Over 500 possible endings for Fallout 3?

[Zeus]

Bethesda's Pete Hines has suggested that in theory there are over 500 different variations that you could experience in Fallout 3. He noted "the) number of variations is over 500, but that's of all the different permutations you can have of each different variable and what you did at each point."

<center><img src="http://www.maxconsole.net/content_img/fall3111.jpg"></center></a>


Bethesda's Pete Hines speaks with TVG about the upcoming Fallout 3. It makes for a quite interesting read, particularly regarding the factors that affect the end game, where Hines shares with us that they are nudging past 500 different variations you could experience in the 20-100 hours of gamplay, "(the) number of variations is over 500, but that's of all the different permutations you can have of each different variable and what you did at each point. I don't know how many of those there are (I haven't gone back to look), but it takes into account any number of things you've done from early in the game, right up to when the game ends - it takes into account all of that stuff," says HInes.


News Source: <A href="http://www.gamerchip.com/over-500-different-endings-for-fallout-3/" target="_blank">Gamerchip</a>

[RDJ134]

Already played a early build of the game last week on a press presentation in Amsterdam, great game with a lot of potential.

[phantomdjp]

500 [...] that's of all the different permutations you can have of each different variable and what you did at each point.

That mean nothing...

10 quests, 3 ways to solve each quest (let's say "good", "bad", "neutral" solution) = 59049 permutations ...

[Xenontc]

Wasn't this basically said about Mass Effect? Basically, no matter what you picked in Mass Effect, the overall outcome was the same...

[sigma8]

That mean nothing...

10 quests, 3 ways to solve each quest (let's say "good", "bad", "neutral" solution) = 59049 permutations ...

Wouldn't that be 30^3 = 27,000 possibilities?

[duncans_pumpkin]

3^10= 59049

[dee]

Somehow I think you have to do permutation calculations or combinations to get the total number of outcomes :confused:

[Xenontc]

LMFAO. The formula for permutation isn't n^k. Holy ****, go back to 9th grade math.

The formula is (N!)/(N-K)!

N being the number of sets, K being the number of objects.

10 Quests with 3 options = 120 options without repetition.