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Quantitative Interview questions and answers

What? I got an answer, but it was complete calculator punching and guess and test.

When you get nested radicals, that series quickly converges into...something, so that something will become readily apparent after a few minutes of punching buttons into a calculator.

I couldn't even come up with a C++ quickie to get that, though, since I have no idea how to code nested radicals. I'm guessing it's some sort of string creation and then a re-cast to a double, but I haven't done that before.
 
quantyst said:
Compute RAD{1+RAD{2+RAD{3+RAD{4+RAD{5+....}}}...}.
Click to expand...

We need to prove whether or not the sequence converges. It should help if we can develop a recursive relation that generates the sequence. In situations like this a bivariate sequence is needed. Here is one:

Define f(m,n)=RAD{(m-n+1) + f(m,n-1)}.

Then f(m,n)=RAD{(m-n+1) + RAD{(m-n+2) + RAD{(m-n+3) + RAD{(m-n+4) + f(m,n-4) }}}}.

Of course, the preceding can be continued until we reach f(m,0) as follows:

f(m,n)=RAD{(m-n+1) + RAD{(m-n+2) + RAD{(m-n+3) + ... RAD{m + f(m,0)} ... }}}.

At this point we can define f(m,0)=0 and then set m=n to get:

f(m,m)=RAD{1 + RAD{2 + RAD{3 + ... RAD{m} ... }}}.

We need to show whether or not the limit of f(m,m) as m approachs infinity exits.

Although there is a lot left to be done, I hope the above helps.
 
You are trapped atop a building 200m high. You have a rope 150m long. There is a hook at the top where you stand. Looking down, you notice that midway between you and the ground, at a height of 100m, there is a ledge with another hook. In your pocket lies a Swiss knife. How do you come down using the rope, the two hooks and the Swiss knife?
 
Hook the rope up to the 200 meter hook.
Climb down and hook the rope at the 100 meter hook.
Continue climbing down to near the end of the rope. Tie a loop around your waist.
Climb up to the 149 meter mark and cut the rope.
The rope should fall down to the 100 meter mark, leaving you right around ground level.
Untie yourself and come down.
 
radosr said:
You are trapped atop a building 200m high. You have a rope 150m long. There is a hook at the top where you stand. Looking down, you notice that midway between you and the ground, at a height of 100m, there is a ledge with another hook. In your pocket lies a Swiss knife. How do you come down using the rope, the two hooks and the Swiss knife?
Click to expand...

Are the hooks movable, transportable? Or are they stuck in their original locations?
 
Greetings Quansters!

Hi It's been a while since I last post a thread in this Forum.
:tiphat:
I'm very much impressed by the high volume of activity here.

Just wanted to say 'hi' to everybody and try to contribute a little to this thriving community
by posting some brain-teasers/problems (some interesting ones) pertinent to Quant Finance.

(I'll post solutions 7 days after):smt024:D

So yea, if you find yourself, on let's say a slow Friday or Saturday night and can't
figure out what to do with yourself.

Feel free to work on the posted problems and
perhaps share some of your TOP 10 favorite interview/brain-teaser questions too.
 
Some Simple problems to get started

Ok here's 4 simple problems to look at.

Good practice problem if you are new to Quant finance.

Just covering the basics here.

Have fun.

You could post the answer in this thread.
 

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World Series Baseball Betting Probabilities

Folks here's a question that I need help on.
I know we need to have binomial trees set up to find the solution.

Let me know what you guys/gals think.
Enjoy
------------------------------------------------------------------------------------------------------------
Imagine that you would like to bet $500 on the outcome of the World Series, say the
Yankees versus the Red Sox. Ignore the possibility of ties. The first team to accumulate four
wins is declared the winner. There are seven games possible. You go to your bookmaker and
say you would like to bet $500 on the Red Sox winning the series. But he says he is only
taking bets on the outcome of each individual game, not the series. How do you structure
your bet so that the net outcome of the bet on each game in the series is equivalent to betting
$500 on the entire series (i.e. Red Sox win the series, you win $500, Red Sox lose the series,
you lose $500). Assume neither team is favored in the house betting line. When you give
your answer, circle your answer showing the amount you should bet on the first game.
​
 
How do I circle my answer on quantnet? ;)

There is a thread with this exact question from last summer with the solution, but not using binomial trees.
 
This is surely wrong

After performing the formal manipulation, does anyone botehred to see whetehr the answer make sense? How do you think taking 1.4141 to the power 1.4141 and the result of that to the 1.4141 and so on we can converge to 2?

bob said:
No, the answer to this is (x=\sqrt{2})

(a=x^{x^{\ldots}}=2)

(ln(x^{x^{\ldots}}) = ln2)

(x^{x^{\ldots}}lnx = ln2)

(a lnx = ln2)

(2lnx = ln2)

(lnx = \frac{1}{2}ln2)

(lnx = ln\sqrt{2})

(x = \sqrt{2})
Click to expand...
 
Wrong again

2 points on circle can define a semicircle in an infinite number of ways - that's the fault in your logic. This problem is related to a standard problem of linear separability in pattern recognition. You can also convert it in to a problem of finding a distribution for the minimum number in a set of N numbers selected from a uniform distribution defined on [0, 1].


Andy said:
Correct. Since the sides are from a unit length, we can calculate that any side of the triangle must be strictly less than 1/2. The prob that the length of each stick being less than 1/2 is 50% and this condition must hold for all three of them. Hence I have 1/8


Not totally convinced they are related or merely coincidence but I thought of the semi-circle as follow:
Any two points on a circle will be on the same semi-circle (just connect one point to the center, the second point must be on one of the two semi-circles. Pick the one that contains two points). The problem now becomes finding the probability that the third point is on that same semi-circle. There are only 2 semi-circles to begin with so the chance is 50%

Welcome any opinions since with 1,2 minutes to come up with an answer, there are pretty good chance that I may have big holes in my reasonings. I hope to learn from my mistakes and not to repeat them again in future occasions.
Click to expand...
 
DamirL said:
After performing the formal manipulation, does anyone botehred to see whetehr the answer make sense? How do you think taking 1.4141 to the power 1.4141 and the result of that to the 1.4141 and so on we can converge to 2?
Click to expand...
Think about it. If you have (x^Y) where (x=\sqrt{2}) and (Y) is the series they give you at the begining that equals 2, then isn't it true that (x^Y=2)
 
Andy said:
Think about it. If you have (x^Y) where (x=\sqrt{2}) and (Y) is the series they give you at the begining that equals 2, then isn't it true that (x^Y=2)
Click to expand...

Correct. ( a^b^c \ne (a^b)^c = a^{(bc)} ), however (a^b^c = a^{(b^c)})

Of all the hundreds of people who have read this thread...
 
Here's one that if a quant applicant can't get, you throw them out the door with the closest thing that passes for a catapult.

Two heterosexual couples are determined to have group sex between all sets of male-female couples, and none want to catch any potential STD, but only have two condoms.

In what procedures is this event possible?


A more challenging follow up would be: what is the minimum number of condoms required for N such couples?
 
QUOTE:
Assume you have (%5Cpi) up to 1M decimal, write an algorithm to find the first matching pair of 3-digit numbers. What is your bigO?
Example: if (%5Cpi)= 3.14159525957... then the first pair is 595


A question:
Suppose (%5Cpi)= 3.141585857026001410089...

What's the answer now? Is it 141 or is it 585?
 
Andy said:
Assume you have (\pi) up to 1M decimal, write an algorithm to find the first matching pair of 3-digit numbers. What is your bigO?
Example: if (\pi)= 3.14159525957... then the first pair is 595
Click to expand...

Let's make an initial observation. The knowledge of one million (1M) digits to the right of the decimal point of (\pi) is way more than necessary. The first matching of two 3-digit numbers occurs within the first 3003 digits. Here's why:

Let (\pi)=3.x(1)x(2)x(3)x(4)...where x(i) denotes the i-th digit to the right of the decimal point (TTROTDP). Now set a(i)=x(3i-2), b(i)=x(3i-1), c(i)=x(3i). Then we can express (\pi) as follows:

(\pi)=3.a(1)b(1)c(1)a(2)b(2)c(2)a(3)b(3)c(3)...

Q: How many different a(i)b(i)c(i) are there?

A: 1000

So, by the Pigeonhole Principle, within the first 1001 of such a(i)b(i)c(i), two must be identical. This means that within the first 3003 of the x(i), we get a matching of two 3-digit numbers. So, the first matching occurs within the first 3003 digits TTROTDP.
 
A LA ANDY:

Assume you have (\pi) up to a sequence S of one million digits to the right of the decimal point. Write an algorithm to find the largest matching pair of n-digit subsequences of consecutive numbers within the sequence S.
 
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