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Jul 30, 2020 at 22:30 history edited Rodrigo de Azevedo CC BY-SA 4.0
added 5 characters in body; edited tags
Mar 15, 2019 at 5:04 answer added Iosif Pinelis timeline score: 4
Mar 15, 2019 at 2:13 comment added Anthony Quas Imagine $p=q=1/2$. Every $n$ steps, you get a chance to jump ahead by 1. After $n^2$ steps, you have had $n$ chances, so you have jumped ahead by $n/2\pm \sqrt(n/4)$. It takes around $n^3$ steps before you would expect to mix.
Mar 15, 2019 at 0:28 history asked Hao Yuan CC BY-SA 4.0