The Coupon Collector's Problem 

Join my discord server! discord.gg/Kj8QUZU Small typo: 7:02 should be 1/p_{n-3}. Credit to viewer Kyro for pointing this out!

Great video Vivek! :)

I know I'm being nitpicky here, but words can often get very confusing when dealing with probability, so it's correct to say the expected value of a random variable, not an event. Great video Vivek!

Nice. Glad I discovered this channel. You've inspired me and given me confidence that I too can master "manim".

Great video, thanks. I think there's an error in the bar graph at the end, though. It appears to just show multiples of 4: 0, 4, 8, 12, 16, 20... The real series starts out with 0, 1, 3, 5.5... then approaches the multiples of 4 for a while.

First , thanks for this instructive video. Just in the span of time 6:17 of the video, it has been mentionned that the having a new coupon is bernoulli, whereas it is geometric, with probabil(5-i/5)

awesome explanation brother, thanks : )

Well made video! I was frankly surprised how card picking leads us to harmonic numbers (not that I collect cards or anything lol)

Great video! What do you do when the probability of drawing each card is unequal?

Great and very informative video! 👍

what if the probabilities are not equal?

Really nice video! There’s a small typo at 7:02 on the p_{n-3}

Great video! I was wondering, how could this be extended to account for multiple cards per draw (eg. drawing 3 cards, at random, with N total cards)? Would it still be reliant on Expected Values?

Could you please explain to me what was done in 7:18? how do you get from one to another. Thanks

When you did the card drawing animation, you should replace the cards you have drawn. If you have drawn 1 of the 5 cards, the probability to get a card you don't already have will always be 1 since it will be impossible to duplicate.

Love your content!

Oh god, I need to learn the manim library

Very elegant!

i had one doubt can someone please tell me how we can denote E of n as E of n-1 +1/Pn-1 at 6:41

A slightly different way to solve: Again we want a recurrence between E_N and E_{N-1}. After we draw the first card, we are hunting for the remaining N-1 cards. This can be normally done in E_{N-1}, but we have a chance of a "garbage card", i.e., the same card that was drawn at the start, reappearing. This card will appear roughly 1/N of the times afterwards, so the expected value jumps to N/(N-1) E_(N-1). Thus we get the recurrence E_N=1+N/(N-1) E_(N-1), and we can solve it to get E_N=N H_N. Of course this is not very rigorous, so the argument can be made better by considering the expected value E(N,M) of drawing N "good" cards from a possibility of N+M total cards, where M of them are "garbage" cards. We can find a recurrence between E(N,M) and E(N-1,M+1), and use the fact that E(0,K)=0 to get E(N,M)=(N+M)H_N.

YAAAAS GIRLLL!

What is the distribution of this random variable? Is there a closed form expression for the probability mass function?

CS 70 flashbacks

2:06 put some parenthesis (p1=0.5) my eyes hurt from seeing 0.5=5

Love you

Incredible

Hi Vivek and math community, my original intuition was that the first new card be drawn in n/n ways, the second new card drawn in (n-1)/n ways... the last new card can be drawn is 1/n ways. This seems to be how the solution is, but i thought that we would multiply these probabilities, not add, since we want to have the probability of this whole event, which requires the first new card and the second and so on. What’s wrong with this logic?

NoiCe

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First!

This sounds like the cereal box problem: mste.illinois.edu/reese/cereal/