Ben Alpert
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 Jan 22 comment a question related to two competing patterns in coin tossing @PEV: So if that's a fair coin, you're saying the probability is $1/32$. Of course, that's the probability that a random flip sequence of length 5 is equal to HTHTH but doesn't seem to answer whether HTHTH or HTHH is more likely to appear first. Jan 22 comment a question related to two competing patterns in coin tossing Sorry, but does this actually answer either of the part of the problem? I'm (not very familiar with Markov chains and am) confused about what the mean time tells you. (And I'm assuming $q = 1 - p$; is that right?) Jan 22 comment a question related to two competing patterns in coin tossing I'm also interested to know how to solve this problem; I've seen variations a few times on contests but never knew how to solve it. Dec 11 awarded Quorum Nov 16 awarded Enlightened Nov 16 awarded Nice Answer Oct 1 awarded Enlightened Sep 16 comment Finding $\lim\limits_{n \to \infty} \sum\limits_{k=0}^n { n \choose k}^{-1}$ Why is the text on this so big? Aug 19 awarded Good Answer Aug 3 comment Solve an equation with linear and exponential functions, $x=10^{x/10}$ Besides, that's not a complete answer. x ≈ 1.37129 works as well. Jul 30 comment Which average to use? (RMS vs. AM vs. GM vs. HM) Isaac: Thanks, fixed. Jul 30 comment Which average to use? (RMS vs. AM vs. GM vs. HM) Qiaochu Yuan: Can you elaborate a little more? I'm not sure you mean by rotational invariance. How are you rotating the values? Jul 30 accepted Which average to use? (RMS vs. AM vs. GM vs. HM) Jul 30 revised Which average to use? (RMS vs. AM vs. GM vs. HM) edited body Jul 30 comment Which average to use? (RMS vs. AM vs. GM vs. HM) (Couldn't decide whether or not this is appropriate for this site; vote to close at will.) Jul 30 asked Which average to use? (RMS vs. AM vs. GM vs. HM) Jul 30 comment Proof that $n^3+2n$ is divisible by 3 Indeed. If you do enough problems like this, then at least tricks like this become second nature. Jul 30 revised Proof that $n^3+2n$ is divisible by 3 edited body Jul 30 comment Counting how many hands of cards use all four suits This is a fantastic solution that I hadn't thought of. Thanks! Jul 30 accepted Counting how many hands of cards use all four suits