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Dec
21
revised Rolling dice problems
added 18 characters in body
Dec
21
revised Rolling dice problems
added 802 characters in body; edited title
Dec
17
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.
Dec
16
asked Rolling dice problems
Nov
26
awarded  Supporter
Nov
26
accepted Proof for a binomial equation
Nov
26
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.
Nov
26
comment Proof for a binomial equation
:Yes, that is it! Thank you!
Nov
26
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.
Nov
26
awarded  Editor
Nov
26
revised Proof for a binomial equation
added 273 characters in body
Nov
26
asked Proof for a binomial equation
Nov
25
awarded  Scholar
Nov
25
comment Light bulb color problem
@Henry: Thank you!
Nov
25
accepted Light bulb color problem
Nov
25
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?
Nov
25
awarded  Student
Nov
25
asked Light bulb color problem