Fraternal Twins

[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.

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.

That's correct. I looked into my answer further and it included identical twins in the probability, making it unrelated to the question at hand.

Given that it's an internet discussion forum, however, I'm going to ignore any facts which contradict my previously stated position and center my argument on the fact that you're all wrong because your mothers are whores.

 

[zorq]

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! )

The problem with your analogy is that it isn't analogous.
The proper analogy is that you are reacing into two seperate drawers simultaneously and pulling one sock from each drawer. Each of the two drawers contains two socks. One black sock and one white sock. (or more accurately if we take into account the actual birth statistics: 100 white socks and 105 black socks) Now that I've fixed your analogy, can you see the difference?

Also, I was kind of upset with the way you phrased your analogy anyway because you can't really compare infinities very easily without further elaboration. For example the set of negative whole numbers is an infinite set and the set of even natural numbers is an infinite set. Now if you threw all these numbers into a drawer and started pulling out numbers randomly, you would be pulling out twice as many negative whole numbers as even natural numbers.

 

[Alcoholic Actuary]

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.

The outcome on the dice is 11. I don't have to roll dice in any order, or mark them A and B for the distribution to change. The number of 11's I observe will be double the number of 12's. The number of boy/girl fraternal twins I observe in the nursery will be double the number of girl/girl fraternal twins (or boy/boy).

You are right, order doesn't matter. That is different than distributing the probability evenly across possible outcomes.

aa

 

[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.

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

Adding

OP '...A woman is about to give birth to fraternal twins. What is the chance that one of the twins will be a girl?...'

50/50. Probability of boy and girl can not both be 3/4.


Put a coin in each hand. If the coin is fair and the tosses randomized after 50 tosses here will be 100 tosses. For a perfect experiment the there will be 50 H and 50 T.


Repeat with one hand tossing for 100 tosses, and you still get 50 H and 50T.

'...1 girl plus the probability of at least 1 boy does not need to add up to 1...' But the total probabilities in the sample space must equal 1.

BB
BG
GB
GG

The probability of a girl does not equal 3/4.

Total probabilities must always add up to 1. That is axiomatic. You can not have two probabilities that add up to more than 1.

The definition of a probability distribution is that the integral or area under the distribution curve must equal 1. For a discrete destruction like coin or die toss, probabilities must equal 1.

Toss a pair of ice.The probability of any one side on of one of the pair die is always 1/6.

 

[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.

That's correct. I looked into my answer further and it included identical twins in the probability, making it unrelated to the question at hand.

Given that it's an internet discussion forum, however, I'm going to ignore any facts which contradict my previously stated position and center my argument on the fact that you're all wrong because your mothers are whores.

With a poster of your caliber, I'd expect nothing less.

 

[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

Adding

.Put a coin in each hand. If the coin isfair and the tosses randomized after 50 tosses here will be 100tosses. For a perfect experiment the there will be 50 H and 50 T.


Repeat with one hand tossing for 100tosses,and you still get 50 H and 50T.

Total probabilities must always add up to 1. That is axiomatic.

The definition of a probability distribution is that the integral or area under the distribution curve must equal 1. For a discrete destruction like coin or die toss, probabilities must equal 1.

Toss a pair of ice.The probability of any one side on of one of the pair die is always 1/6.

No. You were looking at the sample space of outcomes and reporting the probability of at least one girl, and adding it to the probability of at least one boy. That does not have to add up to one.

As has been reported the sample space and distribution of outcomes is as follows:
BB = .25
GG = .25
1g1b = .5



aa

 

[steve_bnk]

I stand on one side of a wall with abucket of coins. In the wall is a small chute.


You stand on the other side of the wallin front oi the chute.


I take coins from the bucket in eachhand, shake, and drop first one and then the other down the chute. Asthey fall to the ground in front of you, you record heads and tails. I do this 50 times for 100 coins.


NXt I pick coins one at a time usingone hand, shake, and drop the coins down the chute. Again you recordheads and tails.


The probability of H or T is the same for both experiments. There is n way for you to detect which process was which statistically.


The two embryos are independent coin tosses as in the above example.


Imagine 100 pregnant women withfraternal twins in a room ready to drop. Same statistical process as the coin example above. The issue is your pick of fraternal twinswhich are defined as independent embryos developing independently. No different than two single embryos in two women.


200 women with single pregnancies, or100 women with fraternal independent embryos carried to term. Sa statistical results.


Someone tells me they ran an experimentto test for the occurrence of A or B. The variable is binary, twostates. I am told that the probability of either one turns out to be75%. I 'd say that is impossible. That would be like saying you have a process that fills a box with red balls and black balls saying the probability of picking one or the other is both 75%. Physically impossible.

 

[Alcoholic Actuary]

I stand on one side of a wall with abucket of coins. In the wall is a small chute.


You stand on the other side of the wallin front oi the chute.


I take coins from the bucket in eachhand, shake, and drop first one and then the other down the chute. Asthey fall to the ground in front of you, you record heads and tails. I do this 50 times for 100 coins.


NXt I pick coins one at a time usingone hand, shake, and drop the coins down the chute. Again you recordheads and tails.


The probability of H or T is the same for both experiments. There is n way for you to detect which process was which statistically.


The two embryos are independent coin tosses as in the above example.


Imagine 100 pregnant women withfraternal twins in a room ready to drop. Same statistical process as the coin example above. The issue is your pick of fraternal twinswhich are defined as independent embryos developing independently. No different than two single embryos in two women.


200 women with single pregnancies, or100 women with fraternal independent embryos carried to term. Sa statistical results.

All this is irrelevant. We're not interested in 100s of outcomes we're interested in 2. If I flip a coin 2 times, what is the probability that I get at least one tails as a result? What is the probability that I get at least 1 heads as a result (out of 2 flips). It may surprise you to know that the probability for both is .75.

The complement of 'at least one heads in 2 flips' is '0 heads in 2 flips', which is the probability of 2 tails, which is (.5)(.5) = .25 - and everything is right with the world because the sum of 'at least one heads' and it's complement 'no heads' is .75 + .25 = 1.

Someone tells me they ran an experimentto test for the occurrence of A or B. The variable is binary, twostates. I am told that the probability of either one turns out to be75%. I 'd say that is impossible. That would be like saying you have a process that fills a box with red balls and black balls saying the probability of picking one or the other is both 75%. Physically impossible.

That's nothing like what anyone is saying. The probability of 'at least one girl' in a set of fraternal twins is .75. The probability of no girl is .25. The probability of 'at least one boy' in a set of fraternal twins is also .75. The probability of no boy is .25.

You were incorrectly adding the probability of 'at least one girl' to the probability of 'at least one boy'. It does not make sense to do so.

aa

 

[rstrats]

Alcoholic Actuary,

re: "That's nothing like what anyone is saying. The probability of 'at least one girl' in a set of fraternal twins is .75. The probability of no girl is .25. The probability of 'at least one boy' in a set of fraternal twins is also .75. The probability of no boy is .25. "


That is absolutely correct.


I like the 2 coins experiment. Take two coins of the same denomination and toss them together 50 times, and count how many times at least 1 tail comes up. I guarantee it will be close to 75% of the time. And interestingly, during the same experiment, a head will also come up close to 75% of the time.

 

[steve_bnk]

All this is irrelevant. We're not interested in 100s of outcomes we're interested in 2. If I flip a coin 2 times, what is the probability that I get at least one tails as a result? What is the probability that I get at least 1 heads as a result (out of 2 flips). It may surprise you to know that the probability for both is .75.

The complement of 'at least one heads in 2 flips' is '0 heads in 2 flips', which is the probability of 2 tails, which is (.5)(.5) = .25 - and everything is right with the world because the sum of 'at least one heads' and it's complement 'no heads' is .75 + .25 = 1.

Someone tells me they ran an experimentto test for the occurrence of A or B. The variable is binary, twostates. I am told that the probability of either one turns out to be75%. I 'd say that is impossible. That would be like saying you have a process that fills a box with red balls and black balls saying the probability of picking one or the other is both 75%. Physically impossible.

That's nothing like what anyone is saying. The probability of 'at least one girl' in a set of fraternal twins is .75. The probability of no girl is .25. The probability of 'at least one boy' in a set of fraternal twins is also .75. The probability of no boy is .25.

You were incorrectly adding the probability of 'at least one girl' to the probability of 'at least one boy'. It does not make sense to do so.

aa

Ok. Maybe I am misreading the question.


I understand the double coin tossprobability. My problem is reconciling that with the physical realityof births.


Given equal male-female probabilitiesfor each embryo over many births sequentially as the babies aredelivered the probability of the next baby being male or female hasto be 50 percent. That is what I showed with the wall example.

Thanks for bearing with me.

 

[rstrats]

steve_bnk,

re: "Given equal male-female probabilitiesfor each embryo over many births sequentially as the babies aredelivered the probability of the next baby being male or female hasto be 50 percent."


Which is exactly the same as the coin toss. Each coin has a 50% chance of coming up heads, but taken together it's 75%. Same with the babies.

 

[steve_bnk]

steve_bnk,

re: "Given equal male-female probabilitiesfor each embryo over many births sequentially as the babies aredelivered the probability of the next baby being male or female hasto be 50 percent."


Which is exactly the same as the coin toss. Each coin has a 50% chance of coming up heads, but taken together it's 75%. Same with the babies.

Given a room full of women with fraternal twin pregnancies the probability of the sex on the next birth is 50%.

Not the same question as what is the probability of one of the pair being a girl in the womb.

 

[rstrats]

steve_bnk,

re: "Given a room full of women with fraternal twin pregnancies the probability of the sex on the next birth is 50%."

Agreed.


re: " Not the same question as what is the probability of one of the pair being a girl in the womb."

Agreed. But I'm afraid I don't see the point you're trying to make.

 

[Loren Pechtel]

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

You can flip a pair of coins a lot more times than you can make a pair of babies.

 

[Loren Pechtel]

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.

The probabilities only need to add up to 1 if they are exclusive.

"Has a girl" and "Has a boy" are not exclusive as we are talking about two babies.

 

[Bomb#20]

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

You can flip a pair of coins a lot more times than you can make a pair of babies.

Well, YOU can. Maybe Keith is one of those crooked fertility clinic doctors...

 

[Bomb#20]

Not the way you stated the problem, there's not. Consider the "truth table":

GB
BG
GG
BB

Now, that looks like four, but is only three, as GB/BG are the same as order is ignored/not specified in the problem, and are thus considered as one possible outcome (one is a girl).

So only three possible and equally probable outcomes...one is a girl, both are girls, or neither is a girl, and the probability that one is a girl is one in three.

I think I disagree.

The problem specifically mentions fraternal twins - which is when two separate eggs are fertilized by two separate sperm. Thus there are two separate chances for a girl. No different than flipping two coins simultaneously (order still doesn't matter). The probability of exactly one girl is .5 and the probability of at least 1 girl is .75.

(Or to say it differently, there are only 3 outcomes, but they are not all equally likely)

aa

If you flip two coins simultaneously, there are three possible and equally likely outcomes if order is not considered - one is heads, both are heads, or neither are heads.

The only way to get the .5/.75 result/truth table is if you specify order is significant.

What makes you think the meiotic process by which sex is determined, and/or the coin flipping process by which heads/tails is determined, is physically capable of taking into account whether the problem statement told you to consider the order? Flip two coins simultaneously a thousand times and you are going to get a certain distribution of results. Now do it again after somebody told you to pay attention to which coin was heads and which was tails. You are going to get roughly the same distribution. You aren't going to get at least one tail 667 times in the first run and 750 times in the second run.

I will gladly play poker with anyone who disagrees.

Let's make it craps; poker is barely math and mostly psychology. I haven't figured out which other player the mark is; ergo, it's me.

 

[doubtingt]

I got caught in my own web of parsing the meaning of "one of them" and in arguing that their are only 3 outcomes that differ relative to the question being asked (rstrats saying there were 4). I wrongly thought in terms of equal probabilities, but those arguing for .25, .25., and .50 for the 3 outcome probabilities and thus a .75 probability of at least 1 girl are correct in terms of the logical prediction.

However, what is logically correct isn't always actually true, since the premises might be wrong. In this case, whether .75 is actually correct assumes the two critical premises that 1) the genders of each fetus are independent from each other and 2) their genders are independent from the likelihood of conceiving fraternal twins. There seems to be reason to doubt both these premises. It seems that even though men as a group have a 50/50 chance of providing sperm that yields a boy or a girl, each individual man does not have a 50/50 chance. Some men are more likely to provide sperm that yield a boy and others more likely to provide sperm that yield a girl. This is evidenced by the fact that you can predict the number of boys a man will produce by the gender ratio among his siblings and among his parents siblings (families tend to be boy heavy or girl heavy). What this means is that the gender of each twin is not independent because they share a dependence upon the father's tendency to produce more males or females. IF one twin is more likely to be male then so is the other. While the aggregate probability of "at least one girl" might still be .75 at the population level, that might not be the probability for many actual people. Many people have either a lower or a higher probability than that. It highlights how statistical averages can be misleading and only reflect what is true about a distribution but not what is true about the individual entities that comprise the distribution.
So that is just the first assumption.

The second assumption might not be valid either. The probability of having fraternal twins is not randomly distributed. It is not equal for each person. Some people are much more likely than others, including older people, Africans (Asian have lowest rate), women who've had more kids, women with more nutritious diets, etc.. Then there are genetic factors too. IF any of these factors happen to be correlated (+ or -) with the factors that contribute to the likelihood of having a female, then having twins and having a female would be non-independent and raise of lower the probability that one of the twins is female. IOW, even though the avg probability of a female in the whole population is .50, it may be more or less than that among the women most likely to conceive fraternal twins. It is even possible that the the kind of men available to or selected by the kind of women most likely to have twins, tend to be men who either produce more boys or more girls (meaning these two issues interact). Another possibility could be a relation to having more viscous vaginal mucus, which appears to impact the probability that male or female sperm get to the egg first. Some research shows that more viscous fluid makes males more likely because Y bearing sperm swims faster in such fluid. More viscous fluid is also related to more difficulty in conceiving in general, thus time to conceive is correlated with probability of having a male. . The odds of having a girl also seem impacted by several other factors, including the timing of intercourse relative to ovulation (and thus frequency of intercourse). I haven't found anything on it yet, but it is certainly plausible that one of these factors could covary with the many factors related to twin probability, thus making the male-female ratios among twins different from the ratio for single births.

Of course this could be settled by simple empirical data on the exact observed % of the different twin outcomes. But I can't find anything that I am confident is based on empirical data rather the same kind of logical assumptions were using here.

 

[Alcoholic Actuary]

My problem is reconciling that with the physical realityof births.

Ok, so births are roughly .5 male and .5 female, similar to a coin flip. Fraternal Twins result when 2 separate eggs are fertilized by 2 separate sperm. In other words, there are 2 flips of a coin that happen simultaneously and the outcome is similar (identical) to two heads (= 2 boys), two tails (= 2 girls), or 1 head and 1 tail (equal 1 boy and 1 girl).

It is also identical to selecting 2 moms at random (who are not having twins or multiple births) and asking what is the probability of getting at least 1 girl out of the 2 moms.

Thanks for bearing with me.

Thanks for your attempts at reconciliation.

aa

 

[zorq]

However, what is logically correct isn't always actually true, since the premises might be wrong.
...
What this means is that the gender of each twin is not independent because they share a dependence upon the father's tendency to produce more males or females....

One additional assumption you made which reduces the significance of this factor in the real life scenario is that fraternal twins do not always have the same father. It does happen that two different men contribute the two sperm that eventually fertilize the mother's eggs. Just something to consider.

Of course this could be settled by simple empirical data on the exact observed % of the different twin outcomes. But I can't find anything that I am confident is based on empirical data rather the same kind of logical assumptions were using here.

If you want real life data that is simply the only way to do it. Logic is a poor substitute for actual science. Just look at how the theists try to prove the existence and qualities of something that they have never seen.