# Eric

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bio website location age member for 2 years seen Mar 19 '12 at 9:38 profile views 10

# 18 Actions

 Dec21 revised Rolling dice problems added 18 characters in body Dec21 revised Rolling dice problems added 802 characters in body; edited title Dec17 comment Rolling dice problems @whuber:That was helpful! Thank you! In fact $w(i) = \sum_{j=0}^{i-1} (-1)^j(i-j)^n{i \choose j}=S(n,i)i!$. And we can proceed from there to write out explicit expressions for subsequent computations. Dec16 asked Rolling dice problems Nov26 awarded Supporter Nov26 accepted Proof for a binomial equation Nov26 comment Light bulb color problem :That's helpful. It's the same formula as here stats.stackexchange.com/questions/18975/… , only in slight different notations. But to complete this rigorous proof, you still have to show $n=2k$ and $n=2k-1$ entail same correct probability. Nov26 comment Proof for a binomial equation :Yes, that is it! Thank you! Nov26 comment Proof for a binomial equation @varty:$C_{2k}^k$ is the situation when there're equal numbers of flashes of red and blue (where each color has k flashes). Left hand side is the probability you guess correctly under best strategy when you decide to observe $2k-1$ times; RHS is the probability you when you decide to observe $2k$ times. Nov26 awarded Editor Nov26 revised Proof for a binomial equation added 273 characters in body Nov26 asked Proof for a binomial equation Nov25 awarded Scholar Nov25 comment Light bulb color problem @Henry: Thank you! Nov25 accepted Light bulb color problem Nov25 comment Light bulb color problem @Henry:"If there is a majority of one colour after 2k flashes then the majority must be even and at least 2" I may have misunderstood your point, but why must it be even? For example if k=10 and red is observed 11 times and blue 9 times, where does evenness come from? Nov25 awarded Student Nov25 asked Light bulb color problem