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Bathrooms at the BBs are about as bad, actually worse in many cases. I work at one so I can vouch for this. If they're like this at respectable banks, can't expect much from a CUNY.

There are arguments on both sides. If you don't publish any stats, some will call you opaque, others will just be intrigued and seek out reviews from alumni. And if there's a forum like this where there are enough people willing to give accurate reviews, then there really is no need for stats...

But keep in mind that placement stats can, and often are, doctored. Either they are incorrect or they leave out a part of the story. You must always condition the stats on whatever information you can gather. In fact, placement stats are usually already conditional on some -- often favorable --...

I think he means Joy was going out and interviewing...?

It's tough, I know. But when one puts himself on a pedestal, one makes himself a target for inquisition, no? I didn't realize you were a public relations officer for Baruch. If I'd known there were openings I would have submitted an application.

There's a difference between advertising and crushing all dissenting opinions.

What saddens me most about Baruch (especially because I went there and the behavior gives me and others a bad name) is how everyone needs to constantly reiterate how great it is and/or put other programs down in order to lift Baruch. This also happened right after Baruch won RITC. This is not...

Advait, if I were you I'd count myself lucky for having been spared the difficulties inherent in actually attending the program.

A wise person once said, "if you bite off more than you can chew, be prepared to spit it back out." And don't forget - great things are already known, they don't need an introduction.

"We are Baruch MFE! Resistance is futile! All naysayers shall be smoked!"

Hmm, what if they looked at your older QN posts and some negativity shined through, as Lyosha said? Because if you say something they dislike, you'll often get "smoked."

Lyosha is right about the pre-MFE route, and there's a very good reason for this.

Welcome to the real world. Everyone promises transparency, very few people actually deliver it.

I believe it is included in the 35K, yes. But it used to be 20K. It almost doubled simply because of an "excellence fee" that was tacked on. And judging from the way CUNY operates, more fees and tuition hikes may be right around the corner. Promises, for example, of "guaranteed" interviews that...

Exactly, you shouldn't care what he says about your use of the word 'hyperbole'. Stick to getting your questions answered. Baruch recently started charging an "academic excellence fee" of a couple thousand a semester, so the ROI of the program isn't what it used to be. This used to be one of...

He's trying to derail you and unfortunately he's succeeding a bit. He is trying to troll you because he thinks you're trolling. Don't fall into the trap!

IMHO, I do believe there is a bias towards Baruch on QN. Also IMHO, hyperboles or not, when there are skeletons in closets, the doors should be nailed shut. But that's just IMHO.

He's just trying to derail you with a petty (and incorrect) correction.

What is your specialty, might I ask?

The Wire, yes!

I was referring to this problem, of course, not approximations of distributions in general. Just staying on topic, that's all. I know plenty of pure mathematics, but I consider myself a problem solver rather than a mathematician. And yes, I am also a quant. Anyhow, I found it ignorant of you...

It really does not take "mathematical depth" to understand your solution or to recognize why one should expand about 1, so relax, you sound like a narcissist. Speaking of being a poor sport, I believe it was you who came in here treating everyone with an air of superiority right off the bat.

Agreed. Or Breaking Bad...

You actually know some non-trivial math; I'm mildly impressed. What I was referring to, though, was not a series of approximations, which isn't hard to come by, but an exact solution. Still, nice job, you've partially redeemed your previous inane comments.

The show is good, but it's sort of one-dimensional. Each character is good or evil, but not much of a mixture of both. And they're all tragically foolish in their one-dimensionality.

Well, I can't speak for Columbia since I didn't go there. But I went to Baruch. Two things should be noted. 1) Most jobs people get from Baruch are in risk management. 2) The employment stats are a bit dressed up. Stats are presented per graduating class, not cohort. What this means is that...

I get what you're saying. However, in your outline above you said "You know that to get to the last roll, you needed to have rolled 50 or less (otherwise, why stop?)." Can you explain this? It seems incorrect to me, as it still disregards the cost of playing. Can you demonstrate that the e.v...

no. it's the average of the outcomes E-1, ..., 100

yes, but it also depends on the cost of playing, not just what you've rolled...

I see what you mean. Upon closer inspection of RRodriguez's argument, I think the misstep is where he says "to get to the last roll, you needed to have rolled 50 or less (otherwise why stop?)" This doesn't take into account the cost of all the rolls up to that point.

Yes, you've got it right, you're quite the quick study, aren't you... only a poorly endowed individual such as yourself would hide behind an alias to insult other people. The fact that you use an alias shows you're uncomfortable being yourself, and you dream of being someone else, someone...

How can I be so sure? Well, what you're describing is equivalent to the recursion equation I posted above, whose solution is 1223/14.

...The exact answer is 1223/14...

I'm sensing some envy here. My sense is you're poorly endowed and dream of being bigger and badder. But you can't have everything. Sorry about that.

Useless pure mathematicians, eh?

Yeah, I don't yet have a solution for $F(x)^{\frac{1}{2}}$. It may be that a solution doesn't exist.

You need to put your exponents between curly braces for them to come out right, if they're more than one character long: F(x)^{0.5}. The only requirements for a CDF are that it be between 0 and 1, and it go to 0 as you go to -infty and go to 1 as you go to +infty. Since powers of F(x) satisfy...

Does it have a specific name? No. Is f(X, Y) unique? That's a good question, I'm actually not sure... Here's another question... We can easily also do (F(x)^3, F(x)^4, ...) using the same approach. Can we also do (F(x)^{\frac{1}{2}})?

Why do we care about causality?

Intuitively it would seem to be 1. The title of this thread is really misleading...

If someone posts a problem here, it's usually because they've thought about it all they can, haven't been able to figure it out, and want a solution.

Why argue instead of just providing a solution? $X\sim F(x)$ means that $P(X\leq x) = F(x)$. Squaring both sides, we may write $P(X\leq x)P(Y\leq x) = F(x)^2$, where $Y\sim F(x)$. Assuming that $Y$ is independent of $X$, this means $P(X\leq x \cap Y\leq x) = F(x)^2$. So we just...

That's the trouble with over-thinking -- it renders everything wrong/right, up/down, left/right all at the same time, which then leads to a kind of madness and/or paralysis. Sometimes you just have to dive right in and go with the flow.

fabricate away!

Truly! Dan's hard work and determination are quite commendable. The students' too. Congrats all!

A bunch of genii, indeed.

You can't just discard these, as they are plausible outcomes that need to factor into the e.v.

It's the correct strategy, but you must be implementing it wrong. I wrote a quick Matlab script with 1e6 MC simulations and got an e.v. of 87.352, confirming my result above.

I don't think anyone is saying it's 87.5. What I was saying above is that it's (\frac{1223}{14} \approx 87.36). What did you get using MC? Note that the recursion we've been discussing assumes that an e.v. optimizing strategy is followed, but it does not tell us what strategy (or...

Nope. This recursion comes from conditioning on what happens on the first roll. If on the first roll you roll more than (E-1) then your expectation is really (\frac{\lfloor E \rfloor +100}{2}) (you walk away after the first roll, and you haven't paid anything).

Isn't the solution to that equation 87.25? The recursion we get for (E) if we want to maximize the e.v. is 1) (E = \frac{\lfloor E-1 \rfloor}{100}\cdot (E-1) + \large(1- \frac{\lfloor E-1 \rfloor}{100}\right)\cdot \frac{\lfloor E \rfloor +100}{2}), if ( E ) is not integral. The solution to...

Welcome back to QN, Joy!

I don't get it.

Yeah, despite popular opinion, doing math will give you a strong foundation for a later (or concurrent) study of stats. You'll be stronger than most other applied math people who often only have a flimsy understanding of mathematics.

Far from it.

It works everywhere. Simplicity is efficiency, and nature is efficient.

Most things in life are interesting. Complexity isn't a prerequisite. Occam's razor, baby.

Irony transcends culture.

A hunting we will go, a hunting we will go Heigh ho, the dairy-o, a hunting we will go A hunting we will go, a hunting we will go We'll catch a fox and put him in a box And then we'll let him go A hunting we will go, a hunting we will go Heigh ho, the dairy-o, a hunting we will go A hunting we...

It never ceases to amaze how often 'smart' people miss irony and sarcasm.

It's a different type of memory. And besides, the proportion of your typical mathematician's inability in chess that's attributable to their inability to understand other mindsets is probably significantly larger. I included lack of memory (particularly for interpersonal situations) as just one...

Uhh, what was this thread about again?

Eh, I am a quant for one of the BBs, and you know how many times I found myself exploiting the underlying math in order to make my code more efficient? In some cases computing time would have been 3x what it was had I not known the mathematical intricacies. Some of these things are subtle. And...

The Chess vs. Math analogy doesn't quite apply to Programming vs. Math for the obvious reason that chess is a game. It requires foresight, putting yourself in the mind of another person, good memorization, etc. No wonder a lot of mathematicians are not good at chess -- for one, many have poor...

DrD asked about quant roles, not developer roles. If you're just a developer, then you probably provide some kind of support to the quants, and don't always know all the mathematical details (but have some general idea of the structure). Quants though, do a bit of both. They do...

Read my above post and read the part in parentheses about what being good at 'maths' means. I've never met a really good programmer who didn't also have good quantitative reasoning skills (which is what I mean by being good at 'math'). Also, every exceptional math person I've ever met has...

With anything quant, yes of course.

Hmm, don't listen to the above posts. Just because you'll be coding most of the time, does not mean that quant work isn't math. If you don't understand the mathematics of the thing you're coding, or if you don't know combinatorics, optimization, etc., you can't expect to be a good coder. A...

"It's sooo basic!"

Sorry, I had written it fast and made a couple of typos; I clarified things a bit. For the covariance matrix, you can get an estimate from a large number of samples.

This is a multivariate distribution, so you're actually generating vectors. 1. Randomly generate from one of (k) categories with probabilities (p_1,p_2,...,p_k). (This is easy enough. Generate a random number in [0,1], etc.) If category (j) is generated, produce the vector ((0,...0,1,0,...0))...

buy the rumor, sell the fact!

suit yourself!

Plenty of people with half that IQ work in finance. Whoop-de-doo. At least the ones of higher IQ are keepin' it real...

What good has that ever done her? Hmm, let's see: Vos Savant was Chief Financial Officer of Jarvik Heart, Inc. She has served on the Board of Directors of the National Council on Economic Education, on the advisory boards of the National Association for Gifted Children and the National...

I'm a quant and mine was recently measured at a whopping 98.

Obviously the two greatest shows of all time are Breaking Bad and The Wire. Obviously.

Why would anyone care for a title? Why not instead care about being a good human being? Then the title might come. Not holding my breath though. ;)

sports: handball (beating a bouncy blue ball against a wall; like tennis), tennis, some gymnastics, strength workouts (mostly calisthenics; there's so much to be done with just your body weight), and planning to get into parkour soon. beer: too many useless liquid carbs, defeats the above.

... full of Joy?

Pears are most definitely the toughest. May Joy be in all of your hearts this Christmas season!

Boxing what, oranges?

Columbia and Harvard are nothing more than names, they don't bring happiness in life :)

You're absolutely right. It's still true, though, that (P(X_1=1)=P(X_2=1)=\cdots=P(X_{99}=1)), by symmetry, so it should actually be (1+99\cdot\frac{50}{99}=51).

indeedably it is! 'twas asked at JP Morgan.

Let's find your probability of winning if you're in position (k) on line. For you to win, the people in positions (1,...,k-1) should all have distinct numbers, and you should have one of those numbers. The probability of this happening is...

(1) For the status number to be (K), the sequence of draws must have the form (A_1A_2...A_K), where each (A_i) is a homogeneous block of draws -- all white or all black -- and these blocks alternate in type. There must be (a=\lfloor \frac{K}{2} \rfloor) of one type of block and (b=K-\lfloor...

We don't need no education, No more thought control.

Ha, are you serious? You were looking for a rigorous solution, and I set you up for one; the rest of it should be right-quick... We have to check the ratio of $\binom{2k}{k}0.45^{k+1}0.55^{k}$ to $\binom{2k-1}{k}0.45^k0.55^k$, which is easily computed as $2\cdot 0.45 = 0.9<1$. This shows...

Yes, those should be 105's; that's what happens when you type fast. So you're saying that the sets $\{1,2\}$ and $\{3, 4, 5, ..., 20\}$ have equal probability? that's obviously not true; the first has the very small probability of $\frac{1}{20}+\frac{1}{20}=\frac{2}{20}$, while the...

If you have the body type, go to Zara. You can get a nice fitted suit for $300-$400, plus it's Euro, so it's added style that you wouldn't get at an American place like Brooks Brothers where everyone shops. Worst thing is to wear baggy pants and look like a ninja; many people unfortunately are...

Important note: A player can have only so many daubs before a bingo is reached. what does that mean exactly?

For $n$ odd, you either win the last round, and you've come out at least even in the previous $n-1$ rounds, or you lose the last round but you've come out on top in the previous $n-1$: \( P_n = 0.55P_{n-1}+0.45 \large(...

Well, no, those two events are not equally likely to occur, unless those sets have the same size, which only occurs for $x=10$. It's not about the probability of winning, so much as it is about the players' expected winnings. For instance, if player 1 plays 12, then player 2 will play 13...

A poet; I like that. We need more of those around here!

Well this is pretty convoluted, and I'm not sure what you're looking for. Can you be more precise? Also, i'm not familiar with Bingo. How does one get a bingo? And presumably, if the number of daubs can be bigger than 1, numbers may repeat within a given column?

No, lobomattu actually has a point... vertical distance may be the same for the uphill and downhill parts, but the road distance may not. I guess you're supposed to assume the hills are symmetric, which is kind of silly. Let's just dismiss this as a bad problem :P

It's not that $X\in\{1,...,x\}$ or $X\in\{x+1,...,20\}$ if Player 1 picks $x$ ($X$ is random and can be anywhere)... it's that Player 1 knows what Player 2 will play after he plays $x$ (see my argument above). Once he knows this, he knows what his expected winnings are because he knows...

$E(winnings) = \sum_{k=1}^{20} E(winnings|X=k)P(X=k)=\sum_{k\in S}\frac{k}{20}$, where $S$ is the set of $X$'s where you win (because for all other $k$'s, your winnings are 0). For instance, if you pick 11 and then your opponent picks 12, you of course win for any $X\in \{1,...,11\}$...

The problem mentions that there are 'level' parts. In completing a full circuit, to go from level ground back to level, you must obviously go a total downhill distance equal to your total uphill distance.

No, it is 20 as I'm assuming the discrete case. And X is actually random from the point of view of the players: the point of view the problem is concerned with.

It's a two-part problem. (1) Figure out Player 2's optimal move for every move that Player 1 can make. (2) Figure out Player 1's optimal move based on what he knows Player 2 will do. This is all assuming each player is rational and seeks to maximize his expected winnings. Let $f(x_1)$ be...

Maybe for him picking one unit higher or lower than me is not optimal. Either way, we should know what happens in the case where X is precisely in the middle of our picks.

What if we're equally close? X = 14, and we pick 12 and 16, say.

First off, is this discrete or continuous?

Hehe. No problem.

Whatever distance he goes uphill he must go downhill, so $x/16+y/12+y/24 = 9/4$ $3x+4y+2y = 108$ $3x+6y = 108$ $x+2y = 36$

Can you justify that? (I don't think that's true.) Also, like Faisal, you are not using the fact that $1\leq a_i\leq 2n+1$, which is crucial. This means that the argument is probably flawed.

Among the $n$ noodles there are $2n$ ends. Picking one end leaves $2n-1$ other ends to match it with.

For each $n\geq 1$, let $X_n=1$ if there is no winner on the $n$-th toss. Then the expected number of tosses is $1+\sum_{n=1}^\infty E(X_n)=1+\sum_{n=1}^\infty P(X_n=1)$. We'll show that $P(X_n=1)=\frac{1}{2^n}$, so that the expectation evaluates to \(1+\frac{1}{2}+\frac{1}{2^2}+\cdots...

Yeah, that's already been established.

The probability of not getting any money doesn't tell the whole story, as it's possible for the cases that do make money to make disproportionately larger amounts. Your expected winnings, as well as the volatility of those winnings, are what matter. Now, between two alternatives with the same...

I was captain of the Stuy math team and a member of the 2nd place ARML team in 2000. Ahh the high school days...

It ain't about googling, it's about sharing and discussing; I'm sure that's why tuanl posts his problems here. Use indicator variables: For each of the first $n-1$ tosses, out of $n$, let $X_1=1$ if that toss and the next are TH, 0 otherwise. Then the expected number of TH's is equal to...

Because sign does not matter. The proportion of the first quadrant where the condition is fulfilled will therefore be the same as the proportion of the entire plane where the condition is fulfilled. It's $\frac{2}{\pi}$ because we're dividing the total angle where the condition happens by the...

It's enough to treat the case $a,b>0$. You need $1\leq \frac{a}{b} \cdot 10^k <2$ for some integer $k$, i.e., $10^{-k} < \frac{a}{b} < 2\cdot 10^{-k}$. The region in the plane described by these inequalities is a windmill pattern of sectors in the first quadrant. The angles of these...

Yes, it's 5. Now prove that it can't be done in 4...

You're right, I switched them. In other words, just reflect everything about the middle. 15 <-> 5. Thanks!

you mean 999; one of them is surely poisoned.

It's even more basic. It's physically obvious: To balance an object on the left of a see-saw of length 20 with an object 3 times its weight on the right, the hinge must be placed 15 units from the left. But here's a mathematical proof for sticklers: If you guess $a$, then your average...

Yeah, I'm aware of the 12 coin problem. But your problem is 81 coins.

Okay, here's a correct proof. Let $p < q < t < u\in \{n^2+1, ..., (n+1)^2\}$ such that $pu = qt$. Note that $(u-p)^2 = (u+p)^2-4pu = (u+p)^2-4qt > (u+p)^2-(q+t)^2$, where the latter follows by the AM-GM inequality. Then $(u-p)^2>(u+p+q+t)(u+p-q-t)>4n^2(u+p-q-t)$. Note that since...

This is obviously wrong, since nowhere are you using the fact that a_1, a_2, b_1, b_2 are in the range [1,2n+1]...

For #2, Let $S(t)$ and $N(t)$ be the sum of the rolls, and the number of dice, respectively, in round $t$. Then $E(S(t)) = \frac{7}{2}N(t)$ $E(N(t)) = \frac{4+5+6}{6}\cdot N(t-1) = \frac{5}{2}N(t-1)$ Taking expectations in the latter recursion and inducting...

4 is most certainly wrong. You can do it in 4 only if you know that the counterfeit coin is, say, heavier.

Or you can just say: if they're all distinct, then their sum is $\geq 1+2+\cdots+n=n(n+1)/2 \geq n$, a contradiction.

This argument is lengthy, and might fool a lot of people, but unfortunately doesn't make any sense.

The article is genius -- serifed, bolded, and punctuated. It bashes incompetence with poetic glee. Talk about metacognition!

hmm... yeah, I think a quant should be pretty good at math.

It's not about white vs. non-white, it's about who can actually communicate in English. you know, that part on your application where you check off your native language.

Bump away.

Besides, semantics aside, here is why 1 should be included among prime powers... When, for example, computing the number of factors of a number that we've prime-factorized, (p_1^{e_1}p_2^{e_2}\cdots p_k^{e_k}), every factor is of the form (p_1^{f_1}p_2^{f_2}\cdots p_k^{f_k}), where each (f_i)...

Why are you still talking about prime powers? Did you not see that I rephrased the solution? Must be those bath salts again...

Oh, sir, you're right, I humbly apologize for the misunderstanding. How could I ever think 1 was a power of a prime! But maybe we can still salvage ferdowsi's solution by rephrasing it... "Any perfect square that does not have more than 1 distinct prime factor."

You've really gotta be kidding me. Of course 1 is a prime power: any prime to the zero. You really think they intended a solution as esoteric as "the basis set for representing positive integers with positive squares is " blah blah blah... ?? You must be high on bath salts.

Why do you need 1 to be the square of a prime? If I define a set of numbers by the condition "...containing non-distinct prime factors," then the number 1 will vacuously satisfy this condition. Semantics, my friend, semantics. You can also take a more careful look at ferdowsi's statement...

Of course ferdowsi 's solution works; think about it a tad bit more. His answer is much more likely to be the intended one and the one that I (most definitely!) agree with. Just consider bifunctional proportionality, followed by a little hedonic calculus, and you're good to go...

that's so... interesting...

You're right, of course. Thanks. :) Now, given how simple the final formula is, there has got to be a prettier solution...

The number of increasing sequences of elements of (\{1, ..., n\}) of length (k) and ending in (m) is (\binom{m-1}{k-1}) (pick the (k-1) elements less than (m) from (\{1, ..., m-1\}) and there's a unique way to order them). There are (m) ways in which the following roll won't improve on the last...

haha, my algebra skills were way off here, you are right. i've fixed it.

See no evil, hear no evil, speak no evil...

Sure, but let's put an 's' at the end of Santa... this be a democracy of Santas.

I get 100 too. Assume there are (n) santas at the party. There are (\frac{n(n-1)}{2}) potential relationships between them. Of these, (\frac{22n}{2} = 11n) are friendships. Therefore, (\frac{n(n-1)}{2}-11n) are non-friendships. For each non-friendship there are (6\cdot 2=12) friendships that...

it'll probably be brainteasers and tech questions, so read Heard On The Street and Frequently Asked Questions in Quantitative Finance.

that's coo

I'm goin' on tilt, all in!

that's awesome

Obviously that's incorrect... only if exactly half of all the bulbs are currently on can we say that the probability of one more on is the same as the probability of one more off. Trust me, the answer is about 21.75.

so?

Has anyone yet mentioned the Samsung UN55D7000L? Gotta love that narrow bezel!

I saw this question and a bulb went off. The probability that a given bulb is on is the probability that it was flipped an odd number of times: (\binom{50}{1}0.02^10.98^{49}+\binom{50}{3}0.02^30.98^{47}+\cdots+\binom{50}{49}0.02^{49}0.98^1). This is equal to...

1/2 - 1/2 + 1/6 1/4 - 1/3 + 1/8 1/6-1/4+1/10 1/8-1/5+1/12 1/10-1/6+1/14 ... Notice how terms cancel diagonally: (1/6+1/6-1/3), (1/8+1/8-1/4), ... the only term that survives is 1/4.

It makes a lot of sense, and I suspect it's not really discrimination. Gaps in salary in general are larger in the financial sector. Everything is scaled up. So given that women's career goals and preferences are likely to be different, their salaries will be different.

I've suggested this before... Writing math here would be a lot more fluid if it required typing only one character, like '$'.

At least go get some free food.

how is that? if 5 crows land on the number line at 0, 1, 3, 5, 6, then 3 isn't being crowed at, while the others are, including the end ones.

The end crows each have probability 1/2 of not being crowed at. The penultimate crows on each end have 0 probability of not being crowed at. For each "interior" crow, the probability of not being crowed at is 1/4. So the expected number of crows that are not crowed at is...

ha, but your solution is easier and probably the one they had in mind.

It's really a combinatorics question at heart. For the first set of coefficients just pick (1,1,1). Whatever the result is, the number of possible triples (x,y,z) is finite -- they're constrained to the first octant because they're natural numbers, and only a finite number of lattice points on...

By AM-GM, the sum must be at least $3\sqrt[3]{72}=12.48...$. But since they're integers, the sum must actually be at least 13. And as Tsotne shows, 13 is attainable.

1) Say there are (n) noodles in the bowl. Condition on whether the first knot results in a loop or just a longer noodle. The probability of the former is (\frac{1}{2n-1}) while the probability of the latter is (\frac{2n-2}{2n-1}). Denote by (E_n) the expected number of loops given (n) noodles...

Yep, it's 2k-1 :) Now I guess you can generalize to (w) women, (k) condoms... Edit: Actually, a better formulation would be: Given (m) men and (w) women, what's the minimum number of condoms needed if each man fancies every woman?

Even more than that.

Nope. More.

Given $k$ condoms and one woman at their disposal, what is the greatest number of men that may have safe sex with the woman?

You've gotta clarify some points (heheh) here... 1) Are the points numbered in order and we start at point #1? 2) Are you asking for the probability that the (N-2)-st distinct point to be hit is either r+1 or r-1?

Why waste a breath on going "hmmm"? Clearly you go with the one that ranks you higher. If Joe ranks me #10 and Mike ranks me #1 I'm going with Mike, sorry.

Conditional probabilities, people, conditional probabilities! All too often people forget to condition!

The Glass Bead Game... yes, absolutely, I second it!

does it really matter? there's obviously a typo; the real point of the joke is the irony at the end; end of story.

Wow, is your job really more important than your safety? Yes, man, hindsight *is* 20/20! And keep in mind that, even though your area may not have suffered much, some areas did get hit pretty hard!

What's with all the criticism? We're quants, not meteorologists. Better to be safe than sorry -- no?

You can't just rearrange them like that. For one, that last sentence is a run-on sentence which either needs to be fixed by rephrasing or inserting some commas. Two, in a construction like "It weaves..., the Panama Canal linking..." It and the Panama Canal should be different entities, but the...

I play tennis! Who's down? I know some free courts up around 137th street, in Riverbank State Park. Also what I do if I can't make it to an actual tennis court is play against a handball wall; they're a lot easier to come by and the game can be just as fun and competitive. By the way, anyone...

Why is it trash? Which one do you have?

Any places still have 'em??

Well, between making millions on Wall Street and spending more time in school, which would you pick?

I know of a couple of such cases at DE Shaw; but they were a few years ago and the people were IMO gold medalists and Putnam Fellows. These places aren't gonna take your average Joe, especially not straight out of undergrad.

'cause those are real easy on ankles...

Hence the subway.

i also take the train from Penn. the quickest way is really by subway; take the N/R two stops.

We need to accept the fact that some things are better left unsaid.

Because I am talking about a volume (specifically, of the part of the sample space where the needles fall properly onto the stick); when you find volume you integrate 1 over the region. Then I divide this volume by the volume of the sample space, $(1-h)^n$. This is essentially equivalent to...

Okay? And? You essentially repeated my solution...

Very funny guys... since when do a couple of steaming hot days imply global warming? come on now. there is also evidence of cooling...

I agree with Alexandre. When after a couple of years your health starts deteriorating because of the crap food you eat 'cause that's all they have in the kitchen down the hall, or you're in too much of a rush to eat a proper meal, and when you start to forget what it feels like to connect with...

The weather is fantastic...

The ridiculous hours people work in the big banks really make a person ask, "What's more important to me - a career, or a life?"

no smiling? and what fun is that?

3 Idiots is the best Bollywood movie I've ever seen. New York is also good but the acting in some parts is kind of bad.

It's not either/or. But I think you started this thread with a personal conception of what "positive thinking" is and you're not allowing for a broader definition. You seem to be defining positive thinking as a wholehearted but possibly blind belief in one's abilities. For instance, "overrating...

I agree, positive thinking is definitely good. And it's certainly better than negative thinking. But, positive thinking by itself won't cut it; it's awareness that matters more. Too many people are just not aware of their flaws and limitations. If they were maybe they could actually start...

who's got which?

the grades don't count, is what I meant, and they're canned as soon as the refresher is over. you only need to pass... better to take things easy; it's summer time, after all :]

It's only the refresher, it doesn't even count...

Near GCT is hardly a great place to go out. And besides, doing this near GCT only makes it convenient for some people, not others.

As far as beer gardens go, the best I've been to is Studio Square in Astoria. It's a huge outdoor area with lots of tables and all sorts of specialty drinks and beer. Great place if the weather is good.

I won't be able to make it to this one, but hopefully the turnout of new Baruch students this time is better than the measly turnout of three last time.

For those of us who frequently communicate mathematics, LaTeX is the best thing since sliced bread.