Fred again.. | Boiler Room: London

Expected value ought to be zero, right? Since given any configuration, it’s equally as probable as the same configuration but with the first two rows swapped, and they have negated determinants.

I guess so, you can find the det by summing the det of all 2x2 matrice in a specific dev, and you can check the event are independent
Then E(det(2x2)) = E(X1*X2 -X3*X4) = E²(X)-E²(X) =0, So Sun(E(det2x2)) = 0, E =0
It mean the value of det have a average value of 0 but we have no info how often it will have exactly 0 as a value.
Also notice if X = [0; inf [ we will have some issue to calculate the E(X = inf) bc of an infinite sum
Maybe if use 1+2+3+4+5+6... =-1/12 we can have some clue that proof is stable in infinite dim.

a much simpler way could be:
minimum no. of turns required=1 (when traveling along the edge only)
max no. of turns = 15 ( when traveling down stepwise)
by symmetry and intuition, we say that she can choose her path containing any no. of turns between 1-15. there are 8 odd nos. between 1-15 (both included). ans= 8/15

Similar to 2024 AIME I #6

this is a back tracking problem

Great Videos bro!

I answered .50 after .50 seconds. Am I hired?

That’s really clever to think of it like a permutation of 8 Rs and 8 Us. I think from there you can just calculate the probability that the permutation starts with a different letter than it ends. Without loss of generality, assume the first letter is U. Then there are 8 Rs and 7 Us which could be the last letter, giving 8/15.

Thanks a lot man. great vdo

I did like any number of V's between two consecutive H's give rise to even number of turns and vice versa. So if I start with H (horizontal direction) I have to end in V to get an odd number of turns and vice versa. So i permute the remaining 7 H's and 7 V's and twice them because I may be starting either H or V.

Okay I got it why quant get paid so much money.

This is literally just a redesign of a recent Putnam problem 😂

Wait that’s it?? Doesn’t seem so bad honestly

I just counted the number of paths with an odd or even number of turns for matrices of size 1x1, 2x2, 3x3 and 4x4, noting that the results satisfy n/(2n-1), where n is the number of rows/columns. For n=8 we get 8/15 (correct solution). Intuitively, it also makes sense that the probability of having an odd number of turns tends to 50% as n gets larger, because the original biais of having one more odd path by following the edge gets swamped in the multitude of other paths. I'm not sure how to show that this is true, so if anyone has an idea, I'd love to hear it.

Suppose that we start going to the right (the same applies in the other case by symmetry). Odd number of changes <=> the final step is up. There are 15 steps to be determined, 8 of which are up, so the probability of this is just 8/15.

Should have specified some limitations on the path. "All paths equally likely" makes me think every step is not necessarily in a direction towards the goal, so we could in theory go left/down if the grid allows it.

i remember this question being on my combinatorics exam in uni

A lot of these quant interview problems are just like the ones you'd get for a Cambridge/ox maths interview which is kind of interesting

I think the easier way to think of it is “what’s the maximum number of direction changes I can make?” (15). What’s the minimum number of direction changes I can make (1). Between 1-15, there are 8 odd numbers, so 8/15. This way is much easier because you can just do it visually

in my calculation the answer to nd problem is 0 cause from formula sum over permutation p (-1)^is_odd(p) * sum a_{i, p_i} gives us an intuition. There are same number of even and odd permutation for n = 4 so ev of that is 0. I even wrote a simulation and for 10^7 iterations the result is 0.026 avg, 2e7 std dev. Also the interpretation of random signed hypervolume, whose vectors are from 1 to 15 is also a convinient intuition. Next this i noticed is that if we swap two rows we get the -determinans so matrixes are symetrical is some sense or equal to 0 if rows are equal.

This was a midterm review question in my probability course at Berkeley.

E[Det[matrix]] = 0

Welp I'll never work a citadel lol

aight I didn't wanna be a quant anyway...

I have no idea how that solution came in but I am glad that someone solved it and cracked the interview lol

None of this matters.

Please talk slowly and explain more details

you can also do this with another cool trick. Assign 1 to going up and -1 to going right. Then a pivot is going from -1 to 1, and we can extrapolate this to a continuous function just by linear interpolataion between the indices. We know if the function starts and ends at the same spot, then it crosses the x axis an even number of times, and if it starts and ends on different values, then it crosses the x axis an odd number of times. we then know that our probability is determined by the number of permutations where the start and end are different (or 2 times the number of permutations of seven 1s and seven -1s) divided by the number of permutations of the big multiset

Is it an AI narrating the video? Just curious. If it is, it works pretty well!

This problem is equivalent to that of finding the number of ways to get an odd number when you add up the outcomes of throwing two 8-sided dice.

It's like hearing that lethargic phrase "when we get on the grid" referring to VICS, when (as humans) we've been on it since birth. trying to smile more on Mondays

Another way, if going up first, then there will be an odd number of paths if last path is going right. The probability of last move is right given first is up is 8/15. By symmetry this is same for right first

P(odd) = 1/2 * p(oddㅣ up first) + 1/2 *p(odd | right first) = 8/15

got it first try letsgo