Yes! That's a good point that I didn't mention.

I appreciate people playing along :)

All the responses to this post made my week. Thanks!

Appreciate it a lot. Thanks!

Yeah, I suppose we agree then.

Sure: 
For a monogamous partner, finding a successful partner has a value of 1
Finding 2 successful partners also has a value of 1, because in a monogamous relationship, you only need one partner.
The same holds for 3, 4, etc partners. All those outcomes also have a value of 1.
So first, let's find the probability of getting a value of 0. Then let's calculate the probability of getting a value of 1.
The probability of getting a value of 0 (not finding a partner):

There is one other mutually exclusive alternative: Finding at least one partner (which has a value of 1)

So we have a 34.9% chance of getting a value of 0 and a 65.1% chance of getting a value of 1. The expected value is:

If you did this experiment a million times and assigned a value of 1 to "getting at least one monogamous partner" and a value of 0 to "getting no monogamous partners," you would get, on average, a reward of 0.651.

For the sake of brevity, I'll skip the calculations for a polygamous partner because we both agree on what the answer should be for that.

Ah, shoot. You're right. Probably not good to use "odds" and "probability" interchangeably for percentages like I did. Should be fixed now.

95% is a lower bound. It's more than 95% for all numbers and approaches 95% as n gets bigger. If n=2 (E.G. a coin flip), then you actually have a 98.4% chance of at least one success after 3n (which is 6) attempts.

I mentioned this in the "What I'm not saying" section, but this limit converges rather quickly. I would consider any $n\ge 4$ to be "close enough"

I think what Justin is saying is that finding a single monogamous partner is not significantly different from finding two, three, etc. For some things you only care about succeeding once. So a 63% chance of success (any number of times) means a .63 expected value (because all successes after the first have a value of 0).

Meanwhile for other things, such as polyamorous partners, 2 partners is meaningfully better than one, so the expected value truly is 1, because you will get one partner on average. (Though this assumes 2 partners is twice as good as one, we can complicate this even more if we assume that 2 partners is better, but not twice as good)

It's a great idea. I ended up bolding the one line that states my conclusion to make it easier to spot.