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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.

Absurdity indeed is a common theme here. If it's so "inefficient," then how about naming a mathematical typesetting language that is efficient.

they call this math envy ;)

we had wi-fi access last year around this time.

Looks like anyone who has an opinion around here nowadays is a troll! What the hell? Open your eyes, people!

much more likely it's 800 on Q. an 800 verbal is a rarity.

The construction industry...

I have a feeling he was being half-sarcastic about all that. That statement is pretty funny though.

He's of the belief not that math is bad for finance, but that how you use it can be. It's like the saying: guns don't kill people, people kill people.

Yike makes great points and puts this all in a pretty clear light. Somewhere else on this forum someone advised another member to look into Taleb's take on overconfidence. When people come looking for help on this forum, they certainly don't hope to get a bunch of replies from wanna-be experts...

wait, for what reason is an MFE mostly irrelevant? I'd hate to think the degree we're currently all doing is pointless.

half these phones I've never even heard of. having an iPhone just spoils a person. heh.

iPhone 4, of course! Nothing beats the retina display! (Angry Birds is amazing on it!) I use roughly 500MB a month in data. That's because I mostly just surf the web. People who stream a lot of videos or music use significantly more.

That's actually a very good point, believe it or not.

Welcome to the world of finance! ;)

The largest of the first i cards has an equal chance of being in any of the first i positions, so the probability that it's in the i-th position is 1/i.

Well as you were trying to point out, it's much more likely that you'll be hitting occasional peaks (points where you've gotten a card greater than all previous ones) than get cards that keep increasing in value. In fact, unless your first cards are A, 2, 3, ... in the second scenario you're...

As always, use indicator random variables. For (k=2:10) let (X_k=1) if the (k)-th card is greater than all the previous ones, (X_k=0) otherwise. The expected winnings are then (10+10\sum_{k=2}^{10}P[X_k=1]). For the (k)-th card to be greater than all the previous ones, it must have a value...

slim fitting suit with a skinny tie - absolutely.

The GRE subject math test is only relevant if you're pursuing a masters or PhD in pure math.

As I said, I love Excel! I think it's one of many great tools at our disposal.

That's funny ;)

I love Excel!

It's sooooo basic!!!

well there are actually ones with 5 runs; here's all 20: 2 runs: BBBRRR, RRRBBB (2) 3 runs: BBRRRB, BRRRBB, RRBBBR, RBBBRR (4) 4 runs: BBRRBR, BBRBRR, BRRBBR, BRRBBR, RRBBRB, RRBRBB, RBBRRB, RBBRRB (8) 5 runs: BRBRRB, BRRBRB, RBBRBR, RBRBBR (4) 6 runs: BRBRBR, RBRBRB (2) and this gives us an...

You obviously once again neglected to read the posts or just stubbornly refused to believe in the tightness of the arguments we presented. Check out my rebuttal three posts ago. Anyway, that's about all i have to say about this. til next time!

What I think you're trying to say is that runs in certain parts of the deck are affected by runs in other parts. Sure, but that's irrelevant because we're after the average number of runs here. Think of it like this. Consider the following two procedures. (1) Shuffle the deck 10000 times and...

Haha, I am really off on this problem... It should be 51, not 49, because there are 51 places where a change can occur. So, for the third time, the answer should actually be \(1+51\cdot \frac{26}{51}=27\). It's also the most aesthetic answer I've had so far, so I may actually get lucky this time ;)

That's where you're wrong. Look carefully at how I've defined (q): It is the conditional probability that, given that the initial roll is odd, the sum of all rolls, including the initial roll, is even. The problem concerns itself with the sum of all rolls, not just the ones after the first...

There's no way to get a sphere because cylinders have straight geodesics in a certain direction whereas spheres have circular ones in all directions.

The number of runs is essentially the number of changes in color from one card to the next, plus 1. This is easy enough to see; just look at Rados' example. Notice that \(X_1+\cdots+X_{51}\) counts the number of changes, so we're looking for \(1+E[X_1+\cdots+X_{51}]\), which by the linearity of...

For the two-shaft case, the region of intersection is comprised of 4 congruent pieces that look somewhat like slices of an orange. Their total volume is (4\int_{-1}^1\frac{\sqrt{2(1-x^2)}^2}{2}dx=\frac{16}{3}\approx 5.333). (The flat surfaces of the orange slices are each half the ellipse...

For each \(n=1:51\), let \(X_n=0\) if cards \(n\) and \(n+1\) are the same color and 1 if they are different. the answer to the question is \(51E[X_1]+1 = 51P[X_1=1]+1 = 1+51\cdot\frac{26}{51}=27\)

I'm aware we're arguing between two plausible definitions here. But shouldn't there be a standard definition? Every time someone sells a call swaption, shouldn't everyone know what that means without having to ask every time?

oh yeah, i heard about you. you're that troll.

the truth is I didn't bother reading your post because I know it's wrong. i'll leave it to other people to confirm it to you that P(even) is in fact 47/82.

it actually looks like a bas relief. as if it's projecting out from under the skin, haha.

@quantyst, the values 4/7 and 3/7 are wrong, as you'd have seen if you'd read the earlier posts; it would have saved you a lot of typing ;) read my solution above; the correct values are P(even)=47/82 and P(odd)=35/82.

First let me say that we're not debating an intrinsic truth here, but rather which of two definitions is more natural. That said, I again have to disagree in favor of the one that to me makes more sense. What a call option does in general is, it gives the holder the right to buy something with...

It can be that the same term refers to different things, but that's usually when they take place in different contexts. But I don't think that's the case here. By definition, in a call option you should be long whatever's variable, not fixed.

Except it's wrong. See my previous post :)

Nope. It should be the opposite. Think about it... when you're long a call, it makes sense to have a long position in the floating rate. In other words, you would exercise if the floating rate is above the fixed rate, because you would be receiving the floating rate and paying the fixed rate...

I first heard that expression from my 5th grade teacher. You'd expect people in industry to have more sophisticated ways of conveying these messages. lol. What are their worries?? Unless they're embarrassed by their numbers, I don't see why they wouldn't make their placement rates public...

LOL

Harvard will take the top students from all over the place. So if you're the valedictorian of your high school, you were on the debate team, you did sports, you did volunteer work, and you were overall a star in your school, Harvard will take you, whether your high school was run of the mill or...

From what I hear there's a lot of bribing going on in these elite private high schools. Parents will pay good money to have their students make up bad exams, rewrite essays, or even have grades inflated without having the kid do anything. If you send your kid to one of these schools, your rivals...

or notice that the two triangles I mention above, with sides (2,\sqrt{3},\sqrt{3}) and (2\sqrt{2},\sqrt{3},\sqrt{3}) both divide into (1\text{-}\sqrt{2}\text{-}\sqrt{3}) right triangles by drawing heights from the vertex angles; this proves it immediately. no trig ;] now we've beaten this poor...

Conditional on the first toss being odd, either 1) the second toss is the same as the first (the 1/6 part) or 2) the second is an odd number different from the first toss (with probability 2/6=1/3) after which the sum starting with the second toss has to be odd (probability 1-q) or 3) the second...

that's not the calculation he was asking for; that's the trivial part. he was asking about the trig. by drawing perpendiculars to the common edge, you can see that the dihedral angle (which you want to show is a straight angle), is made up of the vertex angles of two isosceles triangles, one...

Nope. Because if any left endpoints of the earlier needles are too far to the right, there's no room to place later needles with no overlap.

For it to be 1-h, we'd have to assume that either the sample space of left endpoints is the entire interval or the sample space of right endpoints is. but then the other endpoint can be off the stick, yet there's no reason to distinguish between right and left. The way I understand it, the...

Ha, really? Intuitively, it does seem to be, but I haven't found a solid reason why Alice can't do better.

I thought about this for a bit but don't yet know the solution. One thing is clear: Alice can guarantee an outcome of at least 356, as follows. Label the subtraction problem ABC-DEF. Let Alice start by picking 5's until Bob either picks A or D to fill in. If he picks A first, then have Alice...

I meant to say the probability of *no* overlap.

When n=1, the probability of no overlap should be 1.

Being accepted is a nice feeling, eh. congratulations!

Hahaaa

I neglected to mention: The above argument holds only if \(hn\leq 1\). If \(hn>1\), the probability is of course zero.

No. The lower bound of \(x_1+h\) is enough: the second left endpoint has to be at least \(h\) units ahead of the first left endpoint.

Don't we need to mention dihedral angles here? The reason it's 5 and not 7 is because the dihedral angle between an equilateral face of the tetrahedron and an equilateral face of the pyramid is actually 180 degrees, which can be seen by some trig calculations. This is why you lose two more...

Dropping the needles so that they all fall within the stick is equivalent to randomly picking \(n\) numbers (the left endpoints) each uniformly and independently drawn from \([0,1-h]\). Under this interpretation, the volume of the sample space is \((1-h)^n\). The volume of the region where the...

Are we assuming the stick is one-dimensional, so that the needles can only lie along the length of the stick?

Because of affirmative action, don't women nowadays actually have the advantage in areas that were traditionally male dominated? Men are competing against a much larger pool of candidates. I have even seen a few cases of highly qualified white males being turned down by major universities and...

can you say prosaic?

respective of what? hold your horses.

hah. hoping isn't gonna change anything.

astute observation!

no problem! just visit the other departments...

I doubt it gets more elegant than that.

Explain :)

A cute one: Mr. and Mrs. Jones invite four other couples over for a party. At the end of the party, Mr. Jones asks everyone else how many people they shook hands with, and finds that everyone gives a different answer. Of course, no one shook hands with his or her spouse and no one shook the...

Keep in mind though that this could very well be a case of mistaking correlation for causality. People with sedentary jobs also tend to not watch what they eat, use vending machines, order take-out, etc.

Maybe not quants in general, but many traders tend to gravitate towards poker...

Explanation?

Good questions. I'm surprised no one's replied yet.

Nice! Yes, I meant \(n=2\). Our results do match for n=1 and n=2, so I'm convinced you're right. Can you explain how you got that expression?

It's the probability of defaulting at some point over the first two years. It is one minus the probability of not defaulting in both year 1 and year 2.

This problem is definitely more difficult than the one concerning the minimum. But for \(n=3\) it's not bad. For \(\frac{1}{3}\leq m\leq \frac{1}{2}\), the probability \(P(\max(R_i)\leq m)\) is \((3m-1)^2\) (the region in question is a triangle) and for \(\frac{1}{2}\leq m\leq 1\), the...

You never talk about the probability of the statistic equalling zero, because that's an event of measure 0. Instead, you talk about the probability of your statistic being within or outside of a given interval. Again, the p-value represents the probability of your statistic being at least as...

Oops, yeah, I misread it.

2. P-value is the probability of the statistic being at least as extreme as the one observed. So with a greater P-value, you're more confident that your observation wasn't an anomaly. I think 5% is the correct answer.

Note that given \(R_0+R_1+\cdots + R_n =1\), we need to find the probability \(P(\min(R_0,...,R_n)\leq m)=1-P(\min(R_0,...,R_n)\geq m)=1-P(R_0\geq m,...,R_n\geq m)=1-P(R_0-m\geq 0, ..., R_n-m\geq 0)\), where \(t_0=R_0-m,...,t_n=R_n-m\) satisfy \(t_0+\cdots +t_n=1-(n+1)m\) In general, the region...

Intuitively, over a longer period of time, the probability of default should be greater, not smaller. It should be (1-0.9^2=0.19).

Why, would you like to win a conversation with Dan? Don't you get those for free now? ;]

I remember this problem from a while back. If I recall correctly, the answer is 2/3, and it's 2/3 no matter how many numbers you're pairing.

Are you sure the terms are independent? In the (3\times 3) case, for instance, you have the terms (a_{1,1}a_{2,2}a_{3,3}) and (-a_{1,1}a_{2,3}a_{3,2}) which are not independent...

I really like that differential equation trick! It's the continuous version of another trick you see a lot on this forum... Here's a related question, which can also indirectly be used to solve this one: Pick numbers (x_1, x_2, x_3, ...) in (U[0,1]) as long as the running sum is (\leq 1). Find...

obviously those are bid and ask. he's asking about the column of D's...

doesn't matter... we have the technology to make it real.

Given the presence of \(\pi\) in the answer, do you think a "cute" solution is likely? Probably not.

marvelous piece of advice...

yongge: read the original problem statement..."as long as they keep alternating in size" -- which implies that we shouldn't count the one that breaks the pattern. if it had said "until the pattern is broken," I'd have agreed with you :)

wait... that would mean... no, can't be! someone at NYU or Columbia?!

The cold, hard truth... Still, I don't believe in discouraging people -- just go for it.

congratulations!

Though quant skills are significantly more important in this field, many FE programs also like to see decent verbal skills too, since finance is also about communicating ideas to people.

Sheen would suffice...

I guess firms should start naming themselves after celebrities...

This makes sense, since the expectation is a sum over all N of the probability of the condition being satisfied up to step N... plugging 1 into the probability generating function finds the sum of those probabilities... Here's another classic interview question whose answer is also (e): What is...

Haha -- It has got to be an April fool's prank. But then again...

well your answer can be written as sec(1) + tan(1)... are you maybe only considering alternating sequences that start on an "up"? from your original problem statement, starting on a "down" should be allowed too.

Well chess definitely requires a lot of mental endurance, as do certain sports (like tennis, for example, where matches can be hours long). Mental stress certainly takes its toll on the body. But then we can argue that certain financial engineering jobs are a sport too. Hey, it's true they do...

Haha... looks like Will's a pro. He kept winning at my house party too...

Hehe. In fact, I had just visited that page before posting. But the first definition is primarily the one people go by -- "an athletic activity requiring skill or physical prowess and often of a competitive nature, as racing, baseball, tennis, golf, bowling, wrestling, boxing, hunting, fishing...

Is chess really considered a sport? I have doubts about table tennis being much of a sport -- much less chess ;)

Hi Rados, First off, nice problem! And damn difficult :) I am not entirely sure how to do it yet, but I do know the answer involves trig functions because the first few probabilities (of getting a sequence of n alternating values) that I get -- 1, 1, 2/3, 5/12, 4/15, 61/360 -- are twice the...

well the probability of the first N draws alternating is 1 for N=1,2, and then 2/3, 5/12, 4/15, 61/360 (computed via multiple integrals) for N=3, 4, 5, 6, so the expectation is at least 1+1+2/3+5/12+4/15+61/360=3.5194...

I'll be the first to admit, many of these questions have no direct bearing on finance, and people wonder why they're asked. They're asked because at core many of them mimic the fundamental principles driving financial engineering. Tsotne and others: If I'm not mistaken, the true answer is...

Why is it that people always attack these problems with simulations? Anyone can write a quick program that will give a fairly accurate answer -- that's easy. But it defeats the whole purpose of the question. When you're asked something like this on an interview, you're being tested on your...

good luck Connor ;]

Lyosha, stop scaring the applicants ;)

A good rule of thumb is, make sure you know what's on your resume and know it well. yes, i know, this should be obvious, but people do forget :)

ha. well-played!

I wouldn't say it's easy or hard. It's about what your background is. Time Series was presented from an engineer's viewpoint and for me it was difficult to follow. I personally dislike the way engineers treat mathematics -- I'm a mathematician, so go figure. That being said, I think the...

it should be. I'd be pretty upset if it's not :)

it's basically the same as my solution, only recast as an argument by contradiction. nice though :)

writing isn't creative because only thinking is. hmm. that makes about as little sense as chewing one's own head off. and I don't see how actors enter the picture out of the blue. but since they were mentioned... actors -- the good ones -- are quite creative and talented people! unfortunately...

No. For instance, if \(n=4\), you want the first 2 rolls to have an even sum *and* be different, so you want to multiply the probability of them being different by the probability that their sum is even given that they're different. The latter is \(\frac{12}{30}\neq \frac{1}{2}\).

you're right, the answer is no... in fact, Bob can always block her from winning. this is one of those problems where you have to find the appropriate tiling. the one that works here is, divide the grid into 2x2 squares, then tile each 2x2 square with two dominoes, alternating horizontal and...

One of the flaws is where you say \(P(\text{sum of previous } n-2 \text{ rolls is even})=1/2\) Also, your probabilities are not conditional, so they don't quite multiply that way.

okay, here we go... Denote \(p\): the probability of getting an even sum given that the first outcome is even. \(q\): the probability of getting an even sum given that the first outcome is odd. Then \(p=\frac{1}{6}+\frac{1}{3}p+\frac{1}{2}q\) \(q =...

conditional expectation... \(E(\text{stopping time})=E\text{(stopping time}|\text{first two outcomes are identical)}P(\text{first two outcomes are identical})\) \(+E(\text{stopping time}|\text{second outcome is different from first})P(\text{second outcome is different from first})\) note that...

actually I think getting the probability is the more interesting part -- and we still don't have a correct answer. getting the average stopping time is pretty easy...it lends itself easily to conditioning (how many times have I said that on this forum? lol) conditioning on the first two outcomes...

so it looks like you got that odd is more likely. finding the actual probability isn't quite that easy, but I think if you try some sort of a bijection approach you get that even is more likely.

Thanks Rados :)

you answered the wrong question. the question is about the parity of the sum of the outcomes, not the parity of the number of rolls.

the answer is it's always possible. Number the 8 cubes 1-8 and number each of the cube's vertices 1-8 as well. Let (r_{ij}) denote the number of red faces meeting at vertex (j) of cube (i). Assume WLOG that (r_{i1}\leq r_{i2}\leq \cdots r_{i8}) for all (i). Note that...

This is equivalent to what Bob did. An eigenvalue of 1 corresponds to "stability".

the problem is not symmetric in heads and tails. heads are flipped over to tails, but for tails the coin is tossed and takes the outcome.

Overall GPA isn't as important as having a particular strength relevant to FE, like math or programming. If you're super-good at something mathematical or technical and can prove it, that'll get you through no matter what your GPA in other subjects is.

1) Both progressions are increasing. Then clearly the loser of the first game - say player I - wins the series because he overtakes the other player by larger leaps. Assume player I's scores are (a<a+d<a+2d), where the last two are winning scores. If (a+d>21), player II's scores in games 2 and 3...

That statistic is right in the article above. Though it's probably a bit exaggerated to discourage people from using them (the source is pretty biased).

This sort of problem lends itself well to conditioning. Let \(p\) be the probability that player 1 ends up with all the money. Then conditioning on the first two outcomes for the second player, \(1-p=\frac{1}{3}\cdot 1+\frac{2}{3}\cdot \frac{1}{3}\cdot (1-p)\) solving gives \(p=\frac{4}{7}\)

why break it up? because the two are first of all mathematically non-equivalent statements, and second of all are non-trivially different -- well, because they're not equivalent.

Well now it's certainly clear as day. Here's the proof...easy and cute... Every element of the set {1, ..., n} appears an odd number of times (namely, n) in the matrix. Every time it appears off the diagonal, it appears at its reflection across the diagonal as well, so the number of...

you sent 'em all on a wild goose chase

I think what he's missing from the problem is that the matrix be symmetric (about the main diagonal). Then the main diagonal must indeed be a permutation of 1, ..., n.

sure, you can view it that way too, but the solution i gave above is more elegant :)

(\frac{1}{2}) For each sequence of tosses (S), let (S') be the sequence with heads replaced by tails and tails replaced by heads. Then there is a 1-1 correspondence ((A,B)\leftrightarrow(A',B')) in which if one side has the property (x<y), then the other side has (x\geq y), showing that the...

this is such a classic. it appeared at the IMO many years ago.

I used my Baruch email and never got anything back from them

how does one get an account on wilmott? I signed up for one a while ago and never heard back.

The problem is basically asking for the expected number of cycles in a permutation of 20 elements. Let's do it for (n) elements. For each (i=1, ...,n), define (P_i=\frac{1}{k}) (think of this as the "weight" that each person carries, so that each group's total weight is 1), where (k) is the...

Now generalize. Find (P(B_{\pi(1)} < B_{\pi(2)} < \cdots < B_{\pi(n)})), where (\pi) is a permutation of (1, ..., n).

this is handwaving at its best ;)

this isn't equal to 5

Again, you have to think about what is being revealed exactly. That is what you're conditioning on. In your argument, you're not conditioning on the proper information.

this is different. in this scenario you know *who* tossed a tail. the other guy then has a tail with probability 1/2. this is different from just knowing that one of them got a tail, but not knowing *who*. probability is all about information. the information you have affects your probabilities.

yes, the answer to your coin question is also 1/3, assuming i don't know which coin you showed me... you just showed me a coin and it was heads. because, conditional on this information, i know the possibilities are HT, TH, HH. on the other hand, if you tell me that the first coin came up...

the answer to this question is exactly the same as the first. seeing the child doesn't give you any new information; it only confirms that one child is a boy.

read the earlier posts

yes, we have the correct answer, and it's 1/3.

There are two children, and you know one of the children is a boy. The "other" child is the child we don't know have information about. The problem can be rephrased as, Find the probability that both children are boys, given that you know one of them is a boy.

The question is not ambiguous at all. The answer is clearly 1/3.

yes, and if you read the wiki link, you'll see that, given the phrasing, the correct answer is 1/3. the link even gives some empirical proof of it.

No, AN is right, and from the way the question is phrased, there is only one correct answer (1/3! not 1/2!) for the Sunday question, if the first child is a Sunday boy, there are 14 possibilities for the second child. if the second child is a Sunday boy there are again 14 possibilities for the...

You're right. I missed a 1 when doing it. thanks.

The answer is 4 minutes. The probability that (n) randomly and uniformly chosen points on a circle do not contain the center is \frac{n}{2^{n-1}} (a nice and simple solution can be found here Semi-circle covering n points : Puzzle). Now let (X) be the number of minutes it takes Mulder to be...

Are they i.i.d. (\mathcal{N}(0,1))?

numerically... analytically... well which is it?

discretize, of course!

If they cut you off *and* come in front of you, that could be a big problem. But, if all he does is cut you off, just do this... whisper such unimaginable obscenities just loud enough for the two of you to hear (but no one else) til they are so creeped out they either make a scene and get kicked...

How can you effectively program something if you don't understand the mathematics that governs it? No matter how good a programmer you are. Both programming and mathematics are equally indispensable to a quant.

Love it. It's so raw.

come on, Joy, make it open source.

weiwern is right... according to the hint, if we multiply the data by lambda, we get the rescaling we want. but lambda is unknown, so we instead multiply by an estimate of lambda, which for the exponential distribution is the reciprocal of the sample mean. the mean cannot be zero because our...

ha. my solution is not wrong; check again. all that was wrong was the numbers in the first two equations. I've fixed them and now gotten the same result as you. albeit, in far fewer words. so the problem is definitely not as complex as you proclaim it to be.

Recursion is ideal for a problem like this. First, let (p=P(10 \text{ before } 7)), (q=P(11 \text{ before }7)). Then (p=\frac{3}{36}\cdot 1+\frac{27}{36}\cdot p\Rightarrow p=\frac{1}{3}), (q=\frac{2}{36}\cdot 1+\frac{28}{36}\cdot q\Rightarrow q=\frac{1}{4}) Now, denote (r = P(10 \text{...

szhong, you're right, it is 24. the way to look at my last case is, (\frac{4!}{4\cdot 2}=3), the 4 to account for cyclic permutations, the 2 to account for repetition of A's. Burnside's Lemma is indeed a powerful combinatorial tool. good call!

Now generalize!

Ha. My point was exactly that -- that your way is NOT different. At least not for any effectively beneficial reason. You're just unnecessarily scaling by a factor of 100. It seems like you have a habit of picking fights with people...

Well then why go through sqrt(10) first? If you're going to use a trick like Taylor and write 10=9+1, why not go straight for sqrt(0.1) and write 0.1 = 0.09+0.01?

I got 6m also. Yeah, Taylor series are covered in AP BC Calculus. The types of students these questions are geared towards would most probably know Taylor.

haha...

Why? 11!/4!4!3! is the number of ways to permute 11 objects some of which are identical. But here we're not arranging 11 objects, we're only arranging 4. Here's another approach... If we only use 1 color, there are clearly 3 ways to arrange the table. If there are 2 colors, that's either a)...

Expecting people to memorize numbers like sqrt(10) is absurd. I think this is a better approach: Powers of 2 are much better known. 1024 = 2^10 = 32^2. So sqrt(0.1) is about 0.32.

I was looking at those sums because to me the question seemed to imply that all "handfuls" of marbles are equally likely, small and large ones alike. (How big is your hand??) Of course, that's an unreasonable assumption. To make the question more realistic, we need a reasonable distribution on...

No, that identity is always true. For example, (\binom{2}{1}=\binom{2}{0}+\binom{2}{2}) even though the numbers of terms differ. If 0 marbles are allowed, then the claim in the original post is wrong, and both odd and even are equally likely. If 0 marbles are not allowed, then the...

"But if the pouch contains an odd number of marbles, then we’re more likely to withdraw an odd number, as there’s one more way of choosing an odd number than an even number. For example, if the pouch contains 5 marbles then we’re more likely to draw 1, 3, or 5 than 2 or 4." This is wrong...

1. correct. 2. you're missing an i (sqrt(-1)) in the exponent. 4. correct. 5. no. it's 1/4. Note that 4 and 5 are effectively the same question.

Seems like you're assuming that if the jerk chooses someone else's seat, then the last person necessarily doesn't get his assigned seat. This isn't true. The jerk can choose Johnny's seat, then everything's fine until Johnny comes in and sees that his seat is taken. At this point he can very...

cool. could you elaborate? i'm really curious how you would answer it...

that's like saying there's a fine line between football and gymnastics because they both require athletic skill. How would you get philosophical on the blender question? you mean like, "We are all one. I am the blender and the blender is me. Now I'm beside myself with awe."? There's that...

these questions aren't intended to be philosophical. they're meant to test your resourcefulness and imagination.

I went to CCNY, and that probability course is pretty good. you should also take MATH A7700: Stochastic Processes (a grad course); it would be really useful to any quant program you plan to apply to.

back to the horse race question... the interviewer may ask for an explanation as to why 7 is the minimum. so here goes... say you could guarantee the top three in 6. the first race eliminates 2 horses from contention, and each subsequent race eliminates at most 4 additional horses (and this...

that's not the point of the question...

2) 7 races. Hold 5 races, each with a different set of 5 horses. then race the winners (race #6). now number the initial 5 races 1 thru 5, according to the rankings of the horses in race #6. clearly the winner of race #1 is the overall winner. in race #7, race 2nd and 3rd places of race #1...

you're asking the dealer to pay you 100+1 to enter the game? but he would always lose at least 1, so why would he ever agree to the deal? to be fair, you should ask the dealer for 50 (your expected *loss*) plus (maybe) a little compensation for your risk (if you are risk-averse). if you're...

you're apparently not thinking different...

2 times your money every 2 years clearly means double your money every 2 years. and that's pretty much it. end of story.

according to the phrasing of the question, yes it does

Oh really. So according to you, if I offer you 1x your money every 1 year you'd pick that because you get money sooner? But wait, 1x your money every 1 year means I don't have to increase your salary. Fantastic.

"bollocks"?? Please!

1) "X times your money in X years" seems to mean in general (X^{\frac{t}{X}}) times your money in (t) years. (X^{\frac{1}{X}) is increasing for (X<e), decreasing for (X>e), so the optimal choice would be (e) times your money in (e) years.* 2) what should you pay means what is the fair price...

if \(T\leq 1\) then \(1-e^{-rT}\leq r\), so PCP is violated and the conclusion follows. if \(T>1\), however, then \(1-e^{-rT}>r\) is possible, so the conclusion can't be drawn without additional information. you must be missing something.

i forgot to say that the strategy generalizes to n students and k types of sandwiches, for any n and k: you can always get n-1 students through. :)

you can actually guarantee 9 of the students get internships, as follows. represent hummus by 0, tuna by 1. let S(k) be the sum of the sandwiches that student k sees, modulo 2. the strategy then is this: have student 1 announce S(1). student 2 knows S(2), so he deduces that his sandwich is...

if you're looking for a cure-all in problem-solving, there is of course no such thing. you mostly have to build a feel for how things work through exposure and practice. but there are certain heuristic approaches that are often useful: 1) simplify your problem 2) look at extremes 3) look at...