Find and solve past AP exams.

Suppose S denotes the two S's stuck together. The problem with your reasoning is that you count something like
CUCSSE
and
CUCSSE
as different, when in reality both are the permutation
CUCSSSE.
So you've overcounted. Figure out by how much (there's a nice characterization for the strings you've counted twice) and subtract that from 6! / (2! * 2!) to get your answer.

There are 7! / 3! 2! = 7*6*5*2 = 420 permutations of SUCCESS. Of those, the combinations with no 2 S's adjacent have S's in position:
135
136
137
146
147
157
246
247
257
357
That's 10 ways to position the S's; each one gives rise to 4!/2! = 12 permutations of the remaining letters: 120 total permutations discounted.

That leaves 300 with at least 2 of the 3 S's adjacent.

There are typically lots of ways to get the right answer to questions like this. I'm sure there are more elegant ways than the ones I posted, but it was the first way I thought to do the problem.

Yeah, in my earlier 'answer' I included an extra 2! in the denominator which was the problem. I should have said:
There are 6!/2! ways of permuting the letters in SUCCES, but this overcounts the times S and S are next to each other - in other words, this doublecounts the permutations where SSS appears together. To count these treat SSS as one block; then you're trying to count the number of permutations of SUCCE, and there are 5!/2! = 60 of them. So the final answer is
6!/2! - 60.

Just out of curiosity, Wyman what is your background?
I would guess math grad-student/PhD, but I've meet so many random science professors who are stunningly good at math.

PhD (math) from Michigan, graduated this summer.

In a)

The payoff for p1 is v1-b2 = {b2 = v2 because we assume that p2 bids his valuation v2} = v1-v2 if b1 = v1 > v2 and zero otherwise. 3 cases:

v1 = v2
Here the payoff for p1 is v1-v2 = {v1 = v2} = v1-v1 = 0 no matter what. Assume p1 now plays another strategy x1. What happens? Well either x1 < v2 and the payoff is zero because he doesn't obtain the object or x1 > v2 and his payoff is still v1-v2 = v1-v1 = 0 because we still pay the second highest bid, which is v2 = v1.

v1 > v2
Here the payoff for p1 is v1-v2 > 0 if p1 wins the bid and zero otherwise. What happens if p1 chooses another strategy x1? Well either x1 < v2 and the payoff is zero because we he doesn't obtain the object or x1 > v2 and his payoff is still v1-v2 because we still pay the second highest bid, which is v2.

v1 < v2
Here the payoff for p1 is v1-v2 < 0 if p1 wins the bid and zero otherwise. What happens if p1 chooses another strategy x1? Well either x1 < v2 and the payoff is zero because we he doesn't obtain the object or x1 > v2 and his payoff is still v1-v2 because we still pay the second highest bid, which is v2.

What am I doing wrong?

You planning on staying in academia?

I took a job at a very small quantitative hedge fund right out of school.

tsk tsk, America is wasting its finest

You just need to notice that the bolded is what happens when P1 bids v1 and you're done

ty =)

New problem:



b1: A strategy set for firm 1 is (x,p1) and (y,p2) for firm 2. But, how do I define the firms' payoff functions?

Let m = (x+y)/2.

Case 1 - x =< y
Then all the customers from 0 to m will buy from firm 1 and the rest, i.e. from m to 1 will buy from firm 2. The profit for firm 1 is then m*p1 and (1-m)*p2 for firm 2.

Case 2 - y =< x
Just reverse case 1.

Hence we have that the profit function for firm 1 is

f(x,y) = p1*(x+y)/2 if x <= y
f(x,y) = p2*(1-(x+y)/2) if y <= x

and the profit function for firm 2 is

g(x,y) = p1*(1-(x+y)/2) if x <= y
g(x,y) = p2*(x+y)/2 if y <= x

However, in this model I do not take into account that p1 might not equal p2. And if for example p1 > p2 then firm 1 will lose some of its customers to firm 2 even though firm 1 is closest to the consumer.

Looks good to me. Nice word choice BTW.

Do you use anything that you learned in graduate school? What I mean is that obviously you learned a great deal in graduate school, but most importantly, you learned to think for yourself, aka research. But has any of the information that you learned helped, aka, topological concepts, etc?

You may be overestimating how much I learned in grad school and overestimating the utility of topology in the real world. I'd like to think that I have been able to think for myself for a long time.

In any case, my day-to-day consists of statistics, information theory, inference problems / machine learning, data problems, software engineering, programming, etc.

Networking was what got me this job; our head of research had hired me at a different company for some summer work several years earlier. It didn't hurt, though, that my thesis displayed pretty much all of the qualities that the company was looking for: we considered a difficult problem, broke it into simple pieces, figured out what the theorems should be using a lot of computation, proved the theorems/lemmas using elementary techniques, and combined them to get some much stronger results than people who had been using high-powered machinery had been able to get for the last 100 years.

Re: my choice to leave academia. I miss teaching. I don't miss the paycheck. I certainly don't apologize for my choice, since this lifestyle will allow me to provide better (in terms of money and time, believe it or not) for my family, choose my location to be near my less immediate family, and probably impact more people (or at least make a more meaningful impact) through philanthropy than I would have as a teacher/mentor/scientist in academia. Additionally, my company is made up of *only* scientists (math/cs/physics -- no finance), so my day-to-day interactions are still with very smart, very ethical people, and I am absolutely loving my job. I don't deal with the bureaucratic BS that many of my professor-friends do, and I still solve (very) hard problems. Oh, and when I/we do solve a hard problem, people really care .

OK, you've piqued my curiosity. Can you give a description of the result? Feel free to use terms like "manifold" and "fundamental group."

I have no doubt that you were a great thinker before graduate school. But, I guess my main question is "Do you find any information or concepts that you learned in graduate school useful in the applied setting?" I feel that this may be a road that I go down one day, and was curious about your experience. To my knowledge, your Algebraic research probably didn't enlighten you too terribly much to most of your day to day topics.

Take my case, I highly doubt I'll use limiting formulas for Modified Bessel Functions if I do not stay in academia. Nor will I ever need to find a partition function.

lol

Basically:
We know how to factor polynomials (say, over C) under ordinary multiplication. Every polynomial factors uniquely into linears over C (Fundamental Thm of Algebra).

Instead of multiplication, let's consider composition as the operation we'd like to "factor" with respect to. So we'll break F into

F = f1 o f2 o f3 o ... o fn

making the fi's as small (degree) as possible. There are some obvious reasons that this factorization is nonunique:

F = f1 o f2
= f1 o (x+1) o (x-1) o f2, for example, and

x^6 = x^3 o x^2
= x^2 o x^3.

But there are some not-so-obvious ones as well:

x^6 + 2x^4 + x^2
= (x^3 - x^2) o (x^2 + 1)
= (x^3 + 2x^2 + x) o x^2

In any case, there's some serious work that's been done over the last 100 or so years (starting with Ritt, Fatou, and Julia, who were asking questions about complex dynamics) classifying the extent of this "non-uniqueness of factorization" under composition for polynomials in C[x] (and later replacing C by all algebraically closed fields, then fields of characteristic zero, and more recently finite fields).

The thing is, this question has nothing to do with fields; all you need to talk about polynomial composition is a ring (loosely, a ring is a place where you can add and multiply: think the integers. A field is a ring where you can divide by anything except 0: think the real numbers. You can't divide 1 by 2 in the integers. You'd get 1/2, which is not an integer. So the integers aren't a field).

So we provided a nice characterization of how things work over rings, basically. Anytime we couldn't prove something stronger, we pushed back with a bunch of counterexamples. Sometimes the truth was difficult to find, but with a lot of computation (typically looking for (counter-)examples), we were able to figure things out.

The whole thesis (which contains most, but not all, of our results) is available here (pdf) for the very adventurous. We're still in the process of putting together a paper with all of this and hope that it'll be in a pretty good journal when the dust settles.

My advice is to learn everything you possibly can, even if you think it's useless. There are so many weird connections in mathematics, and often impossible problems can be translated to another language where they become essentially tautologies.

Day-to-day, I don't use Bessel functions or partition functions, that's right. But it's not crazy to think I might someday end up with some differential equation from my model, the solution of which is Bessel. And actually I just read a paper the other day about counting n-cycles in graphs that relied on immanental polynomials, and these are indexed by partitions (really irreducible characters, but anyway...). So yeah, weird connections. You never know what you might need to know.

First, a (pure) strategy in this game is not just a pair of numbers (x,p1), because the price is chosen after observing your opponent's choice y, hence it is a pair (x,p1(.)), where x is an element of X and p1 is a function X -> R (when your opponent has taken location y, you choose the price p1(y) ). This is important for the questions 2 and 3.

For the payoff function, you have done the case p1=p2, the general case should be similar. Start by assuming x<y, and try to find z in [0,1] where the consumer is indifferent to buying from 1 or 2. Note that you won't necessarily have x <= z <= y.

Thanks, I think I solved it:

Assume that a consumer is living M miles from x (hence 1-M miles from y). The total cost of visiting x is then p1+tao*M. The total cost of visiting y is p2+tao(1-M) and putting these equal yields

M = 0.5 - (p1-p2)/(2*tao).

This means that all consumers on M will be be indifferent from choosing x or y. All consumers to the left of M will choose x and all consumers to the right of M will choose y.

That is not correct.

Actually having done some quick calculations, you will need to distinguish some cases :

First see what happens when p1 > p2 + tau (y-x). (see what happens for a consumer located exactly at x, then on the left of x).
Similar thing occurs for p2 > p1 + tau(y-x).

Then look at the case p1 = p2 + tau(y-x), and p2=p1 + tau(y-x).

Then the last case is | p1 - p2 | < tau |y-x|. This is actually the only case where you will find an indifference point z where all the consumers on the left of z will choose x, and on the right will choose y.

Hi all.

I have a statistics problem that I'm currently stuck on..

15 observations are taken ind. from a chi-square dist. with 4-DOG.

I want to find the probability that at most 3 of the 15 observations exceed 7.779.

When I look at a chi-square table under 4-DOG and 7.779 I see .900.

Suppose I'm trying to find out what the probability that at most 1 of the 15 observations exceed 7.779. Then P( X > 7.779) = 1 - P( X <= 7.779) = .100 correct?

How do I do it if I'm trying to find the probability that at most 3 of the 15 exceed 7.779. That is where I'm having difficulties.

Thanks in advance.