<t>Yes, this is why I said the proof wasn't 100% accurate. Even though you stated it without proof, you were right in saying that you can write n = k*p + m*q with m, k >= 0 as long as n >= (q-1)(p-1), but m and k are no longer multiples of n as you implied. Of course, m and k being multiples of n is...

<t>You seem to have given the answer to the different question: given an n x 1 checkerboard and as many 2 x 1 and 3 x 1 dominoes as you want, for which n can you successfully tile the checkerboard? This answer is clearly n >= 2, by obvious intuition or by your lengthy proof (which isn't even 100% ac...

<t>This works if you replace {1,2} by {1,2,3}. In fact, it will work in general for a fixed k, if you advance the cursor by d_i*k^i-1 where d_i is chosen uniformly in {1, 2, ..., k}, and you make it so the probability of choosing candidate i is k times the probability of choosing candidate i-1. This...

<t>467 is prime, so 2^467 = 2 (mod 467). By that token though, 463 is also prime, so 2^463 = 2 = 465 (mod 463).I had kind of assumed Chukchi was looking for a solution the way he worded it (so 2^n mod n = 465, not 2^n = 465 mod n). These are equivalent if n > 465. Otherwise we could have the trivial...

<t>Well, I don't know if this counts as brute force, but you can calculate all 11 of those probabilities fairly quickly as follows.Let p_k be the probability that the kth prospect buys the product. Now let P_k(n) be the probability that we get at least n orders from the first k prospects. Then P_k(0...

<t>Well, I think the point is that if X has mean m and covariance matrix C, then X can be written as dX = m dt + N dB where N N' = C (this can be done because C is positive definite), and B is a vector of independent 1-dimensional Brownian motions. Then N^-1 * (X_T - m * T) = B_T, which we should cl...

<t>Well, I have yet to see a proof that it isn't 1 in the discrete case, but if we approximate the process as Brownian motion we can proceed as follows. If we let H_n be the number of heads received after n flips and T_n be the number of tails, then B_n = T_n - H_n is a rough approximation of Browni...

<t>Yeah, those first two he's saying are equivalent for his purposes. Basically he's just looking at arbitrary trees, not necessarily binomial and not necessarily with a well-defined root node.That last one is not a tree, since it contains a cycle. In finance it's often called a binomial tree, but t...

<t>Ah, okay, so basically you just mean tree, not binomial tree. In that case, yes, the root is not necessarily distinguishable, so levels need not be preserved. But again, it's not a function of n, because for example with the two graphs you gave, the first has 2 automorphisms, while the second has...

<t>In general a graph is a set of vertices V and a map E : V x V -> {0, 1} which specifies whether two vertices have an edge between them or not. An undirected graph satisfies E(u, v) = E(v, u). A homomorphism from the graph (V, E) to the graph (W, F) preserves connectivity, so it is a map f : V -> ...

<t>How exactly do you define a binomial tree with 2 nodes? Also, not all binary trees with n nodes are equivalent and could have a different number of homomorphisms. For instance, the following two binomial trees both have 7 nodes, but the one on the left has only 2 homomorphisms while the one on th...

<t>Well yes, it leads to the solutions -phi, 1/phi, and 1. We want a solution that's between 0 and 1, so -phi is automatically out. Now admittedly we should prove that it's not 1, but assuming for now that the probability isn't 1, that only leaves 1/phi.Note that with a probability of 1, this implie...

<t>It just occurred to me that it's much easier if we cover it like: 1 3 1 3 1 3 1 ...2 4 2 4 2 4 2 ...1 3 1 3 1 3 1 ...2 4 2 4 2 4 2 ...1 3 1 3 1 3 1 ...2 4 2 4 2 4 2 ...... ... ... ... ...Then in this case, every 3x3 square contains either an equal number of 1's and 4's, 3 more 4's than 1's, or 3 ...

<t>Yes, n has to be a multiple of 2 or 3 for it to work. There might be an easier way to prove this, but I did it in the following way. Color the checkerboard like the following:1 3 1 3 1 3 1 ...2 4 2 4 2 4 2 ...3 1 3 1 3 1 3 ...4 2 4 2 4 2 4 ...1 3 1 3 1 3 1 ...2 4 2 4 2 4 2 ...... ... ... ... ...N...

<t>Just thought I'd note that we can calculate the answer to a) exactly (though as everybody has already said, it's approximately 0).Let p be the probability that we go bankrupt starting with only $1 (I don't have a pound symbol on my American keyboard...). Then what's the probability that we go ban...