crush or nemesis: the probability of finding love (or hate) via substack notes

‘What are the mathematics behind sharing a mutual crush with someone on Substack,’ I hear none of you ask. But I am here to answer anyway. Or at least I’ll answer part of that.

Keep your substack crushes close, and your substack nemeses closer. That’s why, prompted by ✧ brooklyn 𓆏 ‘s note, I had to know:¹ who has a crush, and who is a nemesis? So I asked people to like a note if they had a crush or thought they were my nemesis. And then I got to thinking. And then I thought some more. And then I ate a salad. And then all of a sudden I was like wait I wonder what the probability is of me sharing a crush or nemesis status with someone who liked that note? Would my love or hate be requited?

At the time of writing this I have seven likes on my note. So we’ll use that as the total number of possible pairings here.

I’m mostly interested in calculating the probability that I would share a crush/nemesis status with someone. Ok here comes the math part but don’t be scared

\(P(X = k) = \binom{n}{k} p^k (1-p)^{n-k}\)

This is just saying the probability that I get exactly k guesses right is equal to this formula, based on the binomial distribution, relating the number of trials (in this case people who responded so n=7), and p is the probability of correctly guessing whether I am a nemesis (as a value between 0-1.

Let’s assume I have a 50/50 chance of correctly guessing if their like was a crush or nemesis like. So p is 0.5 or 50%, (also equal to 1-(prob of crush)). In that case, the probability of getting exactly two guesses right would be:

\(P(X = 2) = \binom{7}{2} 0.5^2 (1-0.5)^{0.5-2}=0.055\)

Not great! But fortunately that’s not actually what I care about. Instead suppose I think could find an honest to goodness nemesis in that group, so I only need ONE (or more, they can fight it out from there) to be mutual. Let’s assume I see 4/7 people as potential nemesis material, and we treat a nemesis as a ‘success’ but a crush as a ‘failure’ (sorry it’s valentines EVE I get a pass back to loverboy tomorrow). So we’ll narrow the number of “trials” down to 4. Now we’re interested in at least one mutual nemesis status. This new formula is taking the sum of each trial’s probability of being a success, which in turn gives you a cumulative probability of getting at least one. This is also equal to 1 - (probability of 0/4). Aka chance of 1/4 + 2/4 + 3/4 + 4/4 matches.

\(P(X \geq 1) = 1 - P(X=0)\)

or

\(P(X \geq 1) = 1 - ( (1-0.5)^n = 0.94)\)

Fantastic news! I might just have a near 95% chance of finding a nemesis >:)

If we suppose they’re much more likely to have a crush on me (which let’s be honest that’s oooobviously true), we could adjust the probability down to say 1/5 (0.2). Well we still get

\(P(X \geq 1) = 1 - (1-0.2)^4 = 0.59 \)

60% isn’t too bad. Maybe at long last I’ll have a nemesis, or at least a mutual crush.

If you’re a real nerd like me and want to see the plot of the probability of at least one shared sentiment, here you go:

You can see that at 14 likes on my note I’d have a >95% chance, and at around 25 I’m almost certain to find a nemesis.

Or if we’re treating it as a coin flip:

If you read wow thanks for indulging me <3 and I hope maybe you learned a little fun math. Also I’m tired forgive any typos or mistakes

1. don’t even @ me for this colon use it’s tasteful and I don’t care what the rules say