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I adamantly agreed with you. Then I made a simple spreadsheet that proves us wrong: http://cl.ly/kuQE http://cl.ly/kuQE
by sxv 9y ago
I adamantly agreed with you. Then I made a simple spreadsheet that proves us wrong: http://cl.ly/kuQE http://cl.ly/kuQE
- astrocat 9y agoAh, but see... you're counting (B2,B2) as one item because "order doesn't matter", but then counting (G2,B2) and (B2,G2) independently. If (G2,B2) is different than (B2,G2), then (B2,B'2) is distinct from (B'2,B2).
- sxv 9y agoThink of the x-axis as the first child and the y-axis as the second child. One in fourteen chance of choosing a column and one in fourteen chance to choose a row. I fail to see how there could be any additional outcomes or that any square has a greater chance of occurring than another..