Fraternal Twins

[Keith&Co.]

Keith&Co.,

re: "There are two ways to get one girl and one boy (in either order)."


And there are 3 combinations out of 4 of having at least 1 girl.

But the way you phrased the question, two of those combinations are effectively the same. You did not make a distinction between first and second born child. So there are three outcomes that answer the question.

 

[Playball40]

If you really want to get technical, the natural birth ratio is actually 105:100 boys to girls. That said, my fraternal twins are B/G!

 

[Alcoholic Actuary]

By extension, why can't I whittle the outcomes down to 2: Either there is a girl or there isn't? Under the 'equally likely outcomes' hypothesis, clearly the probability is .5 that there is a girl.

aa

 

[Playball40]

Keith&Co.,

re: "There are two ways to get one girl and one boy (in either order)."


And there are 3 combinations out of 4 of having at least 1 girl.

But the way you phrased the question, two of those combinations are effectively the same. You did not make a distinction between first and second born child. So there are three outcomes that answer the question.

Birth order seems irrelevant to the question....what if there was a c-section? Doesn't change the odds of a girl or boy. I agree with you.

 

[Alcoholic Actuary]

If you really want to get technical, the natural birth ratio is actually 105:100 boys to girls. That said, my fraternal twins are B/G!

Ok. Then I change my answer to a .73766 chance of at least 1 girl. (assuming this birth ratio also holds for fraternal twins).

aa

 

[Alcoholic Actuary]

Keith&Co.,

re: "There are two ways to get one girl and one boy (in either order)."


And there are 3 combinations out of 4 of having at least 1 girl.

But the way you phrased the question, two of those combinations are effectively the same. You did not make a distinction between first and second born child. So there are three outcomes that answer the question.

Birth order seems irrelevant to the question....what if there was a c-section? Doesn't change the odds of a girl or boy. I agree with you.

The order is irrelevant. That doesn't really play into the equally likely bit that everyone seems to want to jump to.

aa

 

[beero1000]

Oh dear. If order is discounted there are three possibilities, but they are certainly not equally likely.

Let A be the event that the firstborn twin is a girl, and B be the event that the second twin is a girl. Then we assume A and B are independent events and that P(A) = P(B) = 1/2.

Then, P(at least one girl) = P(A or B) = P(A) + P(B) - P(A and B) = 1/2 + 1/2 - (1/2)² = 3/4.

The more counter-intuitive results show up when we start using conditional probabilities. In that vein, I ask the following more interesting questions:

1. If a woman is giving birth to fraternal twins and the first child is a girl, what is the probability that the second child is a girl?
2. If a woman has given birth to fraternal twins and (at least) one of the children is a girl, what is the probability that the other child is a girl?

 

[Alcoholic Actuary]

Oh dear. If order is discounted there are three possibilities, but they are certainly not equally likely.

Let A be the event that the firstborn twin is a girl, and B be the event that the second twin is a girl. Then we assume A and B are independent events and that P(A) = P(B) = 1/2.

Then, P(at least one girl) = P(A or B) = P(A) + P(B) - P(A and B) = 1/2 + 1/2 - (1/2)² = 3/4.

Mageth wants to play poker with you.

[*]If a woman is giving birth to fraternal twins and the first child is a girl, what is the probability that the second child is a girl?

Show

I think 0.5

[*]If a woman has given birth to fraternal twins and (at least) one of the children is a girl, what is the probability that the other child is a girl?

Show

I think 1/3

aa

 

[Loren Pechtel]

There is much fail in this thread.

75% that there is at least one girl.

Try it with coins.

 

[Keith&Co.]

Right, because with coins, we won't have to hold those difficult concepts of BOY and GIRL in our minds...

 

[Mageth]

Let's look back at the way the problem was stated...

"A woman is about to give birth to fraternal twins. What is the chance that one of the twins will be a girl?" That was later clarified to mean "at least one."

So leveraging the "socks in a drawer" problem, suppose you have a drawer with an infinite (for the sake of argument) number of black socks and an infinite number of white socks. You reach in, grab, and pull out two socks. What is the probability that you have drawn at least one black sock?

There is one event (drawing two socks) that has three possible outcomes. The three possibilities are that you drew two black socks, two white socks, or one white and one black sock. There is no fourth possibility (i.e., one black and one white sock). You don't number or label the socks in your hand as "sock A and sock B" or "first and second socks" and then do probability calculations. It's just two socks. I.e., "order is discounted." Importantly, you don't draw the two socks as two independent events - you draw the two socks as one event that has three possible outcomes - and thus you don't do the probability calcuation considering drawing the two socks as two independent events.

Each possible outcome is equally likely (all things being equal), the probability of each outcome is 1/3, and the probability that you have at least one black sock in your hand is 2/3.

Similarly, in the twins problem as stated, the birth of the twins is one event and not two independent events, since order is discounted there are three possible outcomes to the event, the probability of each outcome is 1/3, and the probability that there is at least one girl is 2/3.

(All said in good argumentative spirit, admitting that I may be wrong, and that I suck at poker! )

 

[beero1000]

Let's look back at the way the problem was stated...

"A woman is about to give birth to fraternal twins. What is the chance that one of the twins will be a girl?" That was later clarified to mean "at least one."

So leveraging the "socks in a drawer" problem, suppose you have a drawer with an infinite (for the sake of argument) number of black socks and an infinite number of white socks. You reach in, grab, and pull out two socks. What is the probability that you have drawn at least one black sock?

There is one event (drawing two socks) that has three possible outcomes. The three possibilities are that you drew two black socks, two white socks, or one white and one black sock. There is no fourth possibility (i.e., one black and one white sock). You don't number or label the socks in your hand as "sock A and sock B" or "first and second socks" and then do probability calculations. It's just two socks. I.e., "order is discounted." Importantly, you don't draw the two socks as two independent events - you draw the two socks as one event that has three possible outcomes - and thus you don't do the probability calcuation considering drawing the two socks as two independent events.

The probability of each outcome is 1/3, and the probability that you have at least one black sock in your hand is 2/3.

Similarly, in the twins problem as stated, the birth of the twins is one event and not two independent events, since order is discounted there are three possible outcomes to the event, the probability of each outcome is 1/3, and the probability that there is at least one girl is 2/3.

(All said in good argumentative spirit, admitting that I may be wrong, and that I suck at poker! )

The probability of each outcome is absolutely not 1/3. That is where you keep making your mistake. A discrete set of possibilities in no way means that each is equally likely.

If you measure the percentage of fraternal twins with at least one girl and ignore order and I measure the percentage of the same group of fraternal twins with at least one girl and keep track of order, all I need to do to get your percentage is to add the boy/girl percentage to the girl/boy percentage. You have already agreed that the probability where order matters is 1/4 each, so you are claiming that 1/4 + 1/4 = 1/3.

 

[rstrats]

Here are the 4 possible combinations:

BABY A. BABY B.

Boy Girl

Boy Boy

Girl Boy

Girl Girl

Out of the 4 combinations, three of them have a girl.

 

[Tom Sawyer]

Actually, it's 67%

Facts about fraternal twins

About one-third of fraternal twins are two girls, one-third are two boys, and one-third are boy/girl.

2/3rds of the time, fraternal twins are the same gender and they're different genders 1/3rd of the time.

 

[Keith&Co.]

Here are the 4 possible combinations:

BABY A. BABY B.

Boy Girl

Boy Boy

Girl Boy

Girl Girl

Out of the 4 combinations, three of them have a girl.

Exactly why, with respect to the original question, does BOY GIRL differ from GIRL BOY?

 

[steve_bnk]

BB
BG
GB
GG

The final check with probabilities is that all probabilities from the sample space must add up to 1.

The apparent probability of a girl appears to be 3/4, but so is a boy. Doesn't add up to 1.

From the definition of fraternal twin there are two independent embryos not a split of a single embryo. Assuming the variables are uncorrelated, the sex of one embryo is independent of the other, I'd guess the probability of having a girl is 50/50.

The probability of having a pair with a girl is not the same probability as the probability of a girl being born.

Put a coin in each hand, shake and toss. The probability of H or T for each independent hand is always 50/50.

 

[beero1000]

Actually, it's 67%

Facts about fraternal twins

About one-third of fraternal twins are two girls, one-third are two boys, and one-third are boy/girl.

2/3rds of the time, fraternal twins are the same gender and they're different genders 1/3rd of the time.

That is wrong, but in an interesting way. The author may have gotten the probabilities wrong, but might also have mixed up the general twin distribution with the fraternal twin distribution.

Around 1/3 of twin pairs are identical, and always have the same gender. That means that the probability of at two girls (if the identical/fraternal distinction is unknown) becomes P(two girls) = P(two girls | fraternal)*P(fraternal) + P(two girls | identical)*P(identical) = 1/4*2/3 + 1/2*1/3 = 1/3. Similarly, the probability of two boys is 1/3 and the probability of mixed sex is 1/3.

If we condition on the fact that they are fraternal twins, the gender split follows the 1/4, 1/2, 1/4 distribution, as I explained above.

 

[Alcoholic Actuary]

Here are the 4 possible combinations:

BABY A. BABY B.

Boy Girl

Boy Boy

Girl Boy

Girl Girl

Out of the 4 combinations, three of them have a girl.

Exactly why, with respect to the original question, does BOY GIRL differ from GIRL BOY?

Would you at least agree that the probability of rolling an 11 (as the sum of 2 dice) is greater than the probability of rolling a 12? The reason is because there are more ways to arrive at an 11. I could roll a 5 on the die #1 and a 6 on die #2, or I could roll a 6 on die #1 and a 5 on die #2. It doesn't mean the outcome is really different, the outcome is still 11 but there are 2 ways to get there. There is only one way to arrive at a 12 (both die #1 and die #2 show a 6).

There are 2 ways in which I can get 1 boy - 1 girl, vs 1 way to get 2 girls and 1 way to get 2 boys.

aa

 

[Alcoholic Actuary]

BB
BG
GB
GG

The final check with probabilities is that all probabilities from the sample space must add up to 1.

The apparent probability of a girl appears to be 3/4, but so is a boy. Doesn't add up to 1.

You are double counting the sample space. The probability of at least 1 girl plus the probability of at least 1 boy does not need to add up to 1.

aa

 

[Keith&Co.]

There are 2 ways in which I can get 1 boy - 1 girl, vs 1 way to get 2 girls and 1 way to get 2 boys.

aa

well, yeah. If we're talking probability.
The OP was chance.

If we aren't present in the delivery room, if we evaluate the outcome from the nursery, we will either see BB, BG or GG. Doesn't matter if the order of birth was BG or GB, we will see the same distribution of pink and blue blankets in those little cribs. Nothing in the OP seems to matter about the order they come out, so BG and GB are the same outcome. SO there are three possible results, two of which have at least one, or 'a' girl. Two chances out of three.