Michael Lewis contends that for families with 6 children, the birth order BGBBBB is equally likely as GBGBBG. Setting aside the sequence, how is possible that in a pool of 50:50 B:G a random sample of 5:1 is just as likely as 3:3?

  • 3
    $\begingroup$ Please explain what you mean by "5:1" and "3:3". Hint: the answer to your question is embedded in an accurate account of that terminology. $\endgroup$
    – whuber
    Feb 19, 2017 at 23:42
  • 1
    $\begingroup$ closely related: stats.stackexchange.com/questions/256563/… $\endgroup$
    – Sycorax
    Feb 19, 2017 at 23:45

3 Answers 3


In general 3:3 is much more likely than 5:1 because there are many more ways to get 3:3 than 5:1. However, a specific 3:3 is the same probability as a specific 5:1.

  • $\begingroup$ Many thanks to w Huber and especially David Lane (who gave me more to work with). I think I get it but will have to think about it some more. $\endgroup$ Feb 20, 2017 at 0:37
  • $\begingroup$ This is the answer to accept. Mine is just an attempt at making it very, very clear after your comment. $\endgroup$ Feb 20, 2017 at 0:57

tl;dr: You're getting tripped up by the difference between the likelihood of the sequence (BGBBBB vs. GBGBBG) vs. likelihood of having five boys in a family with six children.

Let's simplify this slightly and consider sequences of four, rather than six children. The logic is exactly the same for families of six, but it's much easier to write down the sixteen possibilities for a family of four than the sixty-four possibilies for a family of six.

Here are all the possible sequences, grouped by number of boys (or girls):

$$\begin{array} {|r|r|} \hline \textbf{Sequence} & \textbf{# of boys} & \textbf{# of girls}\\ \hline BBBB &4 &0 \\ \hline BBBG &3 & 1 \\ BBGB & 3 & 1 \\ BGBB & 3 & 1 \\ GBBB & 3 & 1 \\ \hline BBGG & 2 & 2 \\ BGBG & 2 & 2 \\ GBBG & 2 & 2 \\ BGGB & 2 & 2 \\ GGBB & 2 & 2\\ GBGB & 2 & 2 \\ \hline GGGB & 1 & 3\\ GGBG & 1 & 3\\ GBGG & 1 & 3\\ BGGG & 1 & 3\\ \hline GGGG & 0 & 4\\ \hline \end{array} $$

Next, we need to establish the probability of seeing one of these sequences. Suppose we start with no children. Then, we flip a coin[*] to choose the child's sex. Regardless of the outcome, we then flip the coin again, then again, and again, until we have four children.

The probability of having a boy (or girl) is equally likely at the beginning, and the process has no memory (having a boy last time neither increases nor decreases the probabilities this time), so boys and girls are equally likely on each subsequent flip. Therefore, each sequence is also equally likely: $\frac{1}{16}$ of families with four children have all boys, another $\frac{1}{16}$ have all girls, and yet another $\frac{1}{16}$ alternated boy, girl, boy, girl. In other words, the specific sequence BGBG is just as likely as the sequence GGGG.

However, there are six sequences that yield families with two boys and two girls, while only one that yields a family with four boys. Thus, when you ignore the order, you find that the outcomes (as measured by the total number of boys/girls) are not equally likely.

The same logic holds for a family of six. There are $\binom{6}{3}$, or 20 possible families with three boys and three girls, making a family with that composition more likely than one of all boys. However, the specific sequences BBBBBB and BGBGBG are both equally likely and occur $\frac{1}{64}$th of the time each.

[*] Here, we're assuming that boys and girls are equally likely. This isn't quite true, but it's close enough--and the overall point stands even if it weren't.

This result, incidentally, is from Kahneman and Tversky's 1972 Cognitive Psychology paper. They argue that people expect random sequences to:

  • Look somewhat like the average "collapsed" outcome (i.e., have equal numbers of boys and girls), and
  • Show lots of alternations (in other words, "BGGBGB" seems more likely than "BBBGGG")

Neither of these are true, of course, but it's a strong cognitive bias nevertheless.


There are $6$ siblings in both cases. To get $3\text{ boys}:3\text{ girls}$ you have to calculate the possible combinations of $6$ choose $3$, or expressed mathematically, $\binom{6}{3}=20.$ The order does not matter, and it makes no difference if we are counting the ways of selecting $3$ boys or $3$ girls out of these siblings - if we pick boys, the others are necessarily girls. If instead we look at the ways of getting $5:1$, we have to calculate $\binom{6}{1}=6$, or the possible ways of picking the girl: older sibling, second sibling, etc.

The probability of getting $3:3$ or $5:1$ as a ratio of $\text{boys}:\text{girls}$ depends on the different number of ways you can end up with this ratio. For example, $5:1$ can be achieved with $\small \text{B,B,B,B,G}$, or $\small \text{B,B,B,G,G}$, etc. There are six possible combinations resulting in this ratio. On the other hand, there are $20$ combinations with a $3:3$ split.

However, each specific sequence, for instance $\small \text{B,B,B,B,G}$, has exactly the same probability, because the $\small \Pr(\text{boy})=\Pr(\text{girl})=1/2.$ Each string has a probability of $\frac{1}{2^6}.$

Now for a ratio of boys and girls you would have to multiply this $\Pr(\text{any given string})$ $\times$ the number of possible strings resulting the particular ratio of boys and girls you are considering (e.g. $20$ in the case of equal split $3:3).$

  • $\begingroup$ Just an extended on the answer, which is nicely and tersely provided by David Lane. Accept his answer. $\endgroup$ Feb 20, 2017 at 0:59
  • $\begingroup$ Very kind of M K and A P to weigh in. Very helpful. $\endgroup$ Feb 20, 2017 at 1:09

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.