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Another math problem

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Originally posted by: Nathelion
I thought I should jump on the bandwagon and provide some additional amusement for those of us who enjoy the math threads. This thread is more fun than business, I already know the answer to the problem in question, so don't feel compelled to post just to be nice/help out.

For the purposes of this discussion, there is always a 50% chance that a woman giving birth will have a boy and a 50% chance of her having a girl (this is not true in reality).

Imagine that you are going to a party. You arrive, and a man and two boys greet you. The man says "These two boys are my sons, but I have three children. The third child is in the back yard playing." You can't see the back yard from where you are standing. You have no prior knowledge of this man's family.

What is the probability that the child playing in the back yard is a girl?
Click to expand...

The answer is 50%. The first two event has nothing to do with the third event since each event is independent of the other.

It's one of the most basic laws you learn in probability class. 🙂
 
I think one reason why this is slightly ambigous is because of the way the example is written. If you read a book and it is written that someone meets "a man that has his two sons next to him and a third kid is playing in the garden", most people will assume that the kid in the garden is the youngest.
If so, the probablility of the kid in the garden being a girl is 50%.

As usual the main problem is how to correctly transferi the problem from the "real world" to the "math world", once that is done the problem is much easier to understand since you get rid of all "implicit" (incorrect) assumptions.




 
Originally posted by: Finny
That's what I said at the end, the basic answer is indeed 75%. Someone else explained it nicely above, so I don't feel like repeating it, nor do I feel like getting into the deeper statistical analyzation, which I did more than enough of last semester 😕
Click to expand...

yeah now that I re-read your post I don't know what I was thinking when I wrote mine.
 
OK... I will attempt to give a proof (extremely informal) by merely using intuition. If you need to, look at the * and ** at the bottom of this post if you cannot reason out the steps yourself.

Given that there are 3 children with 2 options of gender, there are 2*2*2 = 8 possible combinations of children: BBB, BBG, BGB, BGG, GBB, GBG, GGB, GGG. Each combination has probabilty 1/8 of occuring.*

We know that the man has 2 boys. This leaves the four options BBB, BBG, BGB, and GBB. Each has probability 1/8 of occuring, as proven above. So, the probability that there is a girl is (3* 1/8) / (4 * 1/8)** = 3/4.

For the record, this is exactly how the P(A|B) problems work. I am merely breaking it into more logical steps.



* - The probability of having a boy = 50% and the probability of having a girl = 50%. As given, there are 3 children with 2 possible genders, so the probability is .5 * .5 * .5 = 1/8

** - Recall that the definition of a probability is "the number of favorable outcomes divided by the total number of possible outcomes." Favorable outcomes are those with a G in them, and there are 3 of those. The total number of possible outcomes [with 2 boys] is 4. Each has 1/8 chance of occuring. So the final probability is (3 * 1/8) / (4 * 1/8).
 
I have to point out that were not just solving what are the chances of the third child will be male or female, it's the probabilty of have 2 boys and 1 girl, or 3 boys, as compared to the rest of the possibilites with 3 children, aka 3 girls, 2 girls and 1 boy. SO, the possible outcomes are...

-3 Boys
-2 Boys and 1 Girl
-1 Boy and 2 Girls
-3 Girls

Now, we have to determine the chance of each one occuring, which can be done very simply with a probability tree. The probabilities are...

-3 Boys (1/8 or 12.5%)
-2 Boys and 1 Girl (3/8 or 37.5%)
-1 Boy and 2 Girls (3/8 or 37.5%)
-3 Girls (1/8 or 12.5%)

This shows us exactly 1/2 the time at least 2 boys will be in the actual outcome. Since they're are eight different outcomes (this includes boy-boy-girl and boy-girl boy, etc as different outcomes) and in HALF of those outcomes, at least 2 boys are born, we must then look at those 4 outcomes. Only 1 of those 4 (half the original eight) outcomes includes another boy, and 3 of them include a girl instead of a final boy. It may appear at first to some people that already knowing the 2 boys part doesn't change the fact it's a 50% chance of the last child going either way, but as this shows, THIS IS NOT TRUE. When you take into account you do NOT know the order of birth for the children's genders we have to accept that there are not just 2 outcomes, boy or girl, but 4 outcomes, only of which 1 includes a third boy. 1/4 of the time it will be 3 boys, 3/4 it will be 2 boys and 1 girl.

BTW, Excuse my spelling, i look at the keyboard as i type, so spelling may not be accurate!)




EDIT: If i wasn't clear on my final answer....

There is a 75% chance the child playing in the yard will be a girl.
 
Here's a similar problem:

Bag X has 5 white marbles and 2 black marbles. Bag Y has 3 white marbles and 5 black marbles. A bag is chosen at random and a marble taken from the bag. The marble is white; what is the probability that the bag was X?
 
Originally posted by: stupidkid
Here's a similar problem:

Bag X has 5 white marbles and 2 black marbles. Bag Y has 3 white marbles and 5 black marbles. A bag is chosen at random and a marble taken from the bag. The marble is white; what is the probability that the bag was X?
Click to expand...

Let P(A|B) be defined as the probability that the bag was X (A) given that the marble chosen is white (B).

This is equal to P(A and B)/P(B) = (7/15 * 5/7) / (8/15)* = 5/8.


* - P(A and B) is the probability that both the bag chosen is x and the marble chosen is white. There are 15 total marbles and 7 of those are in bag X so the probability of choosing bag X is 7/15 and the probability of getting a white marble from bag X is 5/7. P(B) is the probability that the marble chosen is white, and there are 8 white marbles out of 15 total marbles, so P(B) = 8/15.

A simpler way to reason through this is that we know the chosen marble is white. We don't care about the black marbles anymore. The only thing that matters is the proportion of white marbles in bag X to bag X + bag Y. This is 5 / (5+3) = 5/8.

Still having trouble? Lets change the numbers around: bag X has 1 white and 1 black and bag Y has 1 white and 97 black. You are given that the marble picked is white. All those blacks in bag Y don't matter; you've already picked the white marble! The probability that the marble came from bag X given that the marble is white is 1/2; the probability it came from bag Y is also 1/2.

(edit: I've never taken a stats/probability class before. If I can figure it out, anyone can.) 😉
 
I still vote 50%, I know the statistics of i being 75%, and if the quester were worded as a continuing probability, that would be correct. But assuming a 50% boy/girl child, the chances of it being a girl is still 50%.

The person with the coin example had it perfect, if you flip a coin and it comes up head twice, what is the probability it is tails on the third try? Many of you are saying that is a continuing probability problem when it is not, it is still simply 50% on the 3rd attempt.
 
Originally posted by: MikeyLSU
I still vote 50%, I know the statistics of i being 75%, and if the quester were worded as a continuing probability, that would be correct. But assuming a 50% boy/girl child, the chances of it being a girl is still 50%.

The person with the coin example had it perfect, if you flip a coin and it comes up head twice, what is the probability it is tails on the third try? Many of you are saying that is a continuing probability problem when it is not, it is still simply 50% on the 3rd attempt.
Click to expand...

You're treating the problem like the wife is out back pushing the kid out then and there. The third child may have been born at any of three possible relative ages. For the coin example, we're talking about the occurance of a tails coming up on any of the three tries and the other two coming up heads, not restricting it to the last toss. It's pretty easy to justify yourself that the probability of having three kids all of the same gender has to be less than the probability of having two male and one female.
 
Originally posted by: MikeyLSU
I still vote 50%, I know the statistics of i being 75%, and if the quester were worded as a continuing probability, that would be correct. But assuming a 50% boy/girl child, the chances of it being a girl is still 50%.

The person with the coin example had it perfect, if you flip a coin and it comes up head twice, what is the probability it is tails on the third try? Many of you are saying that is a continuing probability problem when it is not, it is still simply 50% on the 3rd attempt.
Click to expand...

Yes, but we are given extra information. The man has 3 children and 2 of them are boys. This isn't an independent probability, there is now a condition (we are given 2 are boys). There are 4 possibilities of how his children could have been born: BBB, BBG, BGB, and GBB. Each has an equal chance of occuring (there is an equal chance of having a boy or a girl). You can easily see that 3 out of the 4 possibilities include a female child.
 
I don't doubt at all that the probability that it being BBG is 75% after you know they have 2 boys, but I do think that sex of the unknown child is still 50/50 with the way this problem is worded.

I tell you what, if I go and flip a coin 3 times and I tell you that 2 of the times came up heads, would you give me 4/1 odds that the last one is tails? Cause nobody in their right minds would make that bet.
 
Originally posted by: MikeyLSU
I tell you what, if I go and flip a coin 3 times and I tell you that 2 of the times came up heads, would you give me 4/1 odds that the last one is tails? Cause nobody in their right minds would make that bet.
Click to expand...

If you flip three times you get one of the following series

HHH
HHT
HTT
TTT
THH
TTH
THT
HTH

Each series has the same probability (12.5%)

However, if you now tell me that you got at least two heads I know that it must have been one of the following

HHH
THH
HTH
HHT

Hence, the probability that the last one is tails is indeed 1/4...


 
If you have access to Matlab you can try this. 1 and 0 represents boys-girls, head-tails of whatever you want
("encoding" the data this way simplifies things since you can count the number of ones just by summing over the array).

---cut here----

%Number of series
n=10000;
%length of each serie
l=3;
%%At least y elements are one
y=2;
%Create n series with l elements in each. Store in matrix X. Round off to 0
%or 1
X=round(rand(n,l));

%Find all series with AT LEAST y ones, store index in array u.
u=[]; %%Clear array
for k=1:n
if sum(X(k,🙂)>=y
u=[u k];
end
end

disp('Number of series with at least y ones')
length(u)
disp('Number of those with a zero at the end')
r=sum((X(u,3)==0))
disp('In percent')
100*r/length(u)

disp('Number of those with a zero somewhere')
g=sum(sum(X(u,🙂,2)<l)
disp('In percent')
100*g/length(u)




----end----

A typical run looks like this:

Number of series with at least y ones

ans =

5018

Number of those with a zero at the end

r =

1260

In percent

ans =

25.1096

Number of those with a zero somewhere

g =

3751

In percent

ans =

74.7509

Not exactly 25% and 75%. But if you increase the number of series you will see that it does indeed approach the "exact" result.
 
Previous or next children = no impact on probability
medical predispositions = we have no idea about it

male = xy
female=xx

so the probability of xy with 3 times x and 1 time y is 3/4

so probability is 75% in favor of female

am i right?
 
Heh, no. The first cell of a new offspring gets one X from the female, and either an X or a Y from the male. This is not a "two out of four" shuffle (wherein you could end up being a clone of the father or the mother), but a one-out-of-two drawing.

What the female provides doesn't make a difference /at/all/. The sperm is either X or Y, which is the /only/ thing that determines the offspring's gender. 50/50 if you want to keep it simple.
 
Haha... ITS STILL 50%....


Giving the fact these days that men a horn dogs, they could have suduced 3 different women... therefore it would be 50% of a chance that it is a girl.
 
Originally posted by: travisray2004
Haha... ITS STILL 50%....


Giving the fact these days that men a horn dogs, they could have suduced 3 different women... therefore it would be 50% of a chance that it is a girl.
Click to expand...

Even if the children are birthed by different mothers or the same mother, the probability still remains at 75%.
 
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??)

 
Originally posted by: f95toli
Originally posted by: MikeyLSU
I tell you what, if I go and flip a coin 3 times and I tell you that 2 of the times came up heads, would you give me 4/1 odds that the last one is tails? Cause nobody in their right minds would make that bet.
Click to expand...

If you flip three times you get one of the following series

HHH
HHT
HTT
TTT
THH
TTH
THT
HTH

Each series has the same probability (12.5%)

However, if you now tell me that you got at least two heads I know that it must have been one of the following

HHH
THH
HTH
HHT

Hence, the probability that the last one is tails is indeed 1/4...
Click to expand...


I agree completely with the above logic. However, lets add a little twist to it. Rather than you telling me that you got at least 2 heads, instead, you flip three coins and put each under a separate hat. I get to randomly lift 2 hats and reveal two of the coins.

We'll do that 800 times. Now, out of those 800 times, we can expect that you flipped 3 heads in 100 of those times.
And 300 of those times, you get 2 heads and a tails.

100% of the time that you have 3 heads, I'm going to reveal 2 heads. But only 33% of the time that you have 2 heads and a tail will I reveal 2 heads. Thus, 100 times I see 2 heads, the 3rd coin is a head. And, 100 times, the 3rd coin is a tail. And, a whole heck of a lot of times, I'm going to see a head and a tail, or be introduced to Bob's son and daughter, "my 3rd kid is over there" and we wouldn't stop to turn that situation into a probability problem either.
 
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.
Click to expand...

Besides completely overcomplicating things, your table doesn't make sense.

For example: In your second category, how is "bbb" an arrangement of "bbg"?

And another one: If the short lines are the possibilities of having three children, and the long lines are the arrangements, how is "gbg" and "ggb" on the short lines different from each other? Obviously they are the same if the ordering is covered by the long lines.
 
This same proof has already been posted by several people, but here it is again, in a form that might be easier for some to understand.

All possibilities that the man could've had three children (BTW, this are ordered possibilities, e.g. BBB means "boy then boy then boy"):
BBB
BBG
BGB
BGG
GBB
GBG
GGB
GGG

Now, we already know that he has at least two boys, so we can exlude all the possibilities with two or more girls, leaving us with:
BBB
BBG
BGB
GBB

As you can see, three of the remaining possilbities include one girl, which means that there's a 75% that the kid in the backyard is a girl.
 
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??)
Click to expand...

The error in your logic is that you assumed that the two oldest boys opened the door. However, we don't know if the third child is the youngest, middle, or oldest child, so any combination which has 2 B's in it could have opened the door.

Lets fix your table...

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

The bolded part is all possible orders that the man may have had his children given that he has 2 boys. For clarity's sake, recall that the definition of probability is "# of desired outcomes / # of total outcomes." Think of this as the "# of total outcomes" part.

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

Here is a "subset" of the total outcomes that includes all ordered outcomes where the man had 3 boys. This is the # of desired outcomes part of the probability.

By inspection we can deduce that the probability is indeed 1/4 chance that the man had 3 boys.

Also, as you can see, there is no need to list all possible orders of birth...just a BBB, BBG, BGB, etc. would have sufficed.
 
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??)
Click to expand...

It is 50% (or could be, there is still a problem with your reasoning) if we chose someone randomly off the street (or house as you have put it) only knowing that they had three children and after talking with them find out they have at least two boys. The difference here is that there is a chance that the person had up to three girls when we initially picked them out. This is not the case in the OP because he specifically stated that you see the father with the two boys. In this situation, you know from the start that there is no possibility of having more than one girl and thus you only consider permutations of BBG and BBB. The information that you end up with in either situation is the same, there are three children, two at least are boys. However, in the former there is the added probability that you could have had a person who did not have at least two boys. When this added probability is considered then you get the 50% [sic] chance. This was discussed in the link about Marilyn Vos Savant that I had posted previously and I had already stated that the OPs form of the question fortunately removes such ambiguity.

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.
 
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