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Dec
17
answered Question on Expected Value, discrete case
Dec
17
answered Finding density function for uniform distribution
Dec
17
comment “Let $G$ be a planar graph. Show that every pair of vertex-disjoint odd cycles in $G^c$ is connected by an edge.” Can't figure out why “odd” matters.
@ErickWong it seems to be in isolation in this case; can you see anything wrong with the proof above?
Dec
17
comment “Let $G$ be a planar graph. Show that every pair of vertex-disjoint odd cycles in $G^c$ is connected by an edge.” Can't figure out why “odd” matters.
@hbm You're right, I edited it to change slightly. Is what I have now true?
Dec
17
revised “Let $G$ be a planar graph. Show that every pair of vertex-disjoint odd cycles in $G^c$ is connected by an edge.” Can't figure out why “odd” matters.
added 7 characters in body
Dec
17
revised “Let $G$ be a planar graph. Show that every pair of vertex-disjoint odd cycles in $G^c$ is connected by an edge.” Can't figure out why “odd” matters.
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Dec
17
revised “Let $G$ be a planar graph. Show that every pair of vertex-disjoint odd cycles in $G^c$ is connected by an edge.” Can't figure out why “odd” matters.
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Dec
17
asked “Let $G$ be a planar graph. Show that every pair of vertex-disjoint odd cycles in $G^c$ is connected by an edge.” Can't figure out why “odd” matters.
Dec
17
awarded  Good Answer
Dec
16
awarded  Mortarboard
Dec
16
comment Monty hall problem extended.
To add to this answer a little bit, consider the case with $n$ doors where Monty opens each of the remaining doors except for 1 (another, different generalization of the original problem to OP's). Then, again, if you don't switch your probability of winning is $\frac{1}{n}$, and if you do switch, your probability of winning is exactly equal to the probability that you chose incorrectly at first (this is identical to the case with 3 doors). Then your probability of winning is $\frac{n-1}{n}\rightarrow 1$ as $n\to\infty$, so in this case, Monty's help is incredibly useful.
Dec
16
awarded  Nice Answer
Dec
16
answered Monty hall problem extended.
Dec
16
comment variance and generating function - probability hat problem
@user2369869 ahhh I see your confusion here. I'll be honest, I hadn't actually looked at the PDF until right now—I had assumed that you were referring to the moment generating function, which is a common tool in probability, and not just general generating functions. The moment generating function does encode information about a sequence of values: $M_X(t)=E[e^tX]=\sum_{i=0}^\infty E[X^n]\frac{t^n}{n!}$, so it is the [exponential generating function] of the sequence of moments.
Dec
16
revised How to calculate the interest amount per day
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Dec
16
comment How to calculate the interest amount per day
@CarstenSchultz whoaaa thanks
Dec
16
revised How to calculate the interest amount per day
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Dec
16
suggested approved edit on How to calculate the interest amount per day
Dec
16
answered How to calculate the interest amount per day
Dec
16
accepted Prove that if $n\geq\text{lcm}(a,b)$ and $\gcd(a,b)|n$ then $n=xa+yb$ for some integers $x,y\geq 0$