From the Academic Jungle Archives: Probably Handy Fun

So I dug through my old posts over at the defunct Academic Jungle (*cough-cough* that electronic dust is killing my sinuses). 
This one from 2012 (!) makes for some fun Saturday content. Btw, Smurf is now 10 and is left-handed, like his oldest bro. 


Having kids, especially little ones, is a source of enormous joy and tremendous exhaustion. There are many activities that are not particularly fun for anyone involved (e.g. projectile vomiting and associated cleanup) or are fun for the kids, but not so much for the parent (e.g. watching Teletubbies yet again with the 3rd kid, making it roughly my 11,678th viewing. I hate Tinky Winky and Dipsy with the burning passion of a thousand suns. Lala is still cool, though.)

Surfing the web on my phone or simply thinking about work are my favorite passtimes during protracted mind-numbing activities. This morning, while I was nursing Baby Smurf but — alas! — did not have my phone handy, I started thinking about the fact that Baby Smurf seems like he might be left-handed, although it may be too soon to say with certainty. He usually likes to hold two things at the same time, one in each hand, but he is much more likely to drop an old object and pick up a new, more interesting one with his left hand. Our eldest son is a left-handed, our middle one right-handed. Of course, all this leads to some cute probability questions. [1]

Here is what we know: let’s say that 1 in 8 people are left-handed (quick googling puts the percentage between 8 and 15%, and I remember a few years ago and reading somewhere that it’s 1/8=12.5% , so let’s stick with that). [2]

1) What is the probability that a family with three children will have a left-handed child? (meaning at least one)

2) What is the probability that a family with three children will have exactly two left-handed children?

3) What is the probability that a family with three children will have (at least) two left-handed children?

4) What is the probability that someone’s first and third child will be left-handed? (The only assumption is that the person will in fact have at least 3 kids.)

5) If my first child is left-handed and my second child is right-handed, what is the probability that my third child will be left-handed?

6) (added this one a bit later) If my first child is left-handed, what is the probability that I will have (at least) another left-handed child out of three total?

Have fun!

(I will post the solution in a day or so, although I am sure we’ll have the correct answers before that.)

———————-
[1] I admit this post was inspired by this one at Bad Mom, Good Mom, found through a link at Cloud’s.
[2] As badmomgoodmom says in the comments to the original post, handedness is not a completely random trait. Here, however, I just wanted us to have some fun with probabilities, so let’s assume handedness is a discrete random variable, with values L and R having the probabilities of 1/8 and 7/8, respectively.

4 comments

  1. Honestly I don’t remember how to do any of this math, but isn’t there evidence that handedness is heritable?

  2. @omdg “A large study of twins from 25,732 families by Medland et al. (2006) indicates that the heritability of handedness is roughly 24%.[22]” from https://en.wikipedia.org/wiki/Handedness#Genetic_factors

    My son is left-handed, though neither parent is—I think one of his uncles is.

    The calculations using the assumption of independence are trivial, but the calculations assuming heritability are not—there are many different models of inheritance that would give the same “heritability” estimate, but not necessarily the same results for the other calculations.

  3. OK, since no one else will play, let’s post the solution, with a bit of an elaborate explanation for those who may be interested but don’t have much background in probability.

    1) What is the probability that a family with three children will have a left-handed child? (meaning at least one)

    This is effectively the sum of the probabilities that there will be one left-handed kid [three such options, LRR, RLR, RRL, each with the probability (1/8)*(7/8)^2], two left-handed kids [LLR, LRL, RLL, each with a probability of (1/8)^2*(7/8)], and three left-handed kids [LLL, probability (1/8)^3]. You can add all these up and get the right solution, or you can simplify the computation by realizing that since the probabilities for all 8 outcomes [the 7 above plus RRR that has the probability (7/8)^3] have to add up to one, the answer — your cumulative probability to have at least one L kid will be equal to

    1-(7/8)^3=0.3301

    2) What is the probability that a family with three children will have exactly two left-handed children?

    You have to sum up the probabilities for all the outcomes where two kids are left-handed, so

    3*(1/8)^2*(7/8)= 0.041

    If you are looking for exactly n left-handed kids among m kids total, this result would generalize to (m over n)(1/8)^n*(7/8)^(m-n), where (m over n) denotes the binominal coefficient m!/n!(m-n)!

    3) What is the probability that a family with three children will have (at least) two left-handed children?

    We should sum up the probabilities to have 2 and 3 left-handed kids, i.e.,

    3*(1/8)^2*(7/8)+(1/8)^3=0.043

    4) What is the probability that someone’s first and third child will be left-handed? (The only assumption is that the person will in fact have at least 3 kids.)

    Nothing is specified about the handedness of any other kids, so the quantities that vary are the handednesses of the specific two kids, 1 and 3. All the others kids can be allowed to have whatever handedness they like, essentially making their handedness “certain” (unit probability). So of all the possible combinations of handedness, you want to focus only on those that contain L for the 1st and 3rd kid and then sum them up, which comes down to

    (1/8)^2=0.0156.

    (E.g. for 3 kids, that would be LRL+LLL, with probability(1/8)^2*(7/8)+(1/8)^3=(1/8)^2;
    for 4 kids, we’d have these combinations LRLR+LRLL+LLLR+LLLL; if for every L I write 1/8, and for every R I write 7/8, I get
    LRL(R+L)+LLL(L+R)=LRL+LLL=L(R+L)L=
    LL=(1/8)^2)

    5) If my first child is left-handed and my second child is right-handed, what is the probability that my third child will be left-handed?

    Since I fix the handedness of the first two kids, the last kid’s handedness is the only variable. Its probability is the same as for any single individual,

    1/8=0.125.

    6) If my first child is left-handed, what is the probability that I will have (at least) another left-handed child out of three total?

    We have fixed the handedness of the first child, so effectively we are now looking only at the other two kids and finding the sum of probabilities that one or both of them are left-handed. Therefore, this problem is the same as 1), only with two kids instead of three, i.e.

    1-(7/8)^2=0.2344

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google photo

You are commenting using your Google account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )

Connecting to %s