36 world states: let's do the math.
Ok, first of all this thread doesn't intent to analyze deeply the choices we will have to take in TW3, but just how many major decissions will be and, ok, and maybe the general theme in each one of these. We have been told that TW3 will end with 36 world states. So without considering that some of the choices could avoid take others (this makes a bit difficult the calculations, but it is discussed after) let's do the math.
First approach: five binary (2 possible outcomes) decisions would mean 32 world states (2x2x2x2x2) while six differents decisions will results in 64 ( way too much).
Maybe in some of the choices the possible outcomes will be more than two. 2 binary decissions plus 2 more with three possibilities do the math nicely (2x2x3x3=36) . The same with 4x3x3 (but that will mean just 3 major choices in the game; but a lot of options in each one!). Choices with more than 4 results seems unlikely to me.
Ok, so summing up, we have a maximum number of 5 (being 4 more likely) major decisions during the game, some of them tertiary ones.
Now, about which decisions those will be, let's move to the spoiler tag (everything speculative, but just in case)
Anyway, these calculations are just for fun and very basic, and just minor tweaks will result in much more complex scenarios. So, what do you think? Am I forgetting any major conflict you think we should influence on? Do you think this is a decision tree too complex? too easy?
Ok, first of all this thread doesn't intent to analyze deeply the choices we will have to take in TW3, but just how many major decissions will be and, ok, and maybe the general theme in each one of these. We have been told that TW3 will end with 36 world states. So without considering that some of the choices could avoid take others (this makes a bit difficult the calculations, but it is discussed after) let's do the math.
First approach: five binary (2 possible outcomes) decisions would mean 32 world states (2x2x2x2x2) while six differents decisions will results in 64 ( way too much).
Maybe in some of the choices the possible outcomes will be more than two. 2 binary decissions plus 2 more with three possibilities do the math nicely (2x2x3x3=36) . The same with 4x3x3 (but that will mean just 3 major choices in the game; but a lot of options in each one!). Choices with more than 4 results seems unlikely to me.
Ok, so summing up, we have a maximum number of 5 (being 4 more likely) major decisions during the game, some of them tertiary ones.
Now, about which decisions those will be, let's move to the spoiler tag (everything speculative, but just in case)
First of all, we know we have 3 different endigs. That's our first tertiary choice: 1 decission, 3 outcomes. The nature of this election will most likely recognize the main conflict in the story, that appears to be Ciri and the Wild Hunt. So, in the 3x3x2x2 scheme that will let us 3 more choices (1 tertiary, 2 binary). Those could be related with the political situation in the North (sort of: Nilfgaard wins vs North wins), the sentimental conflict (Yenn vs Triss) and another one (the Free Pontar State? the Lodge Status? the "Witcher order"?) There no much more room, unless some choices negates other: one outcome in one particular choice can negate the possibility to take another one. The easiest example I can came with will be the death of Geralt. If Geralt dies in one of the major endings, the sentimental choice could be unavailable. In that case, the number of total choices will be bigger (possibly one more: 5 or 6 choices). For example, I can imagine this chart (major ending in bold characters)
1. Ciri is alive and free and Geralt is dead. North wins vs Nilfgaard wins. Free Pontar State vs Lodge of Sorceresses. Other binary decision = 8 outcomes.
2. Ciri is ruler of the North and Geralt is alive. Free Pontar State vs New Witcher "Order" vs Lodge of Sorceresses. Triss vs Yenn. Other binary decision. = 12 outcomes
3. Ciri is with the Wild Hunt and Geralt is alive. North wins vs Nilfgaard wins. Free Pontar State vs New Witcher "Order". Bad Ciri vs Good Ciri. Triss vs Yenn = 16 outcomes.
8+12+16=36.
Note how a dead Geralt prevents not only the Triss vs Yenn choice, but also de refoundation of the "Witcher Order". Or how a Queen Ciri signing peace between Nilfgaard and the Northern Kingdoms. Or a Ciri in the Wild Hunt negates the possibility that the Lodge of Sorceresees reunites, but add another possibility...
1. Ciri is alive and free and Geralt is dead. North wins vs Nilfgaard wins. Free Pontar State vs Lodge of Sorceresses. Other binary decision = 8 outcomes.
2. Ciri is ruler of the North and Geralt is alive. Free Pontar State vs New Witcher "Order" vs Lodge of Sorceresses. Triss vs Yenn. Other binary decision. = 12 outcomes
3. Ciri is with the Wild Hunt and Geralt is alive. North wins vs Nilfgaard wins. Free Pontar State vs New Witcher "Order". Bad Ciri vs Good Ciri. Triss vs Yenn = 16 outcomes.
8+12+16=36.
Note how a dead Geralt prevents not only the Triss vs Yenn choice, but also de refoundation of the "Witcher Order". Or how a Queen Ciri signing peace between Nilfgaard and the Northern Kingdoms. Or a Ciri in the Wild Hunt negates the possibility that the Lodge of Sorceresees reunites, but add another possibility...
Anyway, these calculations are just for fun and very basic, and just minor tweaks will result in much more complex scenarios. So, what do you think? Am I forgetting any major conflict you think we should influence on? Do you think this is a decision tree too complex? too easy?


