25 (.5 x .5)

So when we eliminate 'Two female' we are left with:

Two male .25 (.5 x .5)
One of Each .25 (.5 x .5)

which have to add up to 1, and as each is half of .5, the new odds are half of 1, which is 50%

The critical mistake we were all kind of making is thinking that a mixed pair is *ever* more probable than two of the same. If we were talking about permutations of a known set, you would be TOTALLY correct. However, we're talking about an *unknown* set, where each outcome is the product of two events coming true. The thing is, the events are binary--if event A is true, event B is false. And think about the puppy problems in terms of events where A=male and B=female EDIT:

Two male: Event A came true once, and then Event A came true again
One of Each: Event A came true once (and therefore Event B was false once), and then Event B came true once (and therefore event A was false once)
Two female: Event B came true once, and then Event B came true again

The information from the Puppy Washing Man is that Event A comes true at least once, and therefore Event B is false at least once.

@notice how you all have the right answer for when we're picking two puppies out of a set of four, where two are male and two are female. When we know we have to pick a male, we wind up with a leftover pool of 1M and 2F--and there is your 33%!

EDIT: I'm thinking there *might* be a reason to say that an mixed pair is more probable than an all male pair once M has come true once--kind of like Monte Carlo simulations. The problem is in thinking that adding one more trial makes it twice as likely that you'll get a mixed pair than a double male pair--the increase in probability has got to be incredibly small, so the answer of getting a male pair is probably something like just under .5, by thousands--if not more--of a point. The Law of Large Number is for, well, large numbers! More likely here is that people are falling for the Gambler's fallacy and/or using math that isn't appropriate for a system without memory.

Doug:

So, breaking it down, we have

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, assuming P(male)=0.5, P(female)=0.5, we get:

Hence:
P(Male and Male) = 0.25
P(Male and Female) = 0.25
P(Female and Male) = 0.25
P(Female and Female) = 0.25

But if we assume the order isn't important, then we have:
P(2 Males) = 0.25
P(1 Male and 1 Female) = (0.25 + 0.25) = 0.5
P(2 Female) = 0.25

EDIT:
Think about what P(1 Male and 1 Female) *really* is, and don't just fill in values from above, and remember: you don't add probabilites of two events together to find out the probability of them both occuring, you multiply them, otherwise the chances of two male dogs would be P(1 Male and 1 Other Male)=P(.5 + .5)=P(1) which is obviously wrong, you agree?

What are the chances of a male? .5 right? What are the chances of a female? .5 again, right? So shouldn't P(1 Male and 1 Female) actually be:

P(1 Male and 1 Female)=P(.5 x .5)=.25

and from there I think you'll see why the chances are actually 50/50.

Just to try and make it more clear, think of a different permutation matrix. Think of albino alligators. Let's say you call the exotic pet store to ask about alligators, and they tell you that they have a pair of alligators, but they are off at the wrasslin' pit, so they don't know if they are albino, green, or a mix. What are the permutations?

Albino / Albino
Albino / Green
Green / Albino
Green / Green

No one would say 'hey, so each combination is equally likely,' right? The problem is, people think they can do that in the case of the puppies, because male and female are equally probable. However, a permutation matrix is only doubles as a probability matrix when there is a known set, like if we were told that there are two albino alligators and two green alligators, since no matter how rare albino alligators are in the wild, in a set of two albinos and two greens, the chances of picking one are 50%.

The thing is, we're not told we have two male puppies and two female puppies--we are just told we have two unknown puppies. So when one of the puppies becomes known--the Puppy Washing Man responding "Yes!" well, then we only have one unknown puppy and one male puppy, which leaves us with this permutation matrix where U stands for Unknown:

M / U
U / M

and since we know puppies can be either male or female, and *have* to be one of either and not the other, we can expand that to (XOR being the exclusive or):

M / (M XOR F)
(F XOR M) / M

so if we want to know the chances of two males, we get:

1 x .5
.5 x 1

Which leaves us with two equally likely options--one of which must occur--where the chances of getting two males is 50%, so we've got one event where the chance of two males is 50%, and if that event doesn't come true, the other event will come true, an event where the chances of two males are...50%!

Everyone starting to get why it's 50%? EDIT: And why this is really hard and you all got confused? I confused myself just then, treating a permutation matrix like all the rows were equally probable, which is why I wrote equally likely. Now, of course they are, but we can't know that just from a permutation matrix when the members of the set are at least partially unknown. But we know the rows of a permutation matrix represent all the possible outcomes. The reason we get 50% is because both outcomes have a 50% chance of yielding a two male puppy pair, and whatever the odds of one or the other, one or the other must come true.

I'll explain as best as I can, using coin flips as an example.

The problem breaks down to this:
Your friend flips 2 coins.
You ask if at least one coin came up heads.
Your friend says "Yes".
What are the chances that both coins came up heads?

Ok, so I think we can all agree that the odds of any one coin flip being heads is 0.5. So:

- Probability of a single flip:
- P(H) = 0.5
- P(T) = 0.5

Thus the probability of the friend flipping heads THEN heads again is 50%x50%, or 25%. Same logic for heads then tails, tails then heads, or tails then tails.

- Probability of each 2-flip combination
- P(H->H) = 0.5x0.5 = 0.25
- P(H->T) = 0.5x0.5 = 0.25
- P(T->H) = 0.5x0.5 = 0.25
- P(T->T) = 0.5x0.5 = 0.25

So, the total probability of 2 heads is 0.25, same as for 2 tails. But the probability of flipping one of each is the sum of the probability of heads THEN tails and the probability of tails THEN heads.

- H T --- Prob.
- 2 0 --- 0.25
- 1 1 --- 0.25+0.25 = 0.5
- 0 2 --- 0.25

Now, since we know at least one coin was heads, we can ignore the third row, leaving us with:

- H T --- Prob.
- 2 0 --- 0.25
- 1 1 --- 0.5

Which, when normalized, becomes:

- H T --- Prob.
- 2 0 --- 0.25 / (0.25+0.5) = 0.33
- 1 1 --- 0.5 / (0.25+0.5) = 0.66

Thus, the probability of both coins being heads is 0.33, or 33%.

Sorry Cheeze, you can argue all you want, but you're still going to be wrong.

FrcknFrckn:
But the probability of flipping one of each is the sum of the probability of heads THEN tails and the probability of tails THEN heads.

Sorry Cheeze, you can argue all you want, but you're still going to be wrong.

You've flipped twice trying to get the possibility of One of Each, while only flipping once for Two Heads or Two Tails.

That's the problem--people are mistaking permutations for actual flips/puppies/etc.

Cheeze_Pavilion:

FrcknFrckn:
But the probability of flipping one of each is the sum of the probability of heads THEN tails and the probability of tails THEN heads.

Sorry Cheeze, you can argue all you want, but you're still going to be wrong.

You've flipped twice trying to get the possibility of One of Each, while only flipping once for Two Heads or Two Tails.

That's the problem--people are mistaking permutations for actual flips/puppies/etc.

Nope - the actual probability of getting one head and one tail is twice that of getting 2 heads.

Try it - flip a pair of coins 100 times. You'll end up with approx. 25 H-H pairs, 25 T-T pairs, and 50 mixed pairs.

Heck, I could write you a program that would do thousands of tests if you'd like, it's pretty simple...

FrcknFrckn:

Nope - the actual probability of getting one head and one tail is twice that of getting 2 heads.

EDIT: Not when you're only flipping one coin--see the original problem: we know one of the 'coins' is heads. We only ever flip one coin--the value of the non-flipped coin is always Heads/Male.

Try it - flip a pair of coins 100 times. You'll end up with approx. 25 H-H pairs, 25 T-T pairs, and 50 mixed pairs.

Heck, I could write you a program that would do thousands of tests if you'd like, it's pretty simple...

Think about it--if the odds are the same after one additional trial as after 100, then there's something seriously wrong:

"It is important to remember that the LLN only applies (as the name indicates) when a large number of observations are considered. There is no principle that a small number of observations will converge to the expected value or that a streak of one value will immediately be "balanced" by the others. See the Gambler's fallacy."

Yeah, we know one is heads. BUT WE DON'T KNOW WHICH ONE. That, believe it or not, is the key factor. Either coin, or both, could be heads. Lets say the coins are different - say, a dime and a penny. There are 3 *equally likely* situations: DH PH, DH PT, and DT PH. Only one of the 3 has 2 heads. 33% chance.

So I went ahead and wrote the program, here's the output of 5 runs of 100000 flips each:

-----------------------------------

2 Heads: 24762 (24.762)%
2 Tails: 25135 (25.135)%
Mixed: 50103 (50.103)%

1+ Heads: 74865 (74.865)%
2 Heads: 24762 (24.762)%
Chance of 2 heads: 33.07553%

-----------------------------------

2 Heads: 24465 (24.465)%
2 Tails: 25270 (25.27)%
Mixed: 50265 (50.265)%

1+ Heads: 74730 (74.73)%
2 Heads: 24465 (24.465)%
Chance of 2 heads: 32.73786%

-----------------------------------

2 Heads: 24428 (24.428)%
2 Tails: 25591 (25.591)%
Mixed: 49981 (49.981)%

1+ Heads: 74409 (74.409)%
2 Heads: 24428 (24.428)%
Chance of 2 heads: 32.82936%

-----------------------------------

2 Heads: 24724 (24.724)%
2 Tails: 25337 (25.337)%
Mixed: 49939 (49.939)%

1+ Heads: 74663 (74.663)%
2 Heads: 24724 (24.724)%
Chance of 2 heads: 33.11412%

-----------------------------------

2 Heads: 24779 (24.779)%
2 Tails: 25257 (25.257)%
Mixed: 49964 (49.964)%

1+ Heads: 74743 (74.743)%
2 Heads: 24779 (24.779)%
Chance of 2 heads: 33.15227%

-----------------------------------

Should be pretty self-explanatory. Let me know if you want the source code.

Jeez, go on a business trip for 24 hours and another hundred posts flow in...

werepossum:

Alex_P:

werepossum:
Scenario #1: Toss a fair coin ten times in a row. If the first nine tosses come up heads, the tenth toss still has a 50-50 chain of being heads. Each coin flip is independent. This is sequential probability.

Scenario #2: Toss a fair coin ten times in a row without looking at the results. Now begin looking at the results for the first nine. They are all heads. What is the chance of the tenth being heads? About one in a thousand. This is set probability.

Your description of #2 sound fishy.

These scenarios are equivalent:
A. I flip a coin and look at it, then I flip another coin and look at it, &c., up to 10 coins.
B. I flip ten coins and then look at them in some specific order (it can be different from the order in which I flipped them).

The probability only changes when you start cherry-picking: searching until you find a male.

-- Alex

Not at all. Remember that the question is not "Is the tenth coin heads?" but rather "Is the tenth coin heads given that the first nine are heads"?

In the first scenario, I'm looking at the possibility of the tenth coin coming up heads KNOWING that the first nine came up heads. At this point, nothing I can do can change the outcome that at least nine of ten are heads. I have selected my set of ten with knowledge that eliminates the vast majority of probability distributions. In fact, only two possible probability distributions remain - nine heads followed by a tails, OR ten heads in a row. In selecting nine consecutive "heads" coin tosses, I have done the lion's share of obtaining ten straight "heads" coin tosses.

In the second scenario, I'm looking at the possibility that I have randomly selected a set of ten fair coin tosses which are all heads. There's a 50% chance the first toss will be tails; therefore half my random sets fail already even if I don't yet know it. I toss the second coin. My chance of being all heads is now 0.50 x 0.50 or 0.25, because both throws have to be heads to meet my criteria of ten heads. I can fail on the first throw, or on the second. It makes no difference whether I look at the values individually or all together at the end; I've selected them as a set. Now I toss the third coin; my chance of remaining all heads is now 0.50 x 0.50 x 0.50 or 0.125 (or 12.5% expressed as a percentage.) Now I toss the fourth coin: my chances of remaining all heads drop to 0.0625. The fifth, sixth, seventh, eighth, and ninth coin tosses take me to 0.3125, 0.15625, 0.0078125, 0.00390625, and 0.001953125 respectively. This number (0.195%) is the chance of a randomly selected set of ten fair coin tosses containing nine heads in any combination. If the first nine tosses were heads, then the tenth has the same base 50% chance to be heads. However, I selected the set randomly. To get the chance that I randomly selected ten coin tosses and they are all heads, I have to multiply 0.195% by 0.50 again - at which point I get 0.00098 or 0.098% - roughly one in a thousand. In this case, I have selected my set with a huge amount of possible permutations and combinations.

I picked ten coin tosses rather than four to emphasize the difference between sequential probability and set probability. If I group items together into a set AFTER I know their value, then the odds of the last item in the set are independent of the other items because I have negated all the possible probability distributions that don't fit that set's values. If I group items into a set BEFORE I know their values, then all possible probability distributions are still valid, and I can only eliminate them as I learn things about the set OR about individual items in the set.

To belabor a point, if I throw nine heads in a row with a coin that I know to be fair, the tenth heads is not remarkable. It's the nine in a row that are remarkable.

EDIT: Forgot to add, the chances are variables in set mathematics. If you alter the percentage chance of a particular permutation occurring, you alter the value distribution of the set. Thus the value of some set items must be changed. Since grouping the items into a set is merely a matter of convenience, it can't change the value of any particular item in the set; set mathematics can only describe probabilities of each particular combination or permutation of values of the items.

Here's the big distinction:

Given N coins,

"At least N-1 of these coins are heads" => 1/(N+1) probability of getting all heads

"At least these N-1 coins are heads" => 1/2 probability of getting all heads

I think your language is obscuring this.

If I say "I looked at three coins out of this set and they are all heads" (nine is too much typing for me), then my only options are:
HHHH
HHHT

I can't have something like HTHH because I already looked at the second coin and know it is not T.

If, however, I say "at least three coins out of this set are heads," then all of these are still kosher:
HHHH
THHH
HTHH
HHTH
HHHT

See what I mean?

-- Alex

FrcknFrckn:
Yeah, we know one is heads. BUT WE DON'T KNOW WHICH ONE.

Sure we do--the one that isn't known.

Take a look at this variation on the Three Card Problem:

http://www.escapistmagazine.com/forums/jump/18.73797.825805

So I went ahead and wrote the program, here's the output of 5 runs of 100000 flips each:

We're not asking about 5 runs of 100000 flips each, we're talking about one run of one flip: you've fallen for the Gambler's fallacy.

Cheeze_Pavilion:
No one would say 'hey, so each combination is equally likely,' right? The problem is, people think they can do that in the case of the puppies, because male and female are equally probable. However, a permutation matrix is only doubles as a probability matrix when there is a known set, like if we were told that there are two albino alligators and two green alligators, since no matter how rare albino alligators are in the wild, in a set of two albinos and two greens, the chances of picking one are 50%.

The subtext of the problem is that puppy nards are fair coins.

-- Alex

Alex_P:

Cheeze_Pavilion:
No one would say 'hey, so each combination is equally likely,' right? The problem is, people think they can do that in the case of the puppies, because male and female are equally probable. However, a permutation matrix is only doubles as a probability matrix when there is a known set, like if we were told that there are two albino alligators and two green alligators, since no matter how rare albino alligators are in the wild, in a set of two albinos and two greens, the chances of picking one are 50%.

The subtext of the problem is that puppy nards are fair coins.

I know, and that is what is screwing everyone up, even me when I wrote that entry.

Just because puppy nads are as fair as fair coins, that doesn't mean a permutation matrix is a probability matrix.

I mean, if the Law of Large Numbers is what predicts that we'll have an equal amount of HH, HT, TH, and HH, well, if it applicable to a number so small as the bare minimum number of coin flips to even have two results, why is it called the Law of Large Numbers? Why isn't it just called, like, the Law of Numbers, Great and Small.

Wise and Wonderful...

Bright and Beautiful...

okay, I'll stop.

Cheeze_Pavilion:

Alex_P:

Cheeze_Pavilion:
No one would say 'hey, so each combination is equally likely,' right? The problem is, people think they can do that in the case of the puppies, because male and female are equally probable. However, a permutation matrix is only doubles as a probability matrix when there is a known set, like if we were told that there are two albino alligators and two green alligators, since no matter how rare albino alligators are in the wild, in a set of two albinos and two greens, the chances of picking one are 50%.

The subtext of the problem is that puppy nards are fair coins.

I know, and that is what is screwing everyone up, even me when I wrote that entry.

Just because puppy nads are as fair as fair coins, that doesn't mean a permutation matrix is a probability matrix.

I mean, if the Law of Large Numbers is what predicts that we'll have an equal amount of HH, HT, TH, and HH, well, if it applicable to a number so small as the bare minimum number of coin flips to even have two results, why is it called the Law of Large Numbers? Why isn't it just called, like, the Law of Numbers, Great and Small.

Wise and Wonderful...

Bright and Beautiful...

okay, I'll stop.

Rephrase. Right now you're kinda just saying "Why is probability valid at all?"

-- Alex

It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.

EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.

gerrymander61:
It's 50%, or slightly higher given that the dogs could be identical twins. Anyone who says otherwise should take a class in basic stats or probability to find out just how STUPID they are.

EDIT: Wow, just checked the poll results. 34.6% of people here are stupid.

Take discrete math, please. It's actually like the best don't-have-to-be-a-math-major college math class: useful and light on memorization.

-- Alex

Doug:
Incorrect - we aren't assuming they choose from a random list of people with at least one male.

They selected from a random list of people, INCLUDING people with FF pairings.

Putting a sample population to it of 1000 people with dog pups, we get:

2xMale, 250 people
Answered: YES
1xMale 1xFemale 500 people
Answered: YES
2xFemale 250 people
Answered: NO

Hence, the resulting population is bias towards MF's

250 MM's vs 500 FM's

Good man. Nice to see that you've changed your mind when faced with evidence. Refreshing.

After this, I think The Escapist would explode if the cs peeps out there put those brain teasers interviewers are so fond of up.

I didn't realize this place was so fuzzy. Feels like peeps are arguing the way they'd argue about religion, politics, or polarizing video games rather than something techie like statistics and probability.

Cheeze_Pavilion:

FrcknFrckn:
Yeah, we know one is heads. BUT WE DON'T KNOW WHICH ONE.

Sure we do--the one that isn't known.

Take a look at this variation on the Three Card Problem:

http://www.escapistmagazine.com/forums/jump/18.73797.825805

So I went ahead and wrote the program, here's the output of 5 runs of 100000 flips each:

We're not asking about 5 runs of 100000 flips each, we're talking about one run of one flip: you've fallen for the Gambler's fallacy.

That's not the gambler's fallacy at all. You post the wikipedia page, but you don't bother to read it. Gambler's fallacy is where, after flipping one coin and seeing it land heads, you assume that the next coin you flip is more likely to land tails.

Problems like this and Monty Hall play on nonexamples of the Gambler's fallacy, making you believe that two events are independent, and are therefore equally probable, when in reality they aren't. For instance, Monty Hall makes you think that out of the two doors, there is an equal possibility that it could lie behind either. This makes you think that because at least one of two dogs is male, there is an equal possibility that the other could be male or female. Both use the opposite of the gambler's philosophy, making people believe that they are victims of the fallacy when they really aren't.

Samirat:

Cheeze_Pavilion:

Samirat:

You flip 4 coins and ask your friend if two of them are heads. He says "yes." What are the chances that the other two are heads compared to the chances that the other two are comprised of 1 head and 1 tails.

By your logic:
KKHH, KHKH, KHHK, HKKH, HKHK, HHKK = 6 possibilities

KKHT, KHKT, HKKT = 3 possibilities

I'm sorry but, it's not working out between us. You just don't get me, respond to quickly, or something, but we just can't get through to each other.

I mean, if you think my logic would be:

KKHT, KHKT, HKKT

and not something more like (just doing this fast now without double checking)

KKHT, KHKT, HKKT, HKTK, HTKK, KHTK, KTHK, HTKK, THKK, KKTH

You either don't understand me, how to do permutations, or both.

Sorry I couldn't be of more help.

Is it just me, or does that look like, I don't know, a listing of Soviet agencies?

Yeah, you're right, there should be 12 of the things. 4 spaces for the T, times 3 arrangements for the other three. Which still isn't correct. The chances are

6 2 heads 2 tails
4 3 heads 1 tails
1 4 heads

I really do want you to justify this to me, Cheeze.

Okay, one big fundamental misconception I keep seeing referenced over and over but never directly addressed. It sounds kinda like this:

"Two females don't satisfy the 'at least one male' requirement, so any experiment that has the potential to produce the two females result is an inappropriate model for this scenario."

That's pretty much gambler's fallacy at work. You're saying "This fair coin was heads this time so it can never be tails!"

Untrue! It's still a fair coin.

And, yes, that means that if you simulate the problem with two fair coins, then the answer to "Is at least one male?" is gonna be "no" 25% of the time. That's how it works. That's okay.

If you do something like setting one coin to make sure that you can never get TT, then what you're actually doing is breaking the "fair coin" rule completely -- your model ceases to be representative of the problem.

-- Alex

Alex_P:

Cheeze_Pavilion:

I know, and that is what is screwing everyone up, even me when I wrote that entry.

Just because puppy nads are as fair as fair coins, that doesn't mean a permutation matrix is a probability matrix.

I mean, if the Law of Large Numbers is what predicts that we'll have an equal amount of HH, HT, TH, and HH, well, if it applicable to a number so small as the bare minimum number of coin flips to even have two results, why is it called the Law of Large Numbers? Why isn't it just called, like, the Law of Numbers, Great and Small.

Wise and Wonderful...

Bright and Beautiful...

okay, I'll stop.

Rephrase. Right now you're kinda just saying "Why is probability valid at all?"

Funny because to me, that's what everyone else is saying--'watch me use the Law of Large Numbers to come to a conclusion about...the smallest possible number!'

Not to mention that fact that the people who claim to have the right answer can't agree on the right method, which is...kinda troubling, don't you think?

Try this out, and see what you think:

Let's say RED represents a male puppy, and WHITE represents a female puppy. So a card with two RED sides represents a male pair, a card with one RED side and one WHITE side represents a mixed pair, and a card with two WHITE sides represents a female pair.

When we learn there is at least one male, that information allows us to remove the card with two WHITE sides, right? So our 'pool' is one card that is RED on both sides, and one card that is RED on one side, and WHITE on the other. What are the chances of picking either card? 50/50, right? So what are the chances of a male pair vs. a mixed pair? 50/50, just like the cards that represent those two possibilities.

That sound right to you? I think even if it doesn't it gets to the heart of this disagreement.