I have a book called “Fifty challenging problems in probability”, and I started with the first one today, and I reflected on what my mind was doing trying to solve it. I don’t know probability at all, so this is a great chance to reflect on fundamental thought processes without being tainted by knowing how to solve a class of problems.

Here’s the problem:

**A drawer contains red socks and black socks. When two socks are drawn at random, the probability that both are red is 1/2. a) How small can the number of socks in the drawer be? b) How small if the number of black socks is even?**

I tried several approaches, and only my last one worked; I’ll go through these, and reflect on why I didn’t immediately arrive at the correct one. I’ll put pointers into significant parts of the thought process where I made a mistake.

### First stab

I tried coming up with an obviously wrong solution, and improving it.

How about there are an equal number of socks in the drawer? Like 1 red, 1 black (A). Then the probability of pulling out a red twice is (1/2 * 1/2) = 1/4. That’s broken, we need more reds.

How about there are 3 socks, and 2 are red? What’s the probability of choosing 2 red? I reasoned that it’s (2/3 * 2/3) = 4/9, though I wasn’t certain that they’d multiply.

I spent quite a bit of time proving this to myself (bottom of the post).

So the equation becomes (p^2 = 1/2), where p is the probability of choosing red, where p=(R/N), as I thought intuitively. I know that there is no integer R and N that satisfy that equation from high school (B). So… there is no solution?

### Figuring out why my solution isn’t working

One thing I did was scatter for tools. I know there is also conditional probability, so maybe somehow I’m missing some subtlety in my reasoning, and the probabilities don’t multiply (C).

I tried thinking of other representations, which turned out to be equivalent to my original solution (D). I thought of some binary encoding schemes. For instance, (1 1 1 0 0) could represent 3 red socks and 2 black socks in the drawer. How do I represent which socks were selected? (0 0 1 0 1) would represent sock 3 and 5 being chosen. The number of possible selections is the number of possible vectors with 2 ones; for a given distribution of reds and blacks, the probability of 2 red socks being chosen would be the number of vectors with 2 ones located on the left side of the vector, before the reds end.

This turned out to also lead to the same answer I got before.

### Final solution

I had been questioning this procedure of pulling out a sock, putting it back, and pulling out a sock again.

Perhaps we don’t put the sock back? I didn’t want to deal with figuring out this totally different problem, with a completely different enumeration. Yet I was curious about how that problem would work (E).

So instead of having R red socks the second time, I have (R-1) socks. And instead of choosing from (N) socks, I choose from (N-1) socks. The equation becomes:

(R * (R-1))/(N * (N-1)) = 1/2

I don’t know how to solve it! It’s 1 equation, 2 variables. But I try to play around with how it changes, because maybe there’s some obvious pattern (F). I try a random pair of reasonably high numbers: 3, 4

(3 * (3 – 1))/(4 * (4 – 1)) = (3 * 2)/(4 * 3) = 6/12 = 1/2… oh fuck.

## Analysis/Lessons

(A) I made a critical assumption here, and wasn’t aware of it, because I didn’t check for full understanding of the problem immediately.

(B) I got lucky to know it, but that just goes to show that part of solving problems is knowing a lot of stuff.

(C) Having too many tools can be distracting, and induce anxiety. It’s best to think from first principles, rather than grasp for tools.

(D) I should have been 100% confident in my original solution. Second-guessing myself was a waste of time.

(E) Curiosity really saved me here.

(F) Again, curiosity worked out well here, trying to “feel” the problem before attacking it with overly powerful tools.

**Why didn’t I arrive at the correct solution immediately**? Had I accidentally been on the other side of the assumption tree, I would have. But fundamentally, I was at first not aware of an assumption I made, which is a critical error in rigorous reasoning.

So that’s my **main lesson** from this exercise: understand the problem statement, resolve assumptions, or be aware of exploring assumption branches.

**Secondary lesson**: curiosity can help get out of dire situations of misunderstanding.

**Tertiary lesson**: I was thinking through this problem as I walked around, with plenty of space. I thought about lots of other things, though I did want to solve this problem. I noticed that I need more focus to sustain my mind thinking about the problem, and at the same time reflect on what my mind is doing. So this is the tool of improving thinking: sustaining a double stream of consciousness – one to solve, one to rip the solving apart. And it’s essential to be able to recall the solution process; I found that I had trouble recollecting the order in which I thought of solutions, because my solution wasn’t very structured. I will try to avoid that, so that recall is easier moving forward.

## Proofs

### When choosing two socks, probabilities multiply

If the socks are ordered 1 through N, and the reds are the first R socks, I can represent “choosing 1 sock” as a list of numbers 1 through N, with 1 of the numbers circled out. Having this language for the 1-sock event, 2 socks can be represented as 2 lists; they can be placed as a table, and every element in the table would represent a pair of socks. In this table, the red socks are in the upper left, and form a square. Its area is R*R, and the total sock area is N*N, so the chance of choosing 2 red socks is (R*R)/(N*N)

I found it fascinating that 2 events like this have this geometric, tabular representation, and probability connects with geometry that way.