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Mar 19, 2016 at 22:41 comment added Geoff Robinson @IlyaBogdanov :OK, thanks, I see what is going on now.
Mar 17, 2016 at 12:25 comment added Ilya Bogdanov @GeoffRobinson: I do not fix the cardinalities of $A$ abd $B$; I simply put every element to $A$ with probability $1/2$, the same for $B$ (all $2n$ choices are independent).
Mar 17, 2016 at 0:19 vote accept Marius Tarnauceanu
Mar 16, 2016 at 16:21 comment added Derek Holt Right - that's easier than I expected!
Mar 16, 2016 at 16:09 history answered Ilya Bogdanov CC BY-SA 3.0