Awesome problem! I earlier saw it in the comments under your another video, and was wondering whether you'd cover it. Thank you! These expectation problems with recursion are very educational!
Thanks for the video! I can propose a problem. U ll see if it ll be interesting for u. You're playing the game of tossing an unsymmetric coin that has a probability of tail =1/3. The game stops when 3rd tail appears. You also can stop the game after 1st or 2nd tail appear and get prize equal to number of tosses between previous tail (not counted) and current (counted). After 3rd tail appears you'd get the prize equal to number of tosses between 2nd and 3rd (counted). The question is how should you play to make the expectance of prize the maximum possible, and what is it equal to.
Example of the game: htthhhht. If u stop after 1st tail, u get 2 conventional units (cu) . If u stop after 2nd tail, u get 1 cu. Otherwise, game stops after 3rd tail and u get 5 cu.
Great Video! But I think it's a mistake using the "the expectation of the sum is the sum of the expectations" property. The property only works in general on independent variables, but in this video x1~x5 are not independent. The property still work in this case due to the fact that it is an uniform distribution. If the combinations have different probabilities(eg. (48,0,0,0,0) vs (46,1,1,0,0)), the "the expectation of the sum is the sum of the expectations" property would be false. Please let me know what you think!
Happy to hear that you enjoyed the video! But, unfortunately, you are mistaked about the lack of general liniarity in expectations. You might be thinking of variances, in which case you are correct that you cannot use this unless they are independent. You can check the link below: brilliant.org/wiki/linearity-of-expectation/ for a short proof in the case of discrete RVs. The proof for continous ones is very similar, swapping sums for integrals. Hope this clears it up!
Hi! I'm sorry this might be a stupid question, but why would the average number of draws not be 13? I assumed that the number of cards pulled before an Ace was a geometric distribution with a probability of success equal to 4/52 or 1/13. Then the expected value of this random variable would just be 1/p or 13 pulls. I can see from your code that this interpretation isn't correct, but I'm not sure why. Would greatly appreciate your insight! Thank you so much :)
Hey just wanted to say tysm for you videos, they are really great. If you have time/want to I would love to see you do some videos on the more open ended questions you get in quant interviews which require you to come up with a 'good enough' strategy@@atypicalquant
@@atypicalquantHey just wanted to say thank you for your videos they are very helpful, also do you know of any good resources to practice the more open-ended questions that are often asked by quant firms these days like the question asked in Jane Street's mock QT interview on youtube? Thanks!
I was able to verify it with induction. But by induction, do you mean you came up with the equations by induction, or proved a solution using induction (and equation came up with pattern matching). I don’t know how I would came up with using induction
You are right, induction can only help you prove the formula once you have it. There is no "one solution" on how to come up with the formula in the first place. For this problem, I considered that it was clear that the values were incrementing with 1/5, so I started from there to write a potential formula for the solution.
thanks for explanation for this problem, but i have a question can you please explain me the part 4:53 in the video, you mentioned symmetry, but i don't understand why?
The five random variables can be reshuffled or relabeled and they will keep their meaning. There is not statistical difference between them (i.e it does not matter that the first variable measure the length of the first split, it can, just as easily mean the third one)
Great video! This might be a stupid question, but can you do it with conditional probability? I've seen this as 'how many flips until first tail' given some set of probabilities for a coin, but is this question different because you are drawing without replacement?
You are right, the difference between this question and one with coin flips is the fact that the probabilities of getting an Ace by pulling cards are not constant. In fact, they are increasing as you extract more cards that are not aces. On the other side, it's not impossible to find the expected number of cards by expanding the probability of getting an Ace as the 1st, 2nd,... up to the 51st card. It would still entail some gruesome computations, and not be nearly as easy as in the case of coin flips.
Hi, i appreciate these videos. I’m an 18 year old greek student currently seating exams and I’d like to ask whether I would be more likely to become a quant if i took a gap year and entered Cambridge math( I’ve already been looking at their entrance exam and I’m pretty good at it) or just study here( Mind you our math department isn’t well known at all but it’s pretty challenging and we know people who went on to become data analysts etc) and keep my money to start building my trading record early on. Also are these questions really meant to be answered in 2-3 minutes seems really hard right now. thank you again for the content.
Hey! Thank you for your comment :). While having a prestigious university in your CV definitely helps, especially in the screening process, it's definitely not a must. Two ways in which you can easily compensate for it: 1. There are quite a few math competitions that you can take part in while studying at your local university that will make up for it in the CV (IMC, Seemous, Ariel's IMIO); some HFs also have some special recruitment pipelines for them. 2. Nowadays some of the HFs have various online opportunities that you can take part in, prior to employment, such as TwoSigma's AlphaStudio. Doing well there will go a long way towards a full time job after graduation. Wrt to having a gap year, why not just start your year there, and decide after? You'll still be able to move after 1 year, dropping your current one if you get admitted, but will know a lot more :). On your last question, while it sometimes is a hard limit (usually on on-line multiple answer test, prior to proper interviews), you won't get a timer on video interviews. But, given the amount of questions you are supposed to answer and their respective difficulties, I tried estimating what your allocated time should be. As you solve more problems, you'll see that it becomes easier :D. As a last note, I don't think you should be concerned about a trading record. No-one has ever asked me about one, nor have I asked anyone in my interviews. It's so hard to check if it's genuine (you could start 100 different portfolios right now and just show your best), really depends on the market, and hardly relates to what happens at your actual job, where your have all the data that you can think of, but have to struggle with auto-impact and meeting various volatility constrains.
Awesome problem! I earlier saw it in the comments under your another video, and was wondering whether you'd cover it. Thank you! These expectation problems with recursion are very educational!
Thank you! Glad to know you are enjoying them!
Genius!! The animation is simply flawless!!
Thank you! Happy to hear!
Thanks for the video! I can propose a problem. U ll see if it ll be interesting for u.
You're playing the game of tossing an unsymmetric coin that has a probability of tail =1/3. The game stops when 3rd tail appears. You also can stop the game after 1st or 2nd tail appear and get prize equal to number of tosses between previous tail (not counted) and current (counted). After 3rd tail appears you'd get the prize equal to number of tosses between 2nd and 3rd (counted). The question is how should you play to make the expectance of prize the maximum possible, and what is it equal to.
Example of the game: htthhhht. If u stop after 1st tail, u get 2 conventional units (cu) . If u stop after 2nd tail, u get 1 cu. Otherwise, game stops after 3rd tail and u get 5 cu.
Btw, it's not my problem, it is from Kolmogorov olympiad
Thanks for the suggestion! Will keep it in mind :D
Superb solutions provided.
Thank you!
Amazing! Thanks for the nice explanation.
Thank you! :)
Great Video! But I think it's a mistake using the "the expectation of the sum is the sum of the expectations" property. The property only works in general on independent variables, but in this video x1~x5 are not independent. The property still work in this case due to the fact that it is an uniform distribution. If the combinations have different probabilities(eg. (48,0,0,0,0) vs (46,1,1,0,0)), the "the expectation of the sum is the sum of the expectations" property would be false. Please let me know what you think!
Happy to hear that you enjoyed the video!
But, unfortunately, you are mistaked about the lack of general liniarity in expectations. You might be thinking of variances, in which case you are correct that you cannot use this unless they are independent. You can check the link below: brilliant.org/wiki/linearity-of-expectation/ for a short proof in the case of discrete RVs. The proof for continous ones is very similar, swapping sums for integrals. Hope this clears it up!
I am convinced now, thanks!🥰
Wow!!!
Hi! I'm sorry this might be a stupid question, but why would the average number of draws not be 13? I assumed that the number of cards pulled before an Ace was a geometric distribution with a probability of success equal to 4/52 or 1/13. Then the expected value of this random variable would just be 1/p or 13 pulls. I can see from your code that this interpretation isn't correct, but I'm not sure why. Would greatly appreciate your insight! Thank you so much :)
If we pulled a card and also shuffled each time then we would expect 13 cards to pull an ace, but since we are doing so without replacement it must be
I think that the explanation @drewkavi6327 provided is a great intuition on why the answer is not 13.
Hey just wanted to say tysm for you videos, they are really great. If you have time/want to I would love to see you do some videos on the more open ended questions you get in quant interviews which require you to come up with a 'good enough' strategy@@atypicalquant
@@atypicalquantHey just wanted to say thank you for your videos they are very helpful, also do you know of any good resources to practice the more open-ended questions that are often asked by quant firms these days like the question asked in Jane Street's mock QT interview on youtube? Thanks!
nicely explained!
Thank you!
I am bit stuck on how to prove n+1/5 by induction. Could you give me hint?
I was able to verify it with induction. But by induction, do you mean you came up with the equations by induction, or proved a solution using induction (and equation came up with pattern matching). I don’t know how I would came up with using induction
You are right, induction can only help you prove the formula once you have it. There is no "one solution" on how to come up with the formula in the first place. For this problem, I considered that it was clear that the values were incrementing with 1/5, so I started from there to write a potential formula for the solution.
Nice solutions
Happy to hear that you enjoyed them!
thanks for explanation for this problem, but i have a question can you please explain me the part 4:53 in the video, you mentioned symmetry, but i don't understand why?
The five random variables can be reshuffled or relabeled and they will keep their meaning. There is not statistical difference between them (i.e it does not matter that the first variable measure the length of the first split, it can, just as easily mean the third one)
Great video! This might be a stupid question, but can you do it with conditional probability? I've seen this as 'how many flips until first tail' given some set of probabilities for a coin, but is this question different because you are drawing without replacement?
You are right, the difference between this question and one with coin flips is the fact that the probabilities of getting an Ace by pulling cards are not constant. In fact, they are increasing as you extract more cards that are not aces.
On the other side, it's not impossible to find the expected number of cards by expanding the probability of getting an Ace as the 1st, 2nd,... up to the 51st card. It would still entail some gruesome computations, and not be nearly as easy as in the case of coin flips.
Hi, i appreciate these videos. I’m an 18 year old greek student currently seating exams and I’d like to ask whether I would be more likely to become a quant if i took a gap year and entered Cambridge math( I’ve already been looking at their entrance exam and I’m pretty good at it) or just study here( Mind you our math department isn’t well known at all but it’s pretty challenging and we know people who went on to become data analysts etc) and keep my money to start building my trading record early on. Also are these questions really meant to be answered in 2-3 minutes seems really hard right now. thank you again for the content.
Hey! Thank you for your comment :).
While having a prestigious university in your CV definitely helps, especially in the screening process, it's definitely not a must. Two ways in which you can easily compensate for it:
1. There are quite a few math competitions that you can take part in while studying at your local university that will make up for it in the CV (IMC, Seemous, Ariel's IMIO); some HFs also have some special recruitment pipelines for them.
2. Nowadays some of the HFs have various online opportunities that you can take part in, prior to employment, such as TwoSigma's AlphaStudio. Doing well there will go a long way towards a full time job after graduation.
Wrt to having a gap year, why not just start your year there, and decide after? You'll still be able to move after 1 year, dropping your current one if you get admitted, but will know a lot more :).
On your last question, while it sometimes is a hard limit (usually on on-line multiple answer test, prior to proper interviews), you won't get a timer on video interviews. But, given the amount of questions you are supposed to answer and their respective difficulties, I tried estimating what your allocated time should be. As you solve more problems, you'll see that it becomes easier :D.
As a last note, I don't think you should be concerned about a trading record. No-one has ever asked me about one, nor have I asked anyone in my interviews. It's so hard to check if it's genuine (you could start 100 different portfolios right now and just show your best), really depends on the market, and hardly relates to what happens at your actual job, where your have all the data that you can think of, but have to struggle with auto-impact and meeting various volatility constrains.
Amazing, how did you creat your animation?
Hey Steve. Thanks for the comment. What particular animation part are you interested in?
How do you add the cards to the animation?
Hey. I'm just using Manim's SVGMobject class and have individual SVGs for each card.