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I think taking the random walk (rather than the birth analogy) is much more helpful in tackling the problem. By breaking the being in position 0 at step 2n to sum of product of being for the first time at zero and returning to zero again in later time, we can derive the probability of first time return to zero.

Yet you didn’t manage to feature someone that is from the UK on this “promo” video? Jokers

The question is asking us to show that the Markov chain of a simple random walk is null recurrent

commodities are over

Pr (1 six) = n * 1/6 * (5/6)**(n-1) Maximize n (5/6)**(n-1) Max at biggest n s.t. (n+1)/n * (5/6) <= 1 two sides are equal at n=5 So you should roll 5 or 6 dies (both have equal value)

Can this be related to the Ballot problem as it is a continuous process with comparison of 2 variables at every stage , and at the end proving a relation between them ?? (Would love to discuss further or other approaches !)

Divide both sides by P_n, then it can be seen that P_6/P_5 = 1

Soft eng @optiver

I'm lost in the theory. Let's see if I can figure it out by actually doing it

if you model it as a markov chain where E_n = 1 + E_(n-1)/2 + E_(n+1)/2 you can solve it with induction, it turns out that E_n = n + n/(n+1) E_(n+1), plugging that in for 1 + E1 which is the total expected number of children you get a divergent series 1+1+1...

Can anyone tell me the specs on the 4 monitor vertical setup? Dell? What size? Thanks.

These are great! Keep em coming :)

or in one line using optimal stopping with martingale (X_n)^2 - n and stopping time T_a : first time to reach a or -a....

That's a good one! Keep posting such problems

damn never ssen vertical monitor that long

Game 1. Bc8 Kramnik🏃🏃🏃

Optiver but chinese

Sounds like a naive joker who stumbled on a scam to rip off pension funds

1:14 did not know that Jared Kushner works at Optiver! Cool!

This video taught me nothing about what they actually do. Could've been anything.

Lisa Su?

hi

I will never get into this firm.

Do people really need 4 big vertical screens? 🤣

These are great! More please :)

im so cooked bruh

I like that 4 screen setup that you have on your floor. What size are the monitors? Thanks.

Scam!

Interesting. I simulated this process and got the same maximum probability when n is 5 and n is 6.

when considering n=5, we are (approximately) choosing a subset of 5 of the set {1, 2, 3, 4, 5, 6}. There are 6C5 subsets possible; only one of these {1, 2, 3, 4, 5} doesn't have the outcome we desire. Therefore 5/6 subsets have the outcome of having at least one six; this makes up for the 5/6 probability term contributed by the extra dice.

1. As we show, it is a binomial distribution, and the peak is actually coming at 5.5 (if n is assumed continuous). Then, we can find 6 and 5 are equal distances from the peak. Considering the symmetry in the distribution of the random variable around the peak( binomial distribution), we can say 5 and 6 have the same probability (maximum if n is discrete, taking only integer value). 2. in the second question, at n=mr & n = mr-1 the probability will be maximum!😉😉 btw I am a big fan of optiver, here in my college ( IITD) you guys hired intern , unfortunately I couldn't appear due to low cg😅😅😅, it's pretty cool to solve problem in this series, keep bringing more. Hope I will get into optiver someday!

I think we can solve this problem in a simpler way using recursion. Let E(x) be the expected number of dice rolls you need to throw to get at least x as sum. Now, just consider the first roll, it can either be 1, 2, 3, 4, 5, or 6. By including all these cases the recurrence becomes E(x) = 1/6 (1 + E(x - 1)) + 1/6(1 + E(x - 2)) + 1/6(1 + E(x - 3)) + 1/6(1 + E(x - 4)) + 1/6(1 + E(x - 5)) + 1/6(1 + E(x - 6)). We know E(x) = 0 if x <= 0, now just find E(6). It is not difficult to see that E(x) = (1 + 1/6)^(x - 1).

aah dream of iitians!

You should roll r*m or r*m - 1 dice (both maximise the probability).

For those interested, this is a hypergeometric distribution problem. en.wikipedia.org/wiki/Hypergeometric_distribution

It's interesting because the probabilistic binomial distribution solutions are the same as the expectation of negative binomial distributions, which is where I assume people get their "intuition" from. (Intuition seems to come much more quickly from expectations). Like on a nuanced level probability maximisation in one distribution shouldn't necessarily be linked to expectation in a different distribution, but intuitively in your mind it feels much more appropriate initially (at least to me) to calculate the expectation.

What type of problems this is called?

The reason 5 or 6 does not matter is because there are two cases where rolling a 6th dice makes a difference to the outcome of our game : 1. the first 5 dice did not contain any 6 2. the first 5 dice contained exactly one 6 For case 1, which happens with probability (5/6)^5, if we roll a six, with probability 1/6, we turn a "loss" into a "win", and therefore increase the probability of winning by (5/6)^5 * 1/6 For case 2, which happens with probability 5/6 * (5/6)^4 = (5/6)^5, if we roll a six, with probability 1/6, we turn a "win" into a "loss", and therefore decrease the probability of winning by (5/6)^5 * 1/6 Notice these are the same amounts! Therefore the "marginal benefit" (difference in probability of success between rolling 5 and 6 dice) is exactly 0.

This is one is quite straightforward…

Love it :) Good luck on my Optiver OA today :)

Actually, for a n*n grid, it takes n-1 steps instead of n steps for you and your friend to meet.

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hhhh good thanks its fun to hear

Bonus question is easy. Intuitively, most people can tell player 2 has the advantage, but you can prove it easily using bounds. Suppose you go first. You win if it takes 1, 3, or 5 rolls to hit 6+. Now for each of these scenarios, you can just assume you hit the lottery and always get the highest possible sum. The highest possible sum if it takes 1 roll is 6. If it takes more than 1 roll, the highest sum you can get is 11 (5 from the previous n-1 rolls + 6 on the nth roll). So for player 1, just assume the best case scenario for all possibilities and that is the upper bound of his expectation = P(1 roll)*6+P(3 rolls)*11+P(5 rolls)*11 = 3.5873 from the formula in the video. Now assume the worst case scenario for player 2, which is that the sum is always exactly 6 and not any higher. Player 2 wins when the game terminates at roll 2, 4, or 6. Lower bound for the expectation of player 2 = P(2 rolls)*6+P(4 rolls)*6+P(6 rolls)*6 = 3.5887, just barely greater than 3.5873. Since the lower bound of P2's expected winnings is higher than the upper bound of P1's expected winnings, you should be player 2.

import numpy as np prior_distribution=np.array([1]+[0]*5) dice = np.array([0]+[1/6]*6) for i in range(6): new_distribution = np.convolve(prior_distribution, dice) payout = np.sum((np.arange(12)*new_distribution)[6:]) prior_distribution=new_distribution[:6] print(f"Expected payoff at round {i+1} is {payout}") Output: Expected payoff at round 1 is 1.0 Expected payoff at round 2 is 4.305555555555555 Expected payoff at round 3 is 1.875 Expected payoff at round 4 is 0.35185185185185186 Expected payoff at round 5 is 0.03137860082304526 Expected payoff at round 6 is 0.0010931069958847735 Even rounds win $1.75 more.

Toss 1st, Ev = 45200/15552. Toss 2nd, Ev = 72449/15552.

There’s also a really elegant way of doing this with Markov Chains - one that gets absorbed to the sum of at least six (the states can be your sum and then you can find the mean absorption time)

Lee George Lee James Young Ronald

yoo who made the melody need some loops

customer support is not working can you give me a number or website that i can contact you