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seen Dec 12 at 5:07

I'm a first year computer science graduate student. My research interests are mathematical logic, complexity theory, algorithms, spectral graph theory, lower bounds, computational economics, algorithmic game theory, and probability.


Dec
9
awarded  Caucus
Nov
18
answered Proving an algorithm returns a random bit
Sep
25
awarded  Nice Question
Sep
24
awarded  Autobiographer
Sep
6
awarded  Popular Question
Sep
5
asked Linear programming with an independent variable
Jul
13
awarded  Yearling
Jul
12
awarded  Good Question
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
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awarded  Enthusiast
May
16
answered Statistics Review
May
15
comment Statistics Review
Not quite, no - I'll post an answer. Thanks for sharing your thoughts.
May
15
comment Statistics Review
Can you post what you've tried so far? (1) is pretty straightforward.
May
9
comment Probability of having her purse
@Winther: that's not right at all, see below. That gives probabilities of $\frac{4}{16}, \frac{3}{16}, \frac{9}{16}$ for stores 1, 2, 3, respectively.
May
9
comment Probability of having her purse
Yes, that's exactly right -- does that clarify things?
May
8
answered Probability of having her purse
Apr
20
comment Basic questions about descriptive complexity
Most of my questions were answered by these gentle slides: msor.victoria.ac.nz/foswiki/pub/Events/ALC2011/LectureSlides/…
Apr
20
asked Basic questions about descriptive complexity
Apr
6
comment Pure vs mixed strategy Nash Equilibria
@BalazsRau : Let's say you're playing a game where the players can pick between $n$ different strategies. In general, there might be mixed equilibria that only mix over $k$ of those $n$ strategies. In this case, there will be some linear combination of my $k$ strategies that will make you like some $k$ of your strategies equally well (and the other $n-k$ strategies less than that). In the pure case ($k=1$), this statement is uninteresting: it just says that you like $1$ of your strategies at least as well as all the others, which is obvious.