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Does your culture really affect the gender distribution?

June 17, 2009

Answering the question “what does a physicist do?” can be difficult.  Even the much more modest question “what do you do?” is pretty hard.  Over the course of my relatively brief physics career, I have studied sonar, traffic patterns, cloud formation, parasite epidemics, low-temperature magnetic structures, and the electrical properties of DNA, to name a few things.  These topics have seemingly nothing in common, but they represent only a small sampling of the kinds of research you could find in the physics department of any major university.

And really, these subjects don’t have much in common.  But that’s okay, because physics isn’t really a topic.  It’s more of an approach, or a set of strategies for problem solving.  Someone trained in physics is not valuable because they have learned a lot of useful information, but because they have acquired a way of reasoning that is useful for solving complicated problems.

Acquiring new ways of thinking is what physics is all about.  In this post I want to illustrate the value of a “physics approach” to a problem by discussing an example where reasoning without it can get you a very wrong answer.



Not long ago I read Matt Ridley’s The Red Queen: Sex and the Evolution of Human Nature.  It was a very interesting book, full of complex and important ideas.  But in my mind too many of its conclusions were based on “common sense”; which can be a dangerous thing.  An idea that seems correct often is, but is certainly not guaranteed to be.  There are plenty of strange and counterintuitive phenomena out there, and their explanations cannot be rooted out by common sense.  Sorting strange fact from sensible fiction takes quantitative reasoning, and that’s what physics was designed for.

Somewhere around the fourth chapter of The Red Queen, the author starts talking about factors that can influence the gender of an unborn child.  During the course of his discussion, the author mentions that “merely by ceasing to breed once they have a boy … people would have male-biased sex ratios at birth.”  At first sight, the idea seems sensible.  After all, if every couple decided to stop having children once they had a boy, then every family would have at least one male child and maybe the result would be a male-biased gender distribution.  But on second thought, maybe the opposite should happen.  If everyone keeps having children until they have a boy, then you could have families with 9 girls and only one boy.  They should bias the gender distribution in the other direction, right?  So which is it?  Will a society produce more boys than girls simply by considering them to be more valuable?

To construct an argument that could be physicist-approved, we need to come up with a quantitative prediction for how many boys and girls would result from a society where couples have children until they have a boy, and then stop.  Let’s enumerate all the possibilities.  For a given couple, the sequence of their children could be {boy}, or {girl, boy}, or {girl, girl, boy}, or {girl, girl, girl, boy}, … and so on.  What is the probability for each of these sequences?  The probability for a given child to be either a girl or a boy is 50% (or 1/2).  So the probability of the outcome {boy} is 1/2.  The probability of the outcome {girl, boy} is 1/2 * 1/2 = 1/4.  For {girl, girl, boy} it’s 1/2 * 1/2 * 1/2 = 1/8 (or 12.5%).  And, in general, the outcome with n children has the probability (1/2)^n.  We can make a diagram of all possible outcomes that looks something like this:

A diagram of possibilities for a couple's children.  Bottom (blue) paths represent a male child, which ends the sequence of children.  Top (pink) paths represent a female child, which cause the couple to continue having children.

A diagram of possibilities for a couple's children. Bottom (blue) paths represent a male child, which ends the sequence of children. Top (pink) paths represent a female child, which cause the couple to continue having children. Each blue line has a percentage indicating the likelihood of that sequence.

Every couple will have one boy.  But how many girls, on average, will they have?  Well, according to our chart, there’s a 50% chance they will have no girls, a 25% chance they will have one girl, a 12.5% chance that they will have two girls, and so on.  Summing all possible outcomes gives

(average number of girls per family) = 1/2 \times 0 + 1/4 \times 1 + 1/8 \times 2 + ...

= \sum_{n=0}^\infty n \cdot (1/2)^{n+1} = 1

You can plug the first eight terms or so into your calculator to get a good approximation to the answer, or you can remember what you learned about infinite series from calculus.  Either way, the answer is exactly one.  That is, nothing happens to the gender distribution: on average, every family has one girl and one boy.

If you think this was a fun problem, you can try to see what happens when you modify the “child birth rules” a little bit.  For example, maybe every couple stops either when they have their first boy or when they have 5 children total.  Or if every family keeps having children until they have two boys (the “tree” of possibilities is a little more complicated here, but you can still draw it out).  You’ll find that nothing changes the inevitable: your cultural values have no effect on the ratio of male to female babies.

This answer seems obvious in hindsight, but it really wasn’t.  We needed quantitative reasoning to be our King Solomon, judging between a number of different sensible arguments.  Of course, I’m not saying that non-quantitative arguments are never valuable or correct; they very frequently are.  It’s just that physics has trained me not to trust them.



And in case you’re wondering, yes, I did have the audacity to contact Matt Ridley and inform him of the error in his book.

14 Comments leave one →
  1. Jill Alvarado permalink
    July 2, 2009 3:10 pm

    Some claim there is a way to predict and/or choose your child’s gender using this chart:
    I don’t know if it is 100% accurate, but when I checked it out for my 2 boys it was accurate. Is there scientific base or psychological base or mere superstition? I dunno.

    • gravityandlevity permalink*
      July 2, 2009 4:47 pm

      Apparently I was supposed to be born a girl. I doubt that there is more to it than tradition and superstition.

      Is it possible that your gender is affected by the time of the year/month that you are conceived? Probably. As my advisor liked to say, “everything depends on everything else, as long as the dependence is not forbidden by a fundamental symmetry of the universe.” But I am skeptical that the effect is noticeable.

  2. August 1, 2009 3:44 pm

    What a terrific post (and not just this one, have discovered your site an hour or two ago and am voraciously reading and clicking).
    You really take the elitist edge off scientific reasoning, and show that it’s just a neat way of breaking down a problem into bits you can solve without making too many assumptions. As an econ student, I’m trying hard to do the same and not get too caught up in ideology, but it’s not easy. Brains like stories….

    • gravityandlevity permalink*
      August 1, 2009 3:48 pm

      I’m glad you like it, Helena!

      I am a firm believer that behind every scientific idea is a story. I have been very fortunate to have had great “story tellers” as teachers. Don’t let anyone make you feel like science is something that can only be understood by the world’s most intelligent and highly-educated men.

  3. Arend permalink
    August 6, 2009 11:16 am

    Great post! Part of the conclusion in your initial scenario, however, hinges on the assumption that the sex ratio is 1/2, which is not quite true. The ratio of boys to girls at birth is 105 boys to 100 girls. Hence asymptotically there would be a bias: 0.95 girls on average (modulo calculation errors on my part).

    Still, your original conclusion stands: the deviation is much smaller than you would naively expect.

    • Arend permalink
      August 6, 2009 1:57 pm

      The problem has got me thinking a little further and I realized that I overlooked the obvious.

      The average number of children is 2 in the initial scenario (sex ratio of 1:1, keep having children until you have a son). If the sex ratio is 105:100 (change of a daughter is 0.488), however, the average family size will be 1.95, so the percentage of daughters will be 0.488. In other words, the scenario has no effect on the final sex ratio in the population of children. (A little math shows this is a general result.)

      In hindsight this makes perfect sense: if you keep throwing a dice until you roll a six, say, you are not affecting the actual probability of throwing a six (or five, or four, or…) in a given throw. Counting over all throws in all repetitions of this experiment, you will still get the proportions of sixes and ones and twos etc. determined by the dice.

      It’s still an interesting question where the misconception creeps in. My feeling is that we’re very good at counting the possible scenarios (a single boy, a boy and a girl, two girls and a boy, three girls and a boy, etc.) but not very good at weighing each scenario appropriately.

  4. August 11, 2009 4:08 pm

    Say you have 100 % parents who can give birth to kids,

    50% of them give birth to boy and 50 % to girl i.e. the ratio is 1:1
    50% of parents with girl give birth to 25% male and 25% female child. The ratio is still 1:1.
    25% of parents with girl child give birth to 12.5% male and 12.5% female child. The ratio is still 1:1.

    Even if they stop at any point i.e. they stop having kids after their 3rd baby, the gender ratio is 1:1. My friend Mike suggested this argument to me, which I found interesting enough to post here.


  5. August 14, 2009 5:07 pm

    Nice. I suspect a more intuitive way to present this argument is as a kind of ergodicity property. Since births are assumed to be independent coin tosses, the identity of the parents doesn’t actually matter, does it?

    So attribute ALL births to one hypothetical parental production queue. Stop-events are effectively meaningless…

    But yes, it takes some effort to see this. Common sense is often wrong, but I think you aren’t “done” until you get to the most intuitive way to see that the right answer IS right (i.e. make it common sense). Not there yet in this case.

    One that I struggled with a really long time (and still do) is Marilyn Von Savant’s famous Monty Hall problem. On Mondays, Wednesdays and Fridays, I believe the correct answer. On Tuesdays and Thursdays I don’t. On weekends I try not to think about it 🙂

    I am planning a post on it at some point. Would be interested in your take on the most intuitive way to understand the right answer.

    • June 13, 2015 5:26 pm

      Having only recently found this blog (and loved it), I realize that I’m commenting on a very old post. But I cannot resist saying that Venkat’s position on the Monty Hall problem captures it better than any that I’ve ever seen (and I’ve seen a lot).

      Brian, thanks for this wonderful blog! Venkat, I’m going to start reading at Ribbonfarm.

  6. August 15, 2009 2:00 am

    Though this isn’t related to the post, Monty Hall problem seems kind of obvious once you see the solution to it. An “intuitive” way to explain that may be

    Say you have 1000 doors and pick one, your probability of picking the correct one is 1/1000. Whereas if I take the rest of the set my probability of winning is 999/1000. If the guy(who knows what’s behind the doors) opens 998 empty doors, my probability of winning is still intact i.e. 999/1000.

    You can enunciate a similar argument for 1 out of 3 doors too.

  7. Lior permalink
    August 17, 2009 12:30 am

    What you’ve run into is the Optional Stopping Theorem. For ease of analysis, let’s assume that the sequence of sexes of your potential children is decided once and for all when you get married: to every couple we associate a sequence Y_1, Y_2, Y_3, … where Y_k = 1 if the “k”th child is a boy, Y_k = -1 if the child is a girl. As in the blog post we assume that each Y_k is chosen independently and uniformly at random.

    Now let X_t be the sum of the Y_k up to child t, so X_t is the excess of boys over girls after t children. The key fact is that since the Y_k have mean zero, X_t is a martingale: given your family at time t, the expectation for the “excess” X at time (t+1) is the excess X_t you already know (after all, there’s a 50% chance of the excess going up by 1, a 50% change of the excess going down by 1).

    Next, let T be any stopping rule only depending on the existing family (such as “T=birth-number of the first boy” or “T=birth-number of the second boy” or “the birth number of the first girl whose number is at least 6”). The kind of rule that is not allowed is a rule depending on the future (“if the next two children are boys then we don’t want to have them”, that is “T=birth order of the first child whose two successors will be boys”).

    Finally, Theorem says that the expectation of X_T (the state of the process at the stopping time) does not depend on the rule T. In particular, taking T to be the rule “stop at time 1”, we see that in our case the expected excess is zero — no matter what stopping rule you use!

    Arend’s analysis is also right, and can be understood using the same theorem. All you need to do if the sex ratios are not 50-50 is to subtract a constant from Y_k so that it has mean zero. This will show that the expectation of the “excess” at time T is exactly the expected number of children times the bias, in other words that the sex ratio is preserved as found by Arend.


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