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Feb 20, 2011 at 17:30 answer added Konstantinos Panagiotou timeline score: 6
Feb 18, 2011 at 14:47 comment added Pradipta That’s great to know, Mike.
Feb 17, 2011 at 23:21 comment added Mike Spivey @Pradipta: Thank you for asking this question! This question and the answers provided gave me the ideas I needed to answer one of my own MO questions from a while back: mathoverflow.net/questions/46777.
Feb 17, 2011 at 16:47 vote accept Pradipta
Feb 17, 2011 at 16:02 comment added Pradipta Yes, I tried that. Didn't get anywhere.
Feb 17, 2011 at 8:43 comment added Gerhard Paseman Looking at f(n+1) - f(n) gives 2^(-n-1) times sum of terms of the form (n choose k)(f(k+1) - f(k)), when n > 0. Have you investigated this path? Gerhard "Ask Me About System Design" Paseman, 2011.02.17
Feb 17, 2011 at 0:56 comment added Douglas Zare The recurrence from mathoverflow.net/questions/11255/… is similar. That recurrence was the probability of a tie.
Feb 16, 2011 at 21:42 answer added Mike Spivey timeline score: 5
Feb 16, 2011 at 16:30 answer added Louigi Addario-Berry timeline score: 7
Feb 16, 2011 at 16:01 answer added Or Zuk timeline score: 3
Feb 16, 2011 at 15:38 answer added Did timeline score: 11
Feb 16, 2011 at 15:33 comment added Pradipta yeah. with probability $\frac{1}{2^n}$ no coin succeeds in round 1, thus you are back where you started.
Feb 16, 2011 at 15:30 comment added Emil Jeřábek Are you sure $f(n)$ appears on both sides of the first equation?
Feb 16, 2011 at 15:24 comment added Pradipta I think its working now.
Feb 16, 2011 at 15:20 history edited Pradipta CC BY-SA 2.5
broken formula fixed; added 2 characters in body
Feb 16, 2011 at 15:20 history edited Igor Rivin CC BY-SA 2.5
changed the math, so it is slightly more readable, but still not great.
Feb 16, 2011 at 15:18 comment added Pradipta something's wrong. I'll fix this.
Feb 16, 2011 at 15:17 comment added Igor Rivin I cannot parse the formula.
Feb 16, 2011 at 15:16 history asked Pradipta CC BY-SA 2.5