Another math problem

[madh83]

I'm sure someone has made a similar argument already but anyway, I remember seeing a similar question about a deck of cards during an interview. However, with the deck of cards you know the events are not independent of one another so you can use conditional probability to calculate the answer. However, with this question, it says that the assumption is that the probability is always 50% for a boy or girl. This indicates that the events are independent of one another. So, it should not be a permutation as above, but actually a combination. With the combination the bottom two rows don't matter from above. So, it's still 50%. But hey, I can kind of see how it can be read as 50% for each case and not necessarily meaning the events are independent.

 

[vanvock]

It seems to me that both answers are correct, depending on how you look at it. If you see the children as a series of events then it's 75%. If you see them as individual events it's 50%. I'll go with 50% because of the statement that every birth is allways 50-50.

 

[newschool]

If you flip a penny. One time. It is 50/50. If you flip it 2 times, I think evreyone would agree that it is still 50/50. Thoses stats are used in many domains, for example in poker odds.

 

[Cattlegod]

Originally posted by: 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?

Bingo. There are two questions here, the question in the OP isn't clear as to which one it is asking.

 

[SuperISD32]

Because what generates the probability is the mother, who cans give a male or a female at 50% indipendently from the earlier sons' sex (as implicitly specified by the text of the problem), the probability the third son (so considered by itself) is male o female is 50%.
Only if the general statistic (about all the popoulation) says that people is circa 50% male and 50% female, then, the probability the third chil is a female would be 66,6_%.

Thank you.
P.S.: this the first time I write in english out the school environment; are the sentences enough normal?

Bye!

 

[SuperISD32]

Originally posted by: lousydood
This is a Word Problem, like you might do in High School.
This has nothing to do with genetics, family, or biology at all. That is all window-dressing for the problem.

Put that out of your head right now.

Let X be the random variable with
P(X=0) being 0.5
P(X=1) being 0.5

read that is "the probability of X equaling 0 is 0.5" etc.

Each row in the following table is a sample of 3 events. There are 2^3=8 possible permutations. Each row in the table has an equal probability of occurrence: (0.5)^3 = 0.125

X1, X2, X3
----------
0 , 0 , 0
0 , 0 , 1
0 , 1 , 0
0 , 1 , 1
1 , 0 , 0
1 , 0 , 1
1 , 1 , 0
1 , 1 , 1

Now consider the condition "two of the samples must be 0" and eliminate rows that do not satisfy it

X1, X2, X3
----------
0 , 0 , 0
0 , 0 , 1
0 , 1 , 0
1 , 0 , 0

The remaining rows show that the probability that none of X1, X2, or X3 being equal to 1 is 0.25 under the condition stated. The complement of 0.25 is 0.75.

Re-interpret the math: "Boy=0", "Girl=1" and 3 samples meaning "a sample of 3 children"

You can't keep any row containing two zeros because their dispositions must be ignored since the problem doesn't specifies them.
There are no differences if the child in the yard is the first, second or third born; what is important is instead that the mother as ALWAYS the probability of 50 per cent to give a male child.

 

[Syringer]

Don't know if this point has been raised yet, as I've only skimmed through this thread..but what if this were extrapolated to the point where the man had 15 kids?

If he said he had 15 kids, and you see 14 of them as boys, then using the logic above it would suggest that, because there's only one way to get 15 boys..and 16 (?) ways to get 15 boys and 1 girl, that the probability that it's a girl would be 15/16. That would make zero sense.

It HAS to be 50/50, that is if you discount that the girl is a loner, boys don't like girls, man is predisposed to having boys BS logic.

 

[SuperISD32]

There is always only one way to get each combinations because the order of the elements (the sons) has to be removed.
Without the order, there is always the probability of 50% the unique unknow son is male or no.

 

[Kyteland]

The answer to this question is definitely ambiguous and depends on the question asked. A way of stating the problem that doesn't depend on introducing the two oldest is:

If a man has three children and says "I'll introduce you to two boys" then the probability is 75% that the third is a girl.

If a man has three children and says "I'll introduce you to two of them" and those two are boys then the probability is 50% that the third is a girl.

One of these questions is like the Monty Hall problem. One of them isn't.

 

[TuxDave]

Yeah I heard of this question and the 3-door variant of it. The question really screws up because of what is random and what is not. For instance of the two boys answering the door. If it was a random factor that in a household of BBG that the two boys answered the door, then it really goes back to 50%. If there was a worldwide rule saying that when two people answer the door, only boys can answer it unless there is only one or no boys in the household. Then the answer gets back to 75%.

 

[Syringer]

Originally posted by: Kyteland
The answer to this question is definitely ambiguous and depends on the question asked. A way of stating the problem that doesn't depend on introducing the two oldest is:

If a man has three children and says "I'll introduce you to two boys" then the probability is 75% that the third is a girl.

If a man has three children and says "I'll introduce you to two of them" and those two are boys then the probability is 50% that the third is a girl.

One of these questions is like the Monty Hall problem. One of them isn't.

Agreed on the second point, but what if we reword the first statement to:

If a man has 100 children and says "I'll introduce you to 99 boys"...

The probability of the 100th child being a girl (not taking into consideration his propensity to have boys and assuming that the chance of a boy/girl is 50/50), is NOT going to be astronomically high. It's still 50%.

 

[Kyteland]

Originally posted by: Syringer
Agreed on the second point, but what if we reword the first statement to:

If a man has 100 children and says "I'll introduce you to 99 boys"...

The probability of the 100th child being a girl (not taking into consideration his propensity to have boys and assuming that the chance of a boy/girl is 50/50), is NOT going to be astronomically high. It's still 50%.

Lets generalize it further to an arbitrary variable:

If a man has N children and says "I'll introduce you to N-1 boys" then the probability is X that the Nth is a girl.

If a man has N children and says "I'll introduce you to N-1 of them" and those N-1 are boys then the probability is Y that the Nth is a girl.

Solve for X and Y

The variables I'm using are:
N = Total children
B = Total boys
P = Probability of a child being a boy. (Assumed to be 1/2.)

These formulas hold true:
p(N,B) = Probability of having B boys given N children
p(N,B) = choose(N,B)*(P)^(B)*(1-P)^(N-B)
p(N,B) = choose(N,B)*(1/2)^N
sum(i=[0,N]) {p(N,i)} = 1

p(N,N-1) = choose(N,N-1)*(P)^(N-1)*(1-P)^(1)
p(N,N-1) = N*(1-P)*P^(N-1)
p(N,N-1) = N*(1/2)^N

p(N,N) = choose(N,N)*(P)^(N)*(1-P)^(0)
p(N,N) = P^N
p(N,N) = (1/2)^N

The man has N boys with the probability (1/2)^N. In this case there is a boy in the back yard.
The man has N-1 boys with the probability N*(1/2)^N. In this case there is a girl in the back yard.

Solve for X

If the man has N boys, the probability that he introduced you to N-1 boys is choose(N,N-1)/choose(N,N-1) = 100%. He must choose N-1 boys.
If the man has N-1 boys, the probability that he introduced you to N-1 boys is choose(N-1,N-1)/choose(N-1,N-1) = 100%. He must choose N-1 boys.

The total probability is then N*(1/2)^N*100% + (1/2)^N*100% = (N+1)*(1/2)^N

N*(1/2)^N / (N-1)*(1/2)^N = N / (N+1)
(1/2)^N / (N-1)*(1/2)^N = 1 / (N+1)

When N = 3 this gives probabilities of 75% girl and 25% boy.
When N = 99 this gives probabilities of 99% girl and 1% boy.

Solve for Y

If the man has N boys, the probability that he introduced you to N-1 boys is choose(N,N-1)/choose(N,N-1) = 100%.
If the man has N-1 boys, the probability that he introduced you to N-1 boys is choose(N-1,N-1)/choose(N,N-1) = 1/N.

The total probability is then N*(1/2)^N*(1/N) + (1/2)^N*100% = 2*(1/2)^N

N*(1/2)^N*(1/N) / 2*(1/2)^N = 1/2
(1/2)^N / 2*(1/2)^N = 1/2

When N = 3 this gives probabilities of 50% girl and 50% boy.
When N = 99 this gives probabilities of 50% girl and 50% boy.

 

[CycloWizard]

Why don't you just walk outside and introduce yourself to the young whippersnapper?