# More on Gender Probability: Twins

#### (An archive question of the week)

Recently we looked at the question of how likely a two-child family with a boy is to have another boy (or, to the contrary, to also have a girl). Searching for those questions turned up another one of interest involving the gender of a pair of siblings: How do things change when they are twins?

Here is the question, from 2006:

Gender Probabilities for Twins

Hi Doctor Math.  I am pregnant with twins--sex unknown.  Since neither is older than the other, what is the probability of having any gender combination?  I remember from college genetics that each birth is mutually exclusive, therefore the probability of any combination of boy or girl is 1/2, i.e. each child will either be a boy or a girl.

My husband is using the bb, gg, bg, gb theory of probability, and that would increase the odds of having a boy and a girl to 1/2, while having a bb or gg would be 1/4.  My point is that bg and gb are the same combination, so his probability isn't correct.  What do you think?  I saw your string on a similar question, but it doesn't address twin births, and I wanted to be certain that I was thinking about this correctly.  Thanks!

I think that Melissa meant that each birth is independent (so that one being a boy doesn’t influence whether the other is a boy), not that they are mutually exclusive (that is, one child can’t be both a boy and a girl!).

Her husband used the same basic model we used in The Other Child: there are four equally-likely possibilities in a two-child family, based on birth order. But Melissa disagrees with this model, because twins are the same age. Is she right?

I replied, first stating my qualifications:

As a twin myself (and my identical twin brother is also a Math Doctor), I HAVE to take this question!

If there were only fraternal twins, then your husband would be right. Even with twins, you can distinguish them (firstborn/secondborn, favorite/nonfavorite, or whatever), so BG and GB are not the same.  As an example, I was expected to be a girl (because my heartbeat was weaker, and they didn't have ultrasound yet).  We could be distinguished, that is, even without knowing which was a boy and which was a girl or what our names would turn out to be.

Never tell a pair of twins (or their mother) that they can’t be distinguished! In fact, never tell them that one isn’t older than the other! My brother Rick is older by only 5 minutes, but we both know he is older (and that much wiser).

So you could make up a table of the four equally likely cases:

Dave
| boy | girl|
----+-----+-----+
boy | 1/4 | 1/4 |
Rick ----+-----+-----+
girl| 1/4 | 1/4 |
----+-----+-----+

That gives probabilities of 1/4 (two boys), 1/4 (two girls), 1/2 (one of each).

(Of course, I’m using our names to stand for “older” and “younger”, regardless of who we turned out to be.)

So Melissa’s husband is right … but only if it’s true that sexes of twins are truly independent. That isn’t quite true, and this is what makes the question interesting:

Now in reality, you also have to bring in the probability that the twins turn out to be identical--in which case they MUST both be the same sex.  (I understand there are other odd possibilities, but I'll neglect those.)

According to what I’ve read, in addition to fraternal (dizygotic) or identical (monozygotic) twins, there are very rare cases like “semi-identical” (sesquizygotic) twins, as well as many variations in the genetics. None of these are really relevant to our question as far as I know.

Suppose the probability of having fraternal twins is F, and of having identical twins is I.  Then, given that you have twins, the probabilities of their being identical is I/(F+I); we end up with this table:

BB     BG     GB     GG

I                    I
Identical  ------               ------
2(F+I)               2(F+I)

F      F      F      F
Fraternal  ------ ------ ------ ------
4(F+I) 4(F+I) 4(F+I) 4(F+I)

The identical twins comprise $$\displaystyle\frac{I}{F+I}$$ of all twins, which I’ve split into two equal parts, boy and girl. The fraternal twins consist of the remaining $$\displaystyle\frac{F}{F+I}$$ of all twins, split into 4 equal parts.

I didn’t assign specific numbers to these yet, in order to separate the (reliable) mathematics from the (empirical and uncertain) statistics, and can choose what numbers to use later, after considering general principles.

So the probability of two boys, or of two girls, is [summing the first or last column]

2I + F
------
4(F+I)

and the probability of one of each is [summing the two middle columns]

F
------
2(F+I)

Now we can look at the realities, which are far less certain:

A couple sites I looked at said that F = 1/125 and I = 1/300.  Another site said that 1/3 of all twins are identical.  Others give different numbers.  If I take that first pair of numbers, we get the probability of two boys, or of two girls, is 0.3235, and the probability of one of each is 0.3529.  So each probability is about 1/3.

That came from a quick search for frequencies of twins, just looking for reasonable numbers without trying for perfection. Repeating that search now, I find that Wikipedia says that, due to fertility drugs, the overall rate of twins “rose 76% from 1980 through 2009, from 18.8 to 33.3 (or 9.4 to 16.7 twin sets) per 1,000 births.” So these numbers are not very stable. Fraternal twin rates have not only increased over time, but also vary greatly among populations, between 6/1000 and 14/1000 or more. (Other sources say 20/1000 or more.) Again, from Wikipedia, identical twins are 3/1000 of deliveries worldwide, with little regional or temporal variation. (Other say 4/1000 or 1/300.) The numbers I used in my 2006 answer correspond to 8/1000 and 3.3/1000, the former apparently being low by current standards, and the latter being about right.

Note that for the overall rate, Wikipedia carefully distinguishes between the rate of twins and the rate of twin sets; one pregnancy with twins produces one twin set, but two twins, so the rate of twins is twice the rate of twin sets. The term “birth” or “delivery” seems ambiguous to me; it might mean either one mother giving birth, or one child being born. So I’m unsure how to interpret the data I’ve found.

For my calculations above, it doesn’t matter which way we take it, as long as F and I use the same definition, since we are looking only at twins. The distinction will be more important for the extra calculations I do below.

The rate given for identical twins appears to be for individuals, but it is very hard to confirm this! The fact that it is not as easy as most of us think to determine whether twins are identical further complicates this; many sources say that 1/3 of all twins are identical, but the site just referenced says instead that 1/3 are opposite-sex twins (for whom there is no question). This is consistent with the claim that 2/3 are fraternal (since half of fraternal twin sets will be opposite-sex). So F should be about twice I – presumably meaning before modern increases, and in a general American context, presumably. So we might take F = 6/1000 and I = 3/1000 as normal, while recognizing that in the right population F might be as much as 20/1000.

Using these numbers, we find that the probabilities are:

$$P(\text{same sex} | \text{twins}) = \frac{2I + F}{2I + 2F} = \frac{2(.003) + .006}{2(.003) + 2(.006)} = \frac{2}{3}$$

$$P(\text{opposite sex} | \text{twins}) = \frac{F}{2I + 2F} = \frac{.006}{2(.003) + 2(.006)} = \frac{1}{3}$$

This confirms what I said about each of the three probabilities (two boys, two girls, boy-girl) is about 1/3. In fact, this will be true regardless of the actual probabilities, as long as F = 2I:

$$P(\text{same sex} | \text{twins}) = \frac{2I + F}{2I + 2F} = \frac{2I + 2I}{2I + 2(2I)} = \frac{4I}{6I} = \frac{2}{3}$$

(Ironically, do you notice that this is what Melissa would conclude from her wrong assumption that BG = GB?)

But taking the highest reported frequency of fraternal twins, 0.020, we get

$$P(\text{opposite sex} | \text{twins}) = \frac{F}{2I + 2F} = \frac{.020}{2(.003) + 2(.020)} = \frac{10}{23}\approx 0.435,$$

bringing the probability considerably closer to 1/2. (Of course, if all twins were fraternal, this probability would be 1/2.)

Now let’s go beyond the question, and find the probability of any two-child family being a boy and a girl. Here, the exact meaning matters. I defined F and I as probability of having fraternal or identical twins, which sounds like twin sets, not individual twins (the probability of being a twin); that’s appropriate for the question we were answering. But my impression is that most of the data refer to individuals, so let’s call it that.

Here is a chart of the probability of all possibilities, broken down by kind of pair and genders in birth order:

               BB     BG     GB     GG   |  Total
---------------------------------------+-------
1-I-F  1-I-F  1-I-F  1-I-F  |
Non-twin   -----  -----  -----  -----  |  1-I-F
4      4      4      4    |
|
I                    I    |
Identical   ---     0      0     ---   |    I
2                    2    |
|
F      F      F      F    |
Fraternal   ---    ---    ---    ---   |    F
4      4      4      4    |
---------------------------------------+-------
1+I    1-I    1-I    1+I   |
Total       ---    ---    ---    ---   |    1
4      4      4      4    |

Not too surprisingly (if you think about it!), fraternal twins have no effect of the probability that two siblings have the same gender; F doesn’t show up in the bottom line.

So the probability of any pair of siblings having the same sex is simply

$$P(\text{opposite sex} | \text{twins}) = \frac{1+I}{2} = \frac{1 + .003}{2} \approx 0.5015,$$

that is, 50.15%. This is barely more than the usual assumption of 50%, which makes sense because identical twins are rare. Compare this to the fact that the ratio of boys to girls at birth is not 1:1, but about 1.05:1, so that about 51.2% of babies are boys. Twins have less effect on the ratios than other causes.

This site uses Akismet to reduce spam. Learn how your comment data is processed.