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seen Aug 4 at 20:59

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.


Jul
13
awarded  Yearling
Jul
12
awarded  Good Question
Jul
2
awarded  Curious
Jul
2
awarded  Inquisitive
Jun
10
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.
Apr
6
comment Pure vs mixed strategy Nash Equilibria
This is off on a bit of a tangent from your original post, though =) the answer to your original question is "a pure strategy is just a mixed strategy over exactly 1 strategy."
Apr
6
comment Pure vs mixed strategy Nash Equilibria
I agree with your NE analysis. The "equally happy with anything" bit of my post referred specifically to mixed equilibria, sorry if that was unclear. It turns out that if $A$ plays a mixed-strategy equilibrium where he has nonzero chance of playing $k$ different strategies, then there is some set of $k$ strategies for $B$ that are all equally good (and therefore $B$ can randomize among them however he likes). In the special case of $k=1$ ("pure strategy equilibrium"), that means there is (at least) $1$ strategy for $B$ that stands up to scrutiny, which is not so informative.
Apr
6
answered Pure vs mixed strategy Nash Equilibria
Mar
25
comment Questions on “simple-connectedness-like” property
One problem with your definition: some disconnected sets will not be "non-simply connected" (because, for example, there do not exist open connected subsets $U, V \subset X$ eith $U \cup V = X$) and therefore it will be defined as "simply connected." I doubt that's what you want.
Mar
25
comment The next number in the series: $1,3,10,33,109\dots$
Or rather, it's $t_n = 3t_{n-1} + t_{n-2}$.
Mar
25
comment The next number in the series: $1,3,10,33,109\dots$
Yup, I think that checks out. No problem!