Difference between revisions of "Talk:216: Romantic Drama Equation"
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This can't be right, even at 50/50, the number of gay pairings far outnumbers the number of straight pairings.[[Special:Contributions/80.235.105.134|80.235.105.134]] 20:10, 28 February 2013 (UTC) <small>Moved from article page</small> | This can't be right, even at 50/50, the number of gay pairings far outnumbers the number of straight pairings.[[Special:Contributions/80.235.105.134|80.235.105.134]] 20:10, 28 February 2013 (UTC) <small>Moved from article page</small> | ||
:Not quite. Consider a cast of 4 with 2 male (A, B) and 2 female (C, D). Possible gay pairings - 2 (A-B and C-D). Possible straight pairings - 4 (A-C, A-D, B-C, B-D){{unsigned ip|122.200.61.203}} | :Not quite. Consider a cast of 4 with 2 male (A, B) and 2 female (C, D). Possible gay pairings - 2 (A-B and C-D). Possible straight pairings - 4 (A-C, A-D, B-C, B-D){{unsigned ip|122.200.61.203}} | ||
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| + | :He says for large casts. For 2000 cast members, with 1000 of each gender, the gay couplings comes out at 999,000 and straight at 1,000,000. Presumably this is the small cross over the diagram alludes to. If you substitute x = n/2 into the equations, then you get (n^2-2n)/4 for the gay combinations and n^2/4 for the straight combinations, so for gender balanced cast size of n, the straight combinations outnumber the gay by n/2 | ||
Revision as of 13:08, 15 November 2013
This can't be right, even at 50/50, the number of gay pairings far outnumbers the number of straight pairings.80.235.105.134 20:10, 28 February 2013 (UTC) Moved from article page
- Not quite. Consider a cast of 4 with 2 male (A, B) and 2 female (C, D). Possible gay pairings - 2 (A-B and C-D). Possible straight pairings - 4 (A-C, A-D, B-C, B-D) 122.200.61.203 (talk) (please sign your comments with ~~~~)
- He says for large casts. For 2000 cast members, with 1000 of each gender, the gay couplings comes out at 999,000 and straight at 1,000,000. Presumably this is the small cross over the diagram alludes to. If you substitute x = n/2 into the equations, then you get (n^2-2n)/4 for the gay combinations and n^2/4 for the straight combinations, so for gender balanced cast size of n, the straight combinations outnumber the gay by n/2
