I’m sure that, while 17k may be mathematically and technically true, they’re going to be virtually indistinguishable. I’m pretty hyped for the game but I also don’t care for the obviously clickbait claim.
It’s nice to know the ending will touch upon your actions thoroughly, but I suspect it will be a lot like the The Witcher 3 technically having 40-something “endings” (but really it’s just 3 major outcomes with some variations on the details).
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.
Since it’s variations of the combined ending, each permutation would count as unique. Meaning that 10 companions with 10 endings each would total 10.000.000.000 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.
I’m sure that, while 17k may be mathematically and technically true, they’re going to be virtually indistinguishable. I’m pretty hyped for the game but I also don’t care for the obviously clickbait claim.
In this ending Shadowheart has a ponytail…
It’s nice to know the ending will touch upon your actions thoroughly, but I suspect it will be a lot like the The Witcher 3 technically having 40-something “endings” (but really it’s just 3 major outcomes with some variations on the details).
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.
Since it’s variations of the combined ending, each permutation would count as unique. Meaning that 10 companions with 10 endings each would total 10.000.000.000 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.
Ten quintillion
planetsendings