Another math problem

[engineereeyore]

Doesn't it depend on how you're asking the question? If you simply want the probability of the child being a boy or a girl, it's 50/50. If you're wanting the probability of having a girl and two boys verses three boys, it's 75/25. Which question is actually being asked?

 

[DrPizza]

bbb bbb bbb bbb bbb bbb
bbg bbg bgb bgb gbb gbb
bbg bbg bgb bgb gbb gbb
ggb ggb gbg gbg bgg bgg
bbg bbg bgb bgb gbb gbb
ggb ggb gbg gbg bgg bgg
ggb ggb gbg gbg bgg bgg
ggg ggg ggg ggg ggg ggg

Ooops, I fixed it in one spot, but not the second spot. I am not claiming that the age of the two children the man meets first matters. We all know, and there's no argument about this point, that there are 8 different possibilities for 3 children. Boy born first, Girl second, Boy third... BBB, BGG, etc.

And, none of us are arguing that if we know that if we're told that a man has at least 2 boys, that there is a 3/4 chance that the third child is a girl.

I AGREE! I completely agree that if all you know is that a man has 3 children and at least 2 are boys, then there's a 75% chance that the third child is a girl.

However, my point is a much deeper thinking of this problem! You've just met a person with 2 boys present. If a million families with 3 children pick 2 of them to go to the grocery store, 100% of the time, the family with 3 boys will have 2 boys with them. This is not true for any other families. Sure, 37.5% of the families have 2 boys and 1 girl(three times as many families as those with 3 boys) BUT, the probability of meeting any of those families with 2 boys present is only 33.3% for each of those families.


Someone stated above that I'm over-complicating the problem. I disagree. They stated I only needed to put BBB, BBG, BGB, and GBB into a hat. Clearly, the probability is 3/4 that the 3rd child is a girl.

What I'm saying is that this simplification is incorrect. What we need are 8 hats. Into 1 hat, we place 3 marbles: BBB, into another BBG, then BGB, then BGG, etc.
Now, we pick a hat at random and draw 2 marbles out. For the BBB hat, we'll select that hat 1/8 of the time. 100% of the time, we'll select 2 B marbles. Thus 1/8 of the time, we are on the BBB hat. 1/8 of the time, we'll be on the BBG hat; 1/8 of the time, the BGB hat, and 1/8 of the time, the GBB hat. For each of these hats, the probability of pulling out 2 B marbles is only 1/3. Thus, 1/3 * 1/8 = 1/24. 1/24* 3hats = 1/8.

Thus, 1/8 of the time when we meet a 3-childed man on the street with two (randomly chosen) of his children with him, he'll have 2 boys with him and the 3rd child at home is a boy. 1/8 of the time we meet a 3-childed man with 2 children (of his, at random) with him, he'll have 2 boys with him and the 3rd child at home is a girl. and 3/4 of the time when we meet a man with 2 of his 3 children with him, he won't have 2 boys with him; so we can ignore those occasions.

See, it all comes down to how the problem is worded. I disagree that the only thing we know from the initial wording of the problem is that a man has at least 2 boys.

*******
There's a similar problem to think about. I put 3 cards into a hat. One card is black on both sides. One card is green on both sides. The other card is black on one side and green on the other. You're going to get to pull out one card and look at one side only.

If I say, I have a card and at least 1 side is black, then we know there's a 50% chance that the other side of the card is black and a 50% chance that the other side is green. There's a 0% chance that we have the card that's green on both sides. Phrasing it this way: "at least 1 side is black" is how you guys are interpretting the original problem in this thread.

However, there's a difference between being told that at least 1 side is black and selecting a card at random and seeing one black side. In the case that you pull out a card and see a black side, there's a 2/3 chance that the other side is black, and a 1/3 chance that the other side is green.

Think about this new problem for a moment, then look back at the original problem. Maybe you can see the similarities.

 

[DrPizza]

Originally posted by: Born2bwire
There is also the problem that you, like everyone else who states 50%, is still forcing a condition upon the relative ages of the children when there is no information given in the OP about their ages. People keep looking at this problem in the same sense that the wife is just about to give birth to the third child when in fact the third child could have been born at any of three relative ages. This increases the probability that we have a girl because the girl has three chances to have been born.

You did not understand my post. See my post above.

 

[DrPizza]

First, let me reitterate: I agree wholeheartedly that if you know a man has 3 children and you are told that at least 2 of the children are boys, the probability that the third child is a girl is 75%. Unfortunately, that's not how the problem was stated. Nor was the problem stated ambiguously. Quite simply, that's the problem that was intended to be stated, but unfortunately, a completely different problem was stated.

I think I came up with the best way to show, by analogy, how the statement of the 3 children problem is incorrect.

I have in a hat 3 cards:
1 is black on both sides
1 is red on both sides
1 is black on one side and red on the other side.

If I state "I looked at all three cards and decided to keep one of them. The card I kept has at least 1 side that's black" then we can deduce that it's either the BB card or the BR card. There's a 50% chance that the other side is black, and a 50% chance that the other side is red.

That's how you guys are interpreting the problem. Unfortunately, I disagree with that interpretation.

If I instead tell you to pull out a card and look at one side only, and you see a black side, then the probability is no longer 50/50 that the other side is red/black. We're now looking at one side out of 6; each of those sides is, of course, attached to another side. In 2 of those cases, we see a black side, the other side is black. In only 1 case out of the 6, we'll see black and the other side is red.

Thus, if we look at one side of 1 cards and see black, then there's a 2/3 chance that the other side is black, and only a 1/3 chance that the other side is red.

Now, suppose we know that there are millions of these cards out there; 1/3 of each type. And, I spot one on the ground and a black side is showing. There's a 2/3 chance that the other side is black.

This is the same situation as with the family - you're not simply told that the man has at least 2 boys. Instead, you see two boys with the man. At first, last night, I wasn't 100% certain of my solution. Right now, I'm 99.99% certain.

***

If you understand the above problem with the cards (It's a fairly well known intro statistics and probability problem), then allow me to make up a similar problem. You have 3-sided cards (shaped like brazil nuts)
On one of those cards, the sides are black, black, and black
On 3 of those cards, the sides are black, black, green
on 3 of those cards, the sides are black, green, green
and on 1 of those cards, the sides are green, green green.

You are told that I have a card and at least 2 of the sides are black. Conclusion: 3/4 chance that the other side is green. (Get it yet? BBG)

However, you are NOT told that at least 2 of the sides are black. Instead, you are told to select a card at random and look at two of the sides. You see black and black. What's the probability that the 3rd side is green, and what's the probability that the 3rd side is black. Assuming you understood the 2 sided card problem, I assume you can calculate that it's 50/50. Or, instead of pulling a card out at random, you see one laying on the street, next to a man. Unfortunately, being shaped like a brazil nut, you can't see the third side. But, you see two of the sides. And, those sides are BB. There's a 50% chance the other side is Boy and a 50% chance the other side is Girl.

***

And lastly, so that I can cross all my t's and dot all my i's, I bothered to look at the link about Marilyn Vos Savant<div class="FTQUOTE"><begin quote>
You meet a woman, and ask how many children she has, and she replies "two." You ask if she has any boys, and she replies "yes." After this brief conversation, you know that the woman has exactly two children, at least one of whom is a boy. When the question is interpreted this way, the probability that both of her children are boys is 1/3, as Marilyn has claimed.

You meet a woman and her son. You ask the woman how many children she has, and she replies "two." So now you know that this woman has exactly two children, at least one of whom is a boy. When the question is interpreted this way, the probability that both of her children are boys is 1/2, as Eldon has claimed.
Because the question does not state how we obtained the information, the question is ambiguous. </end quote></div>

Except, in this thread, it *IS* stated how we obtained the information.
Convinced yet?

p.s. If it makes you feel better, I'll bet that at least some of the original 50% people happened on the correct answer through incorrect logic.

edit:

Originally posted by: Nathelion
I just wanted to watch people mess up and not get it at all for weeks and weeks while I smugly smile into my beard and laugh and feel superior, and you just pop the answer out like that? Horrible manners:evil:

Still feeling superior?

 

[lumbus]

Originally posted by: DrPizza
First, let me reitterate: I agree wholeheartedly that if you know a man has 3 children and you are told that at least 2 of the children are boys, the probability that the third child is a girl is 75%. Unfortunately, that's not how the problem was stated.

What on earth are you talking about? That is exactly how the problem was stated:

a man and two boys greet you. The man says "These two boys are my sons, but I have three children.

You can come up with any other problems you want, but the correct answer to the opening post is still 75%.

 

[DrPizza]

No, it's not stated that way. It's stated that you meet the man with 2 of his boys. If a man has 3 boys, then the probability of meeting him with 2 boys present is greater than the probability of meeting him with 2 boys present if he had 2 boys and a girl. Your error is in neglecting this fact.

 

[Nathelion]

I am somewhat pleased to see that this thread is still going

Originally posted by: DrPizza
bbb
bbb bbb bbb bbb bbb bbb
bbg
bbb bbb bgb bgb gbb gbb
bgb
bbb bbb bgb bgb gbb gbb
bgg
ggb ggb gbg gbg bgg bgg
gbb
bbg bbg bgb bgb gbb gbb
gbg
ggb ggb gbg gbg bgg bgg
ggb
ggb ggb gbg gbg bgg bgg
ggg
ggg ggg ggg ggg ggg ggg

You know, at least at first, I was going to agree with the 75%. Okay, what am I missing here. First, I listed all the possibilities, in order, of having 3 children. These are on the lines with just three letters as in bgb. Then, the next line lists the six possible arrangments of three children.

Lets assume for a moment that we have 48 houses. 6 houses for each way that three children can be born, i.e. girl first, boy second, boy third, or GBB.
Now, in each of these 6 houses, we put the three children into every possible order (six orders) that you can have the three children. So, in the case of GBB, aka Suzie, Tom, Dave, we can have STD, SDT, DTS, DST, TDS, TSD.

Now, when we ring a the doorbell on any house, we know there's a 1 in 8 chance that there are 3 boys, or a 1 in 8 chance of 3 girls, or a 3 in 8 chance of 2 girls and 1 boy, etc.

However, the rule is that the people in the house where we ring the doorbell are going to show us the first two children in the order assigned to them. We see two boys:


bbb bbb bbb bbb bbb bbb
bbg bbg bgb bgb gbb gbb
bbg bbg bgb bgb gbb gbb
ggb ggb gbg gbg bgg bgg
bbg bbg bgb bgb gbb gbb
ggb ggb gbg gbg bgg bgg
ggb ggb gbg gbg bgg bgg
ggg ggg ggg ggg ggg ggg

Now, we can see where the 75% chance in the 2nd post came from. However, if you look at this, you can see that if a family had 3 boys born vs. a family with 2 boys and 1 girl has a greater probability of someone meeting the two boys first.

Notice, out of the 48 homes, in 12 of those homes, you would see two boys first. in 6 of those 12 homes, the 3rd child is a boy. In 6 of those homes, the 3rd child is a girl.


My conclusion: 50% probability that the 3rd child is a girl, 50% probability that the 3rd child is a boy.
My second conclusion: people who know less about statistics are more likely to get this problem correct than people who know more statistics. (That's assuming I'm correct... someone care to poke holes in my logic??)

You are entirely right in stating that the correct solution is 50%, the reason I went back to this thread is because I realized this as well. The original problem is misstated, and the probability of meeting a girl in the yard is indeed 50% in its current form.

You're saying it in a sort of confusing way, but it's still basically correct. Since nothing is stated about how the man in the house would pick the 2 children that he chooses meet you with, it must be assumed that it was done at random. So out of the three children he picks two at random and places them in the house to meet you, and puts the third in the yard.

Now there are four possible families that you can meet that you fulfill the requirements, namely

bbb
bbg
bgb
gbb

So it would seem that the probability of there being a girl in the yard is 75% - but that assumes that it is equally likely that you meet all the families above. But as we will see, the bbb family is three times as likely as the others.

Suppose that you have many houses with 3-child families, and you go into one of them at random. Now if you walk into a house where there is at least one girl meeting you (all the families in the houses always sends exactly 2 children to meet you), you back out again and go to another house. Sooner or later you will be met by two boys.

3 child families:

bbb
bbg bgb gbb
bgg gbg ggb
ggg

Now in the case of the families with less than 2 boys, you will always leave again, so all of those are out. Remaining are:

bbb
bbg bgb gbb

You are equally likely to visit each type of family, which is where the 75% figure comes from. But there is more. Now if you walk into the house with three boys, you will be met by 2 boys 100% of the time, so our conditions are always met in that case. However, if we walk into one of the homes with 2 boys the parents can choose to send 2 kids to meet us in the following ways (A,B, and C are names):
ABC
ACB
BAC
CAB
BCA
CBA

Let's assume the child corresponding to the 3rd letter is left in the yard, and that A and B are boys (it doesn't matter which we pick). We can see that there are 2 cases out of six - all equally likely - where we would be met by boys. So if we walk into a house with two boys, there's only a 1/3 chance that two boys are actually sent to meet us! So two thirds of the time one of the children we meet would be a girl, and we would back out.
Now the probabilities of our criteria being fulfilled (that is, that there are two boys that meet us) is for the four families we can meet:
bbb = 1
bbg = 1/3
bgb = 1/3
gbb = 1/3
In other words, while there are three times as many homes with girls that we can walk into, the homes where there is a girl are inly 1/3 as likely to fulfill our criteria by only sending boys to meet us. So when we have walked into a house and have met two boys, we will be standing in a home with only boys 3 times as often as in a home with any specific of the 3 variants of families with one girl and two boys. The probabilities of us having encountered any particular type of family, then, are:

bbb = 0.5
bbg = 1/6
bgb = 1/6
gbb = 1/6

Now we can clearly see that the chance of us having met a family meets our requirement of having a in the back yard is 50%, not 75%.

 

[Nathelion]

In order to make the probability of there being a girl in the back yard, an addition would have to be made to the problem statement. For example, one could add:
"The neighborhood you are in is culturally homogenous, and in this particular culture families with three children always send two children to meet visitors and if they can they will send only boys".

 

[lumbus]

Originally posted by: DrPizza
No, it's not stated that way. It's stated that you meet the man with 2 of his boys. If a man has 3 boys, then the probability of meeting him with 2 boys present is greater than the probability of meeting him with 2 boys present if he had 2 boys and a girl. Your error is in neglecting this fact.

Wrong. Let's look at your admission again:

Originally posted by: DrPizza
First, let me reitterate: I agree wholeheartedly that if you know a man has 3 children and you are told that at least 2 of the children are boys, the probability that the third child is a girl is 75%. Unfortunately, that's not how the problem was stated.

And the problem:

a man and two boys greet you. The man says "These two boys are my sons, but I have three children.

The key information here is that you know that he has at least two boys. Whether you came by this information by him telling you, you seeing it (him showing you), him writing a note to you, him emailing you etc etc is irrelevant and doesn't change anything.

Figuring out the probability of the man showing you two boys is not part of the problem. It has already happened so the probability of that is 100%.

 

[lumbus]

Originally posted by: Nathelion
I am somewhat pleased to see that this thread is still going
So two thirds of the time one of the children we meet would be a girl, and we would back out.

There is not going to be any backing out, because the fact that we have met two boys is a given.

Originally posted by: Nathelion
In order to make the probability of there being a girl in the back yard, an addition would have to be made to the problem statement. For example, one could add:
"The neighborhood you are in is culturally homogenous, and in this particular culture families with three children always send two children to meet visitors and if they can they will send only boys".

That's not part of the problem. And for your reasoning to make any sense, you would have to list additional conditions, such as the parents pick two out of three kids at random, and each kid as an equal probability of being picked. The reason that this conditions aren't listed is that they aren't part of the problem.

 

[dealmaster00]

Originally posted by: DrPizza
No, it's not stated that way. It's stated that you meet the man with 2 of his boys. If a man has 3 boys, then the probability of meeting him with 2 boys present is greater than the probability of meeting him with 2 boys present if he had 2 boys and a girl. Your error is in neglecting this fact.

I can see where you are coming from, but I don't believe it is correct. The way the problem is stated, it is given that 2 of the children are boys. So the probabilities before that don't matter.

<edit> lumbus already said that above me.

 

[DrPizza]

I again, I still disagree. HOW he finds this out matters. In fact, if you look at the link to the site with a discussion about the problem as posed by Marilyn Vos Savant, you'll see that they make the exact same point.

Maybe you're not reading through my entire post and are just clinging to your belief. Look at the similar problem I posted about 3 two-sided cards. It's the exact same problem. If you know that a card has at least one black side because I told you, it has a different probability that the opposite side is black than if you pull the card out of a hat and look at the one side.

 

[Born2bwire]

Originally posted by: DrPizza
<div class="FTQUOTE"><begin quote>Originally posted by: Born2bwire
There is also the problem that you, like everyone else who states 50%, is still forcing a condition upon the relative ages of the children when there is no information given in the OP about their ages. People keep looking at this problem in the same sense that the wife is just about to give birth to the third child when in fact the third child could have been born at any of three relative ages. This increases the probability that we have a girl because the girl has three chances to have been born.</end quote></div>

You did not understand my post. See my post above.

Yeah you're right, you're thinking along the same veins that I was but I made the wrong conclusion. I think you are right in your distinction, but I choose to ignore your infallibility and instead will now distract myself with kitties.

Heh heh, they're so cute.

 

[lumbus]

Originally posted by: DrPizza
HOW he finds this out matters /..../ just clinging to your belief.

I wonder who's clinging to his belief....

Originally posted by: DrPizzaFirst, let me reitterate: I agree wholeheartedly that if you know a man has 3 children and you are told that at least 2 of the children are boys, the probability that the third child is a girl is 75%.

a man and two boys greet you. The man says "These two boys are my sons, but I have three children.

What exactly is the problem with how you received the information that 2 out of three kids are boys?

 

[lumbus]

Oh yeah, you think you have to calculate the irrelevant probability that you meet two boys out of three kids (which has already happened).

Ok, so let's say that you went to the same house blind-folded and the man said "I have two boys and a third kid". Now all of a sudden the probability of a girl is 75% instead of just 50%... But wait! You have to figure out the probability of the man saying that, versus saying "I have a girl and a boy, and another kid playing in the backyard", right? Oh, and the probability of you going to a house with three kids. And the probability of getting invited. And... actually, all of that happened already. What's the probability of you reading this as you're reading it?

 

[Nathelion]

<div class="FTQUOTE"><begin quote>Originally posted by: lumbus
<div class="FTQUOTE"><begin quote>Originally posted by: Nathelion
I am somewhat pleased to see that this thread is still going
So two thirds of the time one of the children we meet would be a girl, and we would back out.</end quote></div>

There is not going to be any backing out, because the fact that we have met two boys is a given.

<div class="FTQUOTE"><begin quote>Originally posted by: Nathelion
In order to make the probability of there being a girl in the back yard, an addition would have to be made to the problem statement. For example, one could add:
"The neighborhood you are in is culturally homogenous, and in this particular culture families with three children always send two children to meet visitors and if they can they will send only boys".
</end quote></div>

That's not part of the problem. And for your reasoning to make any sense, you would have to list additional conditions, such as the parents pick two out of three kids at random, and each kid as an equal probability of being picked. The reason that this conditions aren't listed is that they aren't part of the problem.</end quote></div>

It is part of the problem, because the problem doesn't specify otherwise. All we know is that we have walked into a house with 3 kids and we have met two boys. But how were specifically the two boys chosen to come meet us? In the absence of other information, we must assume that the fact that specifically two boys came to meet us happened at random. My explanation with many houses above was to explain the reasoning, not because you're actually going to walk into hundreds of houses before you find the right one. But the problem statement is equivalent to walking into a house at random. You know that you walked into a house where two boys came to meet you. But you don't know how that came to be.
It's all in what information you have, and when you got it. If you knew before you walked in that the kids meeting you would be boys, then you have eliminated the cases where a girl meets you beforehand, and the probability of the girl in the back yard being a girl is 75%.
However, you don't know that. You are really just walking into one of those houses at random. Now it turns out that there are two boys meeting you, but there could have been a girl instead. I'm not skilled enough to do it in ASCII, but if you get a pencil and paper and make a tree diagram out of the whole scenario, you'll see that the probability of the 3 cases where there's a girl of the back yard are only one 3rd as likely as the one where there's a boy.

To refer to an earlier thread around here concerning the Monty Hall problem, if you are in a prison camp with 1000 prisoners and before you go to sleep you know that the guards will eventually kill all but one of the prisoners. When you wake up again 998 of the prisoners are dead, there's only you and another guy left. Now there's a 50% chance that you will live and he will die.
However, if the guards, before you go to sleep, tell you "We will kill every person but one in this camp, but you specifically are exempt from getting killed tonight" and you wake up and there's only one guy left, then you know that there's a 999/1000 chance that he will live and still only a 1/1000 chance that you will live.
The situation isn't really analogous to our problem with the house and the boys, but it demonstrates that prior knowledge changes the probabilities of the various outcomes. In our current problem, you could very well have walked into a house with a girl meeting you, because the problem statement doesn't say otherwise.
To put it this way, we could have walked into a house with all girls. Now we didn't, but we could have. The possibility doesn't really interest us, but since we must assume, in the lack of other information, that "our" house could have been any of the other houses, we need to calculate the actual (relative) probability that we have found ourselves in any one of the specific cases. And that probability is lower for the cases where there's a girl in the back yard.

edit: darn italics

 

[Nathelion]

Another way of explaining it:
The quirk is that your acquaintance picked two individuals out of three possible to come meet you. Their gender didn't have anything to do with what two individuals got chosen. There's only one chance in three that two boys got sent to meet you.
However, if you instead were to meet a random man on the street and he said "I have three children, and at least two are boys. What is the probability that the third is a girl?" the answer would be 75%, because there is no choosing two at random without regard to gender and then having them turn out - randomly - to be two boys. Instead, two are "chosen" WITH regard to gender, so you already know that those two will be boys. Do you see what I'm saying?

 

[lumbus]

Originally posted by: Nathelion
Another way of explaining it:
The quirk is that your acquaintance picked two individuals out of three possible to come meet you. Their gender didn't have anything to do with what two individuals got chosen. There's only one chance in three that two boys got sent to meet you.
However, if you instead were to meet a random man on the street and he said "I have three children, and at least two are boys. What is the probability that the third is a girl?" the answer would be 75%, because there is no choosing two at random without regard to gender and then having them turn out - randomly - to be two boys. Instead, two are "chosen" WITH regard to gender, so you already know that those two will be boys. Do you see what I'm saying?

This is getting a bit ridiculous.... Cleary specified by the problem is the fact that two boys HAVE ALREADY COME OUT TO GREET YOU. It has HAPPENED. The probability of that happening is 100%.

Figuring in that probability is as silly as figuring out the probability of "a random man on the street" saying "I have three children, and at least two are boys" after it has already happened. Why don't you make some tables for all the combinations in which he could mention his kids?

 

[DrPizza]

Lumbus, I'll try to keep it simple.

First of all, do you understand:

I have 3 cards,
1 black both sides
1 black one side, red one side
1 red both sides

I pull a card at random and only look at one side. I see black. What's the probability that the other side is black? Answer: 2/3

We'll start with this problem and go from there. Otherwise, I don't know how to convince you that you're wrong about the original problem posted in this thread. *sigh* This is like trying to convince people that .999... = 1. Maybe I'm wasting my time.

 

[Nathelion]

Speaking of which, correct me if I'm wrong, but wouldn't you have to say that the limit of 0.9[1]9[2]9[3]...9[n] approaches 0! (which is defined as 1) as n approaches infinity for the statement to even make any sense? Or is there something obvious that I'm missing here? The brackets are supposed to indicate a subscript.

 

[Lorax]

either 100% or 0%. the child in the backyard is already born and has a gender.

if you were to guess, you'd have a 50% chance of getting it right.

 

[lousydood]

either 100% or 0%. the child in the backyard is already born and has a gender.

But you don't know it. It's the same case as if the mother was pregnant and you didn't know the gender of the fetus. It's information, and in either case, it's unknown.

if you were to guess, you'd have a 50% chance of getting it right.

If you had to guess, and there was no other information available.

But there is information available. You know the family has 3 children. You know 2 of them are boys.

The chance of the family having 3 boys is lower than the chance of having 2 boys and 1 girl.

This is because there are multiple ways to have 2 boys and 1 girl, but only one way to have 3 boys. (not distinguishing personal identity)

BBG, BGB, GBB vs BBB

so, 75% chance of the other child being a girl.

 

[Peter]

Still no. At the time of conception, the probability for having a girl was 50%, and that didn't suddenly change anywhere down the road.

Independent events.

All surrounding information is totally irrelevant. Absolutely the same as saying "there's a kid in the back yard, guess its gender". Nathelion has posted a long winded and complete demonstration of why all the padding in the story is irrelevant.

 

[Nathelion]

Well, in this particular case it is irrelevant. If the problem were to be worded just slightly differently, it wouldn't be irrelevant.

 

[jman19]

Originally posted by: DrPizza
Lumbus, I'll try to keep it simple.

First of all, do you understand:

I have 3 cards,
1 black both sides
1 black one side, red one side
1 red both sides

I pull a card at random and only look at one side. I see black. What's the probability that the other side is black? Answer: 2/3