I agree and I suspect companions are carrying a lot of the weight for this calculation.
Hypothetically, if there’s 10 companions with 10 individual endings each you’d get 100 endings right there. Add in 10 main endings and you get 1000, add in 4 major side quests and 4 variations each and you’re at 16,000 ending variations.
I’m not entirely sure why, but I don’t think that adds up. 10 companions with 10 different endings is a total 100 endings, however there are apparently 1.7x1013 combinations if you were to pick any 10. I don’t entirely know if I did the math right there.
So you don’t have 1000 endings from the 100 total companion endings and 10 main endings, you have 110 total endings.
Either 17,000 is a figure from the various combinations (compare that to Fallout 3’s purptorted 300 endings) or there are 17000 total ending “slides.” The former is much more likely.
Unless I’m getting the math wrong myself, for any “pick 1” combination set like this we’re dealing with just multiplying the combination sets together. Technically we’re multiplying by the factorial of the sample size, but 1!=1.
We’re not picking any 10 from within the subset of 100; you cannot pick both ending 1 and ending 4 from companion A and then no ending at all for companion C. I’m assuming each individual sub-ending is mutually exclusive with the rest of its sample space. That difference of assumptions is what led to your 1.7x1013 combinations.
You’re probably more right than me here, but I’m just not following. Does it matter whether or not to pick 1 from 10 sets of 10 or to pick 10 from a single set of 100? We don’t care about what set each individual item came from, just that it’s unique and the number of possible combinations.
Edit: Probably just best to dismiss this. I clearly have no idea what I’m talking about.
