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Sep 26, 2013 at 15:48 comment added Jan-Christoph Schlage-Puchta Right. I was assuming that f at g(1) and g at f(1) are independent, which they are not. However, taking these probabilities to be $<2/n$ still suffices to give an exponential bound.
Sep 25, 2013 at 11:39 comment added Brendan McKay Unfortunately it is not true that the probability that $f(g(1))=g(f(1))$ is $1/n$ (at least, not exactly). Both by hand and by computer I find 10 out of 16 pairs for $n=2$ have that property. See my answer.
Sep 25, 2013 at 9:38 history answered Jan-Christoph Schlage-Puchta CC BY-SA 3.0