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Sep 25 at 12:43 comment added Will Sawin @IlyaBogdanov Clever trick! That shrinks the gap back from $1/p(p-1)$ to $1/p^2$.
Sep 25 at 7:07 comment added Ilya Bogdanov @Connor Does it make any difference? If there is an uncovered point, you can move it to $0$ by a linear shift, thus making all $s)i$ and $t_i$ to be nonzero.
Sep 20 at 21:27 comment added Will Sawin @Connor Sure, this slightly alters the bound.
Sep 20 at 21:27 history edited Will Sawin CC BY-SA 4.0
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Sep 20 at 14:49 comment added Connor Those $s_i,t_i$ are non-zero, which makes the probability not $(1-p^{-1})^n$.
Sep 19 at 15:11 history answered Will Sawin CC BY-SA 4.0