Rehwyn

Member
Apr 10, 2024
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Wow, madness indeed.

Some quick math for the fun of it.

Assuming that:
  • No scene occurs if you're not on any of their paths
  • No scene occurs if just in a solo path with one of the LIs
  • A LI has two possibilities: excluded or included
The possible combinations is 2^n-(n+1), where n is number of possible LI participants. So here, 2^4-(4+1) = 16-5 = 11.

Now let's assume each LI has three possibilities: excluded, included not pregnant, included pregnant.

Combinations becomes 3^n-(2*n+1). Or for a potential four LI scene 3^4-(8+1)= 81-9 = 72 variations.

Whew! :oops:

P.S. This is why most devs have limitations on group sex combinations and pregnancy content within them.

A hypothetical scene of up to 10 LIs that each could be excluded, included not pregnant, and included pregnant would need up to 59,028 variations.
 
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