Basically, I saw a photo of two people with unknown approximately equal heights and was struck with the thought "How likely is it that they're both tall?". That's the problem I want to answer.

I concede it's possible that I'm making an error in my thinking but I think it's just I don't know how to go about the last step.

The way I thought about that question mathematically is that we can assume Normals for height, i.e. $X \sim N(\mu_X,\sigma^2_X), Y \sim N(\mu_Y,\sigma^2_Y)$, and thus my question is:

$$\mathbb{P}(X = x_1 = Y = y_1 \pm \frac{\epsilon}{2} \space \cap \space x_1 > \mu_X \space \cap \space y_1 > \mu_Y)$$

where $\epsilon$ is some small positive constant which represents the limits of visual equivalence, e.g. if John's 181cm and Henry's 182cm are you really going to be able to tell they're a different height when standing next to each other? I think not, therefore $\epsilon$.

Now, while I suspect two people being photographed together don't have actually independent heights, I'm fine with assuming that they are, so therefore:

$$\mathbb{P}(X = x_1 = Y = y_1 \pm \frac{\epsilon}{2}) \cdot\mathbb{P}(x_1 > \mu_X)\cdot \mathbb{P}(y_1 > \mu_Y)$$

I believe those last two can be calculated simply as $\mathbb{P}(x_1 > \mu_X) = \mathbb{P}(X > \mu_X)$ = 1 - rnorm(mu, mu, sigma) (using R) but I'm not sure how to do the first one, especially in the situation where $X = Y$, which is an assumption I'd make if the two people were both male or both female. In particular, it seems that the starting point is something like:

$$\mathbb{P}(X > Y) => X - Y \sim N(\mu_X - \mu_Y, \sigma_X^2 - \sigma_Y^2) $$

but that will obviously be $X - Y \sim N(0, 0)$ when $X = Y$, which I'm not sure how to work with. I have even less of an idea about how to incorporate my $\epsilon$ wide interval of visual equivalence.

  • 1
    $\begingroup$ I would have thought "approximately equal heights" and "both tall" require some quantification. In addition, is it possible to have them approximately equal heights but one tall and the other not? $\endgroup$
    – Henry
    Commented May 16 at 13:18
  • 1
    $\begingroup$ How do you assume two different distributions of height?? // To be answerable, your question needs to make a (strong) assumption about independence or lack thereof in the heights of pairs of people in photographs. If you assume independence and model the heights with a single common distribution, your question can be framed as "when $(X,Y)$ is a standard binormal variable and $c$ is the threshold of tallness, what is the conditional probability that $X\gt c$ and $Y\gt c$ (ie., "tall") conditioned on $X=Y$ (equal height) or, alternatively on $|X-Y|\lt \epsilon$ (approximately equal height)?" $\endgroup$
    – whuber
    Commented May 16 at 13:36
  • $\begingroup$ Sorry, I forgot about this. Henry, I would approach approximate equalness as whuber does i.e. the |X−Y|<ϵ idea. If that was unclear in the question, my bad. If the issue is the fact ϵ could be anything; that's because I was interested in a generalisable answer. As to the question of tallness itself. I just went with the means of the Normals above but my real thinking is again as whuber suggests, with X>c and Y>c, or rather X>c and Y>k, where c can be but is not necessarily equal to k. Tallness is a rather subjective concern. $\endgroup$
    – Vonvorv
    Commented Jun 12 at 5:27
  • $\begingroup$ Also apologies to you, whuber for forgetting about this. The reason I have two different distributions of height was to allow for the case that we're interested in, say, a man and woman of approximately equal height. This is also why c and k wouldn't necessarily be equal. In my ignorance I assumed having two Normals wouldn't pose a threat to the, er, answerableness of the question. I would agree that the assumption of independence is a strong assumption, but as I said for an idle concern (I see no use case to this question) I am comfortable making it. $\endgroup$
    – Vonvorv
    Commented Jun 12 at 5:32

2 Answers 2


I assume you define "tall" as height being in 50th percentile or higher. The chance that the first person is tall is 50% The second person has nearly the same height, so the probability of being tall (if the first is tall) is nearly 100%. So the chance that a pair of near-equal height people are "tall" is nearly 50%.


The joint probability density of the two heights is $f_X(x)f_Y(y)$, assuming that the two parsons are unrelated (e.g., they are not relatives). Note that it makes sense to take different parameters (same $\mu,\sigma^2$) for the two, only if we believe that their heights are governed by different distributions - e.g., if one is Asian or the other is Scandinavian.

We also need a definition of a person being tall - e.g., that they are taller than some specified height h (which could be related to them belonging to a certain percentile - e.g., being among the tallest 20%.) We now need probability that $$|X-Y|<\epsilon, \frac{X+Y}{2}>h,$$ where $\epsilon$ is the allowed difference in heights - e.g., no more than one or two inches.

The required probability is then $$ \int\int f_X(x)f_Y(y)\theta(\frac{x+y}{2}-h)\theta(\epsilon-|x-y|)dxdy, $$ where $\theta$ is the Heavyside step function. For normal distribution this integral can be evaluated in terms of error function.

  • $\begingroup$ The evaluation is more complicated than that: it's a definite integral of the error function, not the error function itself. $\endgroup$
    – whuber
    Commented May 16 at 15:44
  • $\begingroup$ @whuber infinite integration limits are assumed, of course - the step functions do the rest. $\endgroup$
    – Roger V.
    Commented May 16 at 15:57
  • $\begingroup$ I don't think they do. This is twice the integral of a bivariate Normal distribution over a strip terminated by a triangle. It's not expressible exactly as an error function. $\endgroup$
    – whuber
    Commented May 16 at 20:16

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