Board Thread:General Discussion/@comment-26907577-20191002153041/@comment-32182236-20191002202233

First, we have to consider that the Ball Game only contains seven colors.

...Actually, we don't.

It could be that the missing extra colors aren't a part of the game, because the monsters do not know of them, as they did not fall into the Underground, and any records earlier than the war would have been on the Surface-lost to the monsters now.

So that doesn't mean there must only be 7.

We also must consider that the red trait is related to the other six. The six combine together to make red. Or.. At least, that's what the monsters who created the Ball Game think about SOUL traits.

How is Frisk's scenario unique? I don't see any reason to call it unique.

The only reason why I made them unique in my timeline is because we KNOW when Frisk fell, so any random arrangement where Frisk didn't fall at the very end must be excluded from the possibility space.

But there's no reason to assume that Frisk's fall is unique.

So, let's try again.

What are the odds of getting exactly one pair out of 8, when there are 7 colors? What about 8 colors? 9? n?

We know there must be at least one pair if there are 7 colors and 8 results. (Or three of a kind! Though you could also say there are three pairs as well in the case of 3 of a kind.)

But we're not interested in the odds of getting at least one pair, we're interested in the odds of getting exactly one pair!

There are C(8, 2) ways to arrange the pair, 7 possible values for the pairs, and 6*5*4*3*2*1 (Or 6!.) ways to arrange the colors that are not paired to any other colors.

C(8, 2)= 8!/(2!)(6!)

Now, this gives us a total of 7*6!*8!/(2!)(6!) permutations.

Let's just cancel out the 6!'s. 2! is just 2.

7*8!/2 permutations exist which contain exactly one pair.

There are a total of 7^8 total permutations. To get the odds, we have to divide the number of possible ways to get exactly one pair, to the number of possible colors entirely.

So we get 7*8!/2(7^8), which simplifies to 8!/2(7^7).

This gives us 0.02447959609, or about a 2.448% chance.

Now let's try to generalize it.

When there's one pair, there's ALWAYS C(8, 2) ways to arrange the pair, since there are always a total of eight souls.

If there are n colors, then there are n ways to determine the value of the pair, that is, which specific color we have a pair of.

The number of ways to arrange the colors that are not paired with any other colors is equal to n-1*n-2*n-3*n-4*n-5*n-6 (because there are six colors that are not a part of the pair, the sequence is always exactly six long. Unless n is 6 or less, in which case exactly one pair is impossible.)

This can also be written as (n-1)!/(n-6)!.

As a result, we always have n*(n-1!)*8!/[2(6!)(n-6)!] permutations with exactly one pair.

...That's quite the mess. Let's do some simplifying.

6! is 720, and 8! is 40320. 2*720 is 1440, and 40320/1440 is 28.

That simplifies the equation into 28n*(n-1!)/(n-6)!

Still messy, but at least it's better.

The number of possible total permutations is always n^8.

So the total odds are equal to 28n*(n-1!)/(n-6)!*n^8).

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...Let's just hope I didn't make any mistakes in this calculation..