Pat Browne

Three probability problems

How much do I really understand?

Jul 3, 2025

Yesterday, I tried (and failed) to derive both the density function and the median of the lognormal distribution. I added these facts to my Anki deck when reviewing an older formula sheet, but broke the first rule of a good memory practice: do not learn if you do not understand.

There are many other skeletons in this closet (what of the gamma distribution, when I don’t remember much about the gamma function outside of factorials?). Instead of rooting them all out, I gave myself a quick test this afternoon. How much probability do I understand?

I selected three problems from William Feller’s probability textbook, “An Introduction to Probability Theory and Its Applications”, Volume 1. They were arbitrary, but all questions I could have answered somewhat easily when studying the subject actively.

Question 1 (Chapter IV, “Combination of Events”)

  1. Quadruples in a bridge hand. By a quadruple we shall understand four cards of the same face value, so that a bridge hand of thirteen cards may contain $0, 1, 2,$ or $3$ quadruples. Calculate the corresponding probabilities.

Things that went well:

Things that did not go well:

I got all answers to this question correct except $\Pr(X=1)$, 1 where I didn’t pay attention to detail and only accounted for $12$ of the $13$ cards. I also decided to subtract the probabilities of $1$ quad and $2$ quads by the higher-numbered probabilities, but can’t remember if I discarded that before peeking at the key. The correct answers (as written by Feller) are: $$ \begin{aligned} \text{Let } p^{-1}=\binom{52}{13}\text{. Then} \\[1.5ex] \Pr(X=3) &= 40 \cdot \binom{13}{3} \cdot p \\[1.5ex] \Pr(X=2) &= \binom{44}{5} \cdot \binom{13}{2} \cdot p \\[1.5ex] \Pr(X=1) &= \binom{48}{9} \cdot 13 \cdot p \\[1.5ex] \Pr(X=0) &= 1 - \sum_{x=1}^{3}\;\Pr(X=x) \end{aligned} $$

Overall, I didn’t fare too well here. I need to relearn how to count (and practice it much more). I have Anki cards for permutations and combinations, but some card additions that might help include

A better strategy would be to read through the chapters on combinatorics and conditional probability in several passes, pulling out understanding from each pass to build intuition. It will be a good sign when I don’t question myself on why the number of potential hands is $\binom{52}{13}$. Not sure what I’ll do yet, but I’ll keep this in mind as I encounter more of this type of Anki card.

Question 2 (Chapter IX, “Random Variables”)

  1. Birthdays. For a group of $n$ people find the expected number of days of the year which are birthdays of exactly $k$ people. (Assume $365$ days and that all arrangements are equally probable.

Ah, the birthday paradox he exclaimed! as I proceeded to blunder through this problem. Again relying on pattern matching, I remembered that there was something unique in the birthday paradox, but couldn’t remember what that thing was. I also misunderstood the problem, as the birthday paradox asks about the case where $k \geq 2$.

Reeling, I considered the binomial distribution: we’re given variables with the names $n$ and $k$, and there’s a $p$ ($\frac{1}{365}$). Applying this intuition, I came up with $$ P(X=k)=p^k(1-p)^{n-k} $$ however, as the astute might notice, this is not the probability mass function of the binomial distribution. That is instead $$ P(X=k)=\binom{n}{k}p^k(1-p)^{n-k} $$ which includes an actual binomial term to account for the ways in which these events can be ordered. Counting matters quite a bit in probability.

In the end the correct answer (again, courtesy of Feller) is $$ \binom{n}{k}364^{n-k}365^{1-n} $$

Evaluating what I had originally (and adding the binomial term), $$ \begin{align} &= \binom{n}{k}\left(\frac{1}{365}\right)^k\left(1-\frac{1}{365}\right)^{n-k} \\ &= \binom{n}{k}\frac{1^k \cdot 364^{n-k}}{365^k \cdot 365^{n-k}} \\ &= \binom{n}{k}364^{n-k} \cdot 365^{-n} \\ \end{align} $$ I’m missing $365$ in the numerator. I did not account for another part of this problem: we are not calculating a probability, rather,

the expected number of days of the year…

(strike two from the first problem, attention to detail)

Since this is an expectation, we multiply each day by its probability and sum the days up. In mathematical terms, $$ \begin{align} \mathbb{E}[X] &= \sum_{x=1}^{365}\;1 \cdot \Pr(X=x) \\[1.2ex] &= 365 \cdot \Pr(X=x) \end{align} $$ giving us the missing $365$ from above.

Overall, not great: making the same mistakes, relying on pattern recognition rather than intuition, not understanding the question.

Question 3 (Chapter IX)

  1. Find the covariance of the number of ones and sixes in n throws of a die.

I did not give this problem a full effort. I saw the term covariance, attempted to recall the formula from memory, and gave up when I could not.

This problem attacked another area of probability (properties of random variables) that I studied but can no longer recall. It wasn’t too hard to answer this with the formula for covariance in front of me: $\mathbb{E}[XY] - \mu_x \mu_y = 0 - \frac{n}{6\; \cdot \;6} = -\frac{n}{36}$ (the throws are assumed to be independent). My unfamiliarity with covariance is once again not great.

What have I learned? What do I understand?

While answering these questions, I felt like someone returning to a place they hadn’t been to in awhile. Though it’s only been six months since my last class in statistics, the gaps in my knowledge have grown and eaten through much of the intuition I used to have. Twelve months ago these problems would have been much easier than they were today.

This indicated the same fundamental issue: I have learned, but it is clear I no longer understand. I can tell you that $\text{Var}[X] = \mathbb{E}[X^2] - \mathbb{E}[X]^2$, but if asked follow-up questions, there would be significant gaps in my knowledge that come up.

Where do I go from here? Do I attempt to relearn probability from the ground up, or take my half-built crooked tower of knowledge with me as I go? Or is the choice binary?

The key point is that I now have an explicit choice. I know where my knowledge stands currently better than if I kept going through Anki cards sporadically and haphazardly. As a result, I can be more intentional about my memory practice and what I choose to learn in general. I’ll consider tradeoffs with, e.g., technologies I need to know for work, but now I know better what I… know.


  1. Aside: I am undecided on what to use for $\LaTeX$ probability notation. When writing, I often use $P$ and $\Pr$, but prefer $\mathbb{P}\text{r}$ (without the double line in the bend of the P). I haven’t found a symbol for that P yet and would want to turn it into an Emacs/$\LaTeX$ macro as well if it’s complex. ↩︎