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.
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.
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 purported 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.
Yes, it matters. If you’re picking 1 out of 10 each from 10 different sets, you get 100 combinations. This also limits the sample space to what is possible.
For simplicity’s sake so we can do math that we can intuitively figure out, look at it as picking one from binary choices, with three companions. So you have companions A, B, and C. With possible endings A1, A2, B1, B2, C1, C2.
If you pick 1 from A, 1 from B, and 1 from C you get 2*2*2 possible outcomes, or 8.
If you pick any 3 from the set of 6 (A1, A2, B1, B2, C1, C2) you get 6!/(3!*3!) possible outcomes, or 20.
With the former, you always get one ending for each companion. Every companion has an option selected, and every companion does not have multiple endings selected. With the latter, you might get 1 from each companion. Or you might get A1, A2, and B1 — with no endings for companion C, and two endings for companion A.
How can ending A1 “A lived happily ever after” and ending A2 “A died midway through the player’s journey, never having found happiness” both happen? They cannot. We need to use a system that limits the sample space to exactly 1 per companion, even if that option itself might be “doesn’t show up in the end slides.”
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’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…
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.
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 purported 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.
Yes, it matters. If you’re picking 1 out of 10 each from 10 different sets, you get 100 combinations. This also limits the sample space to what is possible.
For simplicity’s sake so we can do math that we can intuitively figure out, look at it as picking one from binary choices, with three companions. So you have companions A, B, and C. With possible endings A1, A2, B1, B2, C1, C2.
If you pick 1 from A, 1 from B, and 1 from C you get
2*2*2possible outcomes, or 8.If you pick any 3 from the set of 6 (A1, A2, B1, B2, C1, C2) you get
6!/(3!*3!)possible outcomes, or 20.With the former, you always get one ending for each companion. Every companion has an option selected, and every companion does not have multiple endings selected. With the latter, you might get 1 from each companion. Or you might get A1, A2, and B1 — with no endings for companion C, and two endings for companion A.
How can ending A1 “A lived happily ever after” and ending A2 “A died midway through the player’s journey, never having found happiness” both happen? They cannot. We need to use a system that limits the sample space to exactly 1 per companion, even if that option itself might be “doesn’t show up in the end slides.”
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).
Ten quintillion
planetsendings