1. You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She says she has two children. You ask if she has any girls. What is the chance that at least one child is a girl?

*Replay with slight change *

2. You are at the park, pushing your daughter on a swing. You meet a woman with her daughter and start a conversation. She says she has two children. What is the chance that both children are girls?

*Rewind, again*

3. You are at the park, pushing your daughter on a swing. You meet a woman and start a conversation. She says she has two children. You ask if she has any girls. She says yes. What is the chance that both children are girls?

Examples from Randmoness by Deborah J. Bennet

Answers, if I got this right, are 3/4; 1/2; 1/3

Now this is pretty low level stuff, but that is what I think is so interesting. Its like sleight of hand slowed down. Makes me wonder what is going on metaphysically as the new information arrives? Is that what information does to the world -- changes the likelihood of outcomes?

Very good indeed. And the answer to "Is that what information does to the world -- changes the likelihood of outcomes?" is the trivial one: without info your smoked, with some, you might have a chance.

This is the best example of cond. prob. I've seen - strips it down to the very basics. I had to think a while about what separates 2. and 3. before it got clear that in 2. you have all the information in 3. *plus* the info that a randomly selected child in one example had happened to be a girl - which should mark up the 2-girl probability.

Posted by: Mats | November 15, 2003 at 01:25 PM

Are you sure that there is a difference between #2 and #3? In #2 you see that one of her children is a girl, in #3 you are told so. Where's the additional information?

The only thing I can see is that in #2 you know that she is not lying about having a daughter. But that seems outside the scope of the problem.

Posted by: Anonymous Coward | November 15, 2003 at 06:43 PM

It helps if you picture the combinations:

Scenario 1 has the following possibilities

B, G

G, B

B, B

G, G

So at least one girl in scenario one has the odds 3/4

In scenario 2 the combinations are:

The kid you see: G, The kid you don't: G

The kid you see: G, The kid you don't: B

There are only 2 permutations -- so the odds of two girls are 1/2

In number 3 you know she has 2 kids and least one is a girl

(so the combination can't be B,B) but the remaining possibilities are:

G, B

B, G

G, G

So the chances of 2 girls is 1/3

It is weird. The question doesn't ask which age is the boy or girl, but the sequence changes the probabilies. Think of like this: would you rather guess the sum of two dies before you throw them both, or after you throw the first one?

It actually took gamblers centuries to realize that sequence (BG, GB) and set (2 kids: 1 B, 1 G) were two different things.

Posted by: joshp | November 15, 2003 at 09:35 PM

Your analogy to the dice is flawed. The sum of the throws is not an independent variable. Still, I would probably wouldn't care about the first throw and always guess seven. It's the most likely result in the absense of knowledge of the first throw (there's a 1 in 6 chance of getting a seven). And after the first throw, there's still a 1 in 6 chance of getting a seven--no matter what the outcome of the first throw.

My analysis goes something like this:

I will assume that there is a 50/50 chance of a child being a boy or a girl. My understanding is that the odds of conceiving a boy are slightly greater but that boys are a bit more fragile and so it ends up that the probabilities shift so that the odds of actually giving birth to a girl end up being higher.

You know the woman has two children. There are four possible sequences, each equally likely

(B,B); (B,G); (G,B); (G,G)

With no other information, we note that 3 in 4 have a girl. More interisting for the other problems, there is a 1 in 4 chance that both are girls.

Now you are introduced to her daughter. From this we note that (B,B) is not a possibility. So the odds of having two girls now jump to 1 in 3.

We still don't know if the woman brought her older child or the younger one. If she tells us that this is the older one, then we have eliminated (B,G). So now the chance of two girls is suddenly 1 in 2.

Posted by: Anonymous Coward | November 15, 2003 at 11:04 PM

You aren't getting the impact of the observation of the girl at the park. Sorry if the dice example was unhelpful, but it is a mistake to say that the observation of the girl doesn't change the probable outcome.

Agreed that in example 1 there are 4 outcomes:

B, G

G, B

B, B

G, G

But in example 2 you have the additional factor of observing her daughter at the park. Now, assume that it is equally likely for a son or a daughter to be at the park with their mother.

A sample space for this has 8 outcomes (not 4) that are triples whose first element is either B or G for the sex of the elder child, B or G for the younger child, and whose third coordinate is E or Y indicating whether the elder child or younger child was at the park.

For example, (B, G, Y) is the outcome that the elder child is a boy, the younger child is a girl, and the girl was at the park. Here are the 8 outcomes:

1. (B, G, Y)

2. (B, G, E)

3. (G, B, Y)

4. (G, B, E)

5. (B, B, Y)

6. (B, B, E)

7. (G, G, Y)

8. (G, G, E)

Now, given that you saw the girl at the park, what are the possible outcomes?

1. (B, G, Y) - Yes

2. (B, G, E) - No

3. (G, B, Y) - No

4. (G, B, E) - Yes

5. (B, B, Y) - No

6. (B, B, E) - No

7. (G, G, Y) - Yes

8. (G, G, E) - Yes

Now take just the possible outcomes:

1. (B, G, Y) - Yes

4. (G, B, E) - Yes

7. (G, G, Y) - Yes

8. (G, G, E) - Yes

I think you can see from that the answer is 1/2 as G, G is an outcome in 2 of the 4 possible outcomes.

Here it is more formally:

(a) Let T be the event that the household has two girls, and O be the event that a girl was at the park. List the outcomes in T and O.

Solution. T = {GGE,GGY } ,O = {GGE,GGY,GBE,BGY } �

(b) What is the probability Pr {T | O}, that both children are girls, given that a girl was at the park? Solution. 1/2

Here's a link to a similar problem from some MIT homework:

http://ocw.mit.edu/NR/rdonlyres/A71684EE-2C65-4B27-9C03-E98F6E750054/0/cp10Fsol.pdf

Posted by: joshp | November 16, 2003 at 01:54 PM

See - you disagreed exactly on what differed between 2. and 3. "before it got clear that in 2. you have all the information in 3. *plus* the info that a randomly selected child in one example had happened to be a girl"

Posted by: Mats | November 16, 2003 at 02:45 PM

Yep - you had it figured out at the start!

Posted by: joshp | November 16, 2003 at 05:56 PM

What bothers me about problems like this is that the answers to the questions are not real answers with any level of certainty, but still just guesses. They represent a compromise between knowledge (of possibilities) and uncertainty (about other factors that could play a part, like genetic disposition (if such a thing exists), preference of the parent (if they happen to be adopted), effect of the water they're drinking from, etc). The answer to all of them could just as easily be 1/10, and the only way to prove that wrong is to make assumptions about the universe in which they live that you can't prove).

In other words, we are not getting any closer to being certain about what the unknown information is by giving it a probability... we're merely putting a name to a face. We call this type of uncertainty 3/4ths, this type 2/3rds, and this type 1/2, just like we give birds names but learn nothing about the bird by its name. So, in reference to your question about whether or not information changes the likelihood of outcomes, it doesn't, since nothing has actually changed.

But maybe that's an old fashioned way of thinking about this. Real math (of which I'm greatly ignorant) seems to say that probabilities are a lot more real than they seem at first glance. If that's the case, I'd like to learn more about it.

Posted by: Erik Benson | November 17, 2003 at 12:37 AM

Ah, I see. But your question #2 didn't say that the mother took a random child to the park. Without that key piece of data, I don't think there's any difference between #2 and #3.

By the way, the MIT problem is a lot clearer, because it doesn't leave open these ambiguities.

Posted by: Anonymous Coward | November 17, 2003 at 01:22 PM

It is probably always a better idea to get your probability homework assigned by a professional than collect it from random weblogs.

Posted by: joshp | November 17, 2003 at 03:26 PM

You're wrong on 3, it's 50%, i am pretty positive. By saying at least one is a girl, you are basically taking that one out of the equation and the question becomes "what are the odds the unidentified one is a girl". 50%

Posted by: chet | November 18, 2003 at 09:01 AM

I don't understand why the woman admitting to having at least one girl (#3) gives less information that seeing the girl.

Posted by: Michael McDaniel | November 18, 2003 at 10:03 AM

Chet,

It doesn't say at least one. You ask "Any girls" and the woman answers "yes". Though I don't think that would change the probability at all.

This is what I mean by people getting probability wrong at low levels. Let us agree that there are 4 combinations for two kids?

B, G

G, B

G, G

B, B

When the woman has "at least one" girl, we can eliminate B, B as an option. You are left with 1/3 as the odds that both kids are girls. The confusion for you here is that the sequence matters, not just the set. You might not care if the elder or younger kid is a girl, but the fact that there are 2 and there are multiple sequences changes the probabilities.

Posted by: joshp | November 18, 2003 at 10:03 AM

The kid at the park is another variable. It could have been any kid, but it was a daughter. That is not equiprobable, so you can work out the conditional probabliity for that event -- distinct from the other 2 examples.

Posted by: Josh Petersen | November 18, 2003 at 10:05 AM

I have question.

It seems that what's philosophically interesting about all of this is the relationship of subjective information to the outcomes of objective probabilities. In other words why does there seem to be a relationship? It's like the quantum "problem of measurement" on a macro scale.

This leads me to my quesion: in statistics (which I now very little about), at what point do you have to "collapse" the probabilities so that the question you are even asking correlates to a physical result?

In Question #1, everyone agrees that the probability of a girl is 3/4 after looking at this distribution:

(B,B) (B,G) (G,B) (G,G)

However, I didn't see it mentioned above that it's also equally true, given that distribution, that the probability of a boy is ALSO 3/4.

Obviously this can't correlate to the results of an actual experiment. You can't run this experiment over and over again 100 times and come up with girls 75 times and boys 75 times. So, what gives?

If I'm asking for how many girls, then I'll get an answer that seems to makes sense: 3 out 4, but it only makes if I toss out the possibility of even asking for the number of boys which looks like it also can occur 3 out of 4 times. So I can't ask both at the same time? In other words it seems like I have to "collapse the probability function" by choosing to ask for only the probability of a girl.

I know I'm missing something very basic here, but after shying away from statistics in college only to find out that I'm interested now, what is it that I'm missing?

Posted by: Jeff Brownell | November 23, 2003 at 12:08 PM

Doh! never mind my previous question.

Two 75% probabilities is no problem when there are two children to spread the probabilities over.

Note to self: don't enter debates about statistics until you have read a LITTLE bit about it.

Posted by: Jeff Brownell | November 23, 2003 at 11:41 PM

Hey Jeff, since there are two children that we're making guesses about in the first question, it's possible for there to be at least one boy 75% of the time and at least one girl 75% of the time. In other words, it's possible to satisfy both of your questions at once--when there's one boy and one girl (B,G or G,B), there's no conflict. Therefore, the probabilities should somewhat closely match the results of actual experimentation.

Posted by: Erik Benson | November 23, 2003 at 11:53 PM

On the the woman with two children at home, with at least one being a girl. What is the chance that the other child is also a girl?

Let's look at the example before, where you see the girl on a swing. You have G(swing) and then either B(not on swing) or G(swing) and G(not on swing). 1/2 or 50%

Now let's try the 2nd case (as another writer also figured out.)

We know the lady has a girl. So all we know is that

G(unseen).

G(unseen) B(unseen) or G(unseen) G(unseen). Again that is 1/2 or 50%.

If you try to use B(unseen) G(unseen) and get 1/3 or 33% you are in affect saying that the child you know absolutely nothing about it's sex-chromosomes has twice as much chance as being a boy as a girl. And in reality its a 1/2 or 50% chance. You do eliminate the BB condition, but you must also eliminate the BG situation, since you already know one child is a girl.

The probability that one child is a girl already is 100%. You can not use that child in the probability of the second child's sex. That child's sex probability will be 1/2.

Posted by: Bob | November 04, 2004 at 06:18 AM

In my conclusion I should have also stated, that the probablity of one of the children being a girl is 100%. The probability of the other child being a boy is 50%. So

100% times 50% equals 50% or:

1/1 times 1/2 = 1/2

Posted by: Bob | November 04, 2004 at 06:41 AM

Change my last sentence to: The probablity of the other child being a girl is 50%. (sorry)

Posted by: bob | November 04, 2004 at 06:43 AM

the calculations are correct, and it is a great example. If you don't get it, think harder and speak out your butt less. You will benefit, and so will everyone else.

Not to pick on this guy, but...

"Obviously this can't correlate to the results of an actual experiment. You can't run this experiment over and over again 100 times and come up with girls 75 times and boys 75 times. So, what gives?"

Answer: yes you can. You are NOT PICKING ONE SPOT, YOU ARE PICKING PAIRS. (how likely is it that, given 100 PAIRS of children, the oldest is a girl OR the youngest is a girl, OR both are girls? The answer is 75/100 - in actuality) Think, and DO NOT assume a solution is obvious. It is obviously not obvious. I have found in life that the best way to learn is to admit when you don't know what you are talking about, and simply get somebody who does to explain it to you til you do!

"The probability that one child is a girl already is 100%. You can not use that child in the probability of the second child's sex. That child's sex probability will be 1/2."

Answer: Yes you can. You DO NOT KNOW WHICH ONE (OF THE PAIR) THAT THIS CHILD IS!

"You're wrong on 3, it's 50%, i am pretty positive. By saying at least one is a girl, you are basically taking that one out of the equation and the question becomes "what are the odds the unidentified one is a girl". 50%"

Answer: You are not taking it out of the equation. You are taking BB out of the equation.

I could go on but I see no point. Very few people seem to want to admit that they suck at probability. The "probably" should approach the subject with a little more fascination and humility, because it is extremely difficult and rewarding.

Here's some great advice (really):

make a probability tree diagram. look it up. they work, they will give you the right answer if you set them up right, and it won't hurt your brain! Sorry I sound like a jerk right now, but please people, let's not go accusing this poor guy that his example is wrong when you know you dont know what you're talking about. Am I right? (yes) That's what they call a rhetorical question.

Posted by: nate | December 13, 2004 at 02:08 PM

Conditional probability:

P(B|A) = P(a and b)/P(a)

For #2:

P(2 girls given this one is a girl)= P(1 is a girl and both are girls)/P(child at park is a girl)

=P(both girls)/P(child at park is a girl)

=0.25/0.5

=1/2

P(2 girls given 1 is a girl)= P(1 is a girl and both are girls)/P(1 is a girl)

= P(both girls)/P(at least 1 girl)

= 0.25/0.75

= 1/3

Posted by: Ryan | September 02, 2009 at 06:44 PM

Probaility but it doesn't change the variable that there are something that it affects women to have more boys that girls. If I ain't wrong there are a town that most of the newborns are women

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If I ain't wrong there are a town that most of the newborns are women

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