The Coupon Collector's Problem (with Geoff Marshall)

[@standupmaths]

Ok, many are suggestion I should have stood up to reveal an even bigger table next to me. Great concept, but ideas like that require some serious resources. *cough* patreon.com/standupmaths

[@johnchessant3012]

Hi

[@_wetmath_]

second

[@ScientiaHistoria]

…and there was the recursive “first a sense-check”before we start the sense-check. As usual, I wish I had undertaken another layer of sense-check before watching a Matt video.

[@Eli-su6ql]

Nobody noticed the "diverges" was fixed in post Matt. good job.

[@ScientiaHistoria]

@@Eli-su6ql I did but figured it was his math autocorrect tool.

[@ryanparker260]

You were right, we all knew there was a second even SMALLER miniature table prop

[@b0nce]

And that makes us very happy :)

[@darkshoxx]

I was kinda expecting him to go out a layer as well, and standing up from the table between a giant clock and calendar prop

[@bl4cksp1d3r]

I was thinking, he wouldn't have stopped with one 1/10 scale model, and I knew it, I was very happy to see that

[@Avodroc42]

and it was absolutely worth it

[@tandemcart1234]

I legitimately laughed out loud with relief when the smaller one came out. The pause where he should have got it was just a smidgen too long. Perfection!

[@IMacar]

Recursive tables was definitely the pro-KZbinr move.

[@Anonymous-df8it]

I would like this, but it's at 420 likes so...

[@Anonymous-df8it]

Guess I'll have to wait until 669 likes!

[@mattduffyw99]

The second layer got me. Earned the thumbs up

[@koenschaper8821]

It reminded me of something Vsauce would do. Who, by all means, is a certified pro-KZbinr.

[@joshuascholar3220]

The third table got an instant up-vote!

[@itsmattnelson]

Thank you for having me as a guest! My official parkrun time was confirmed to be still *one* second out 😭

[@wordzmyth]

Thank you for sharing this! A little shame you couldn't have texted him on the day. Statistically, even sandbagging it should take a few attempts, so you prove the point

[@chonchjohnch]

Subbed, I need motivation to get back into cardio

[@monkeycigs4762]

It's been a few months, have you gotten your time?? Fingers crossed for you!

[@djadj_]

showcasing your prop ability whilst explaining probability, what a beautiful moment

[@charliebalfour4051]

Nice. A djadj joke.

[@zyaicob]

+

[@wishiwasabear]

The way Matt could read our minds with the third level of recursion was a very neat trick.

[@vigilantcosmicpenguin8721]

Recursive patterns are predictable, but not as predictable as people making jokes about recursive patterns.

[@michaeldirmeyer11]

@@vigilantcosmicpenguin8721 People making jokes about recursive patterns are predictable, but not as predictable as people making jokes about people making jokes about recursive patterns.

[@clarencelam1907]

You can't conclude that the harmonic series diverges just because the expected time goes to infinity. The expected time reaches infinity because n goes to infinity. n being finite doesn't mean that the harmonic series goes to infinity; it just so happens that both n and nth harmonic number go to infinity. If the n out front were replaced with a *constant*, then you could conclude that. As an example, consider the function f(n) = n(1+1/2+1/4+...+1/2^i+...+1/2^n). f(n) reaches infinity as n goes to infinity, but clearly (1+1/2+1/4+...+1/2^i+...+1/2^n) doesn't diverge; it's always less than 2. So the argument here doesn't work.

[@hOREP245]

Parker divergence of a series

[@jfb-]

Parker proof

[@standupmaths]

I think you’re right: that lead n breaks my divergent observation. I suspect the result may be salvageable but not in any intuitive way.

[@fejfo6559]

I think the argument can be saved if you observe the average time needed to collect a coupon ( n(1+1/2+...+1/n)/n ) diverges as the number of coupons goes to infinity.

[@jordanlinus6178]

@@fejfo6559 The problem is, that is not intuitive. The first coupon always takes one try, the one in the middle on average 2. Sure, the last one takes on average n, but that might be negligible among the n coupons. It's not that hard to prove that the harmonic series diverges, but I don't think the park runs can give an easier explanation.

[@RolandWolf]

A park run special, as opposed to a Parker run, where you give running a go, but don't really get the result you wanted.

[@plaguey23]

I was going to make a similar joke but take my like instead.

[@unvergebeneid]

It's sad Matt doesn't run anymore. He could've earned himself the nickname "Park Run Parker"! You know, basically the opposite of "Run, Forrest, run!"

[@SpassNVDR]

@@unvergebeneid Wow, I got to laugh three times at this, understanding one little detail at a time :D

[@unvergebeneid]

@@SpassNVDR 😄😄😄

[@pmoncr]

@@unvergebeneid Is a parkrun parker someone who turns up at parkruns and doesn't get out of their car? Matt could then be the parkrun Parker^2, rearranging would make him the park^3 runerer.

[@SellusionStar]

This recursion joke was no joke. It's a nerd's duty.

[@nitehawk86]

This recursion joke was no joke. It's a nerd's duty.

[@MartinJab]

This recursion joke was no joke. It's a nerd's duty.

[@TlalocTemporal]

I hate to break it to you guys, but due to how YT comments work, you can only do one recursion. All the rest would be iteration jokes. This technically correct joke was no joke, It's a nerd's duty.

[@zyaicob]

@@TlalocTemporal thank you i knew something was off

[@Kaepsele337]

I don't think the seconds would be uniformly distributed even when you're not trying. That would require your time to fluctuate much more than a minute and I think most people run more consistent times. Also, while training you gradually increase your time and might "scan" through a minute, so that way you'd need less runs than if it was randomly distributed.

[@tth-2507]

Hi, runner here. Of course I run a consistent time (when lucky, even slightly increasing), but not that consistent. A variation of +/-1min is to be expected - at least in my case. Additionally one has to take different terain features across locations into account.

[@alimanski7941]

If you scan over a minute, there's not an insignificant chance of missing a seconds value. If you converge on a run time, which is a reasonable assumption for most runners, then your chances of achieving previously skipped times are much, much lower, thereby increasing the number of runs necessary. So, even though I agree with your modelling, I think a uniformity assumption is still a safe approximation.

[@Kaepsele337]

@@tth-2507 Yeah I was thinking about time per kilometer, which is pretty consistent for me (basically between 4min 20 and 4min 30 every time). I forgot that you have to multiply the spread by 5 for 5km obviously. It would still cluster, but less than I had in mind.

[@viniciusfriasaleite8016]

It would be cool to see the time distribution of a runner on the park run

[@kane2742]

Matt's time (Runderground Matt, not Matt Parker) was around 22 minutes. At that pace, a variation of a minute is less than 5%. That seems reasonable, especially given variable weather and terrain - some parks are going to be hillier than others, for example.

[@ALMX5DP]

I was so pumped to start this challenge, knowing I had a 60/60 chance of getting my first 'coupon.' Little did I know that you actually had to finish the run to do so...

[@karl9840]

As someone writing my Bachelor's on this exact problem (and the Poisson Process) this was a gem to watch.

[@viniciusfriasaleite8016]

Luckier than all those runners!

[@ajschlem]

What are you majoring in?

[@DonReba]

By "this exact problem" you mean tables with unnecessary props, right?

[@karl9840]

@@DonReba I wish!

[@karl9840]

@@ajschlem Well, technically I'll be a maths and physics teacher, but I do get the swedish equivalece of a bachelors in maths (and physics if I just write the thesis since im eligible for it).

[@johnchessant3012]

There's actually a recursive solution to this problem. Let f(n) be the answer for n coupons. Your first coupon is guaranteed to be a new one, after which you're left with n-1 coupons to collect, except, you have probability 1/n of getting your first coupon again so only (n-1)/n of your attempts matter. So f(n) = 1 + n*f(n-1)/(n-1). Divide both sides by n to get f(n)/n = f(n-1)/(n-1) + 1/n. Thus f(n)/n is the harmonic series up to 1/n, as expected.

[@bigmoneysam8820]

The recursive tables gag really put the 'stand-up' in 'Stand-up Maths'.

[@joelluber]

Puts the sit down in stand-up math. Lol

[@SomeRandomDevOpsGuy]

Is 2 layers even enough to deduce recursion?

[@tymo7777]

Really upset you missed the “run the numbers” pun!

[@pembrokeshiredan]

Not to mention the Parker Run pun

[@bill_and_amanda]

I came here to say this

[@thegreatmup]

No he didn't, 1:18

[@Whatwhat3434]

He says it 10:57 as well

[@onebronx]

16:03 Matt -single-handedly- bi-pedally saved the narrative of this video.

[@lunasophia9002]

4:59 I love you, Matt. I was hoping for it, wishing in my heart, and you did it!

[@mathmachine4266]

The mean value would be n*(1+1/2+1/3+...+1/n). In that case, that would be 60*(1+1/2+1/3+...+1/60), or 280.7922. As you already mentioned. The variance, however, would be n²(1+1/2²+1/3²+...+1/n²) minus the mean. In this case, that would be 60²(1+1/4+1/9+1/16+...+1/60²) - 280.7922, or 5581.4676. That means the standard deviation is the square root of that, or 74.7092. So, for him to get so far under the expected value is not really that out of the ordinary.

[@gmalivuk]

Yeah, I just ran a bunch of simulations, and the complete set occurs by run 229 a bit under 28% of the time.

[@driwen]

isnt that the average value is n*(1+1/2+1/3+...+1/n), but the mean value should be lower shouldnt it? The distribution of 1 out of 60 will be 1 to infinite. Which pulls the average tries needed to higher number than the mean. edit: sorry got confused with median. But I'm curious if the average or mean is the value people are really interested in. Or the value at which 50% of the people would have completed it

[@gmalivuk]

@@driwen The mean is exactly the expected value calculation done in the video. That's usually what we mean by average. The median is more complicated to calculate, but ends up being 267.5.

[@TheMetallerik]

So I've run 1 milion loops (simulations). Average got pretty close: 281.78, min: 103 max: 1146

[@driwen]

@@gmalivuk yeah as i said after my edit i got the median and mean confused. But this shows that we wont see a bell curve around 281 but before 267.

[@PsiVolt]

The recursion bit was incredible, I might have to use that! This video is giving me discrete math flashbacks

[@Illumas]

Me, "But you didn't make a tinier table prop for your tiny table prop." Mat, "You know I did!" Me, "Yay"

[@ulriksteenandersen4215]

Love the jokes and props; never stop, Matt : )

[@melglobus]

Two of my favourite KZbinrs together again! The platform 0 video made me subscribe here. Loved the Choose Corrour T-shirt too!!

[@smor729]

So what you are saying is that to run every single possible trailing decimal amount of seconds, all I have to do is run 1/12th of one park run backwards? This should be easy!

[@mijkolsmith]

-60/12

[@ghislainbugnicourt3709]

I might have missed something, but the -1/12 or -60/12 joke would have worked only if there was the (1+2+3+...) series instead of the harmonic one, right ?

[@David94spc]

@@ghislainbugnicourt3709 joke worked fine since you got it 😘

[@_wetmath_]

11:40 the camera man awkwardly walking past the two other guys talking was hilarious but completely relatable

[@anfanta2010]

I just want to validate that the extra effort to build out the props was absolutely worth it. I was laughing out loud by myself 🤣

[@TheInternetHelpdeskPlays]

This reminds me of the old seaside Fascination games where you had to sink balls in holes, 1 in each. At the start youd get loads but as you get closer to the end it'd get harder and harder to get the final ones.

[@sbyrstall]

Thanks for giving the parkrun a shout out. I now have to cross post this in the Global Running Channel. They would probably get a kick out of it. I didn't know that there was a Parkrun Bingo.....in do now.

[@BobberWCC]

Harmonic series discovered from park runners. Amazing.

[@LukeSumIpsePatremTe]

I love the 10:45 "We've managed to prove that harmonic series -converges- *DIVERGES* "

[@zachrodan7543]

I feel like a more modern name for this problem might be the (unweighted) lootbox completion problem... (the weighted lootbox problem would be where different outcomes have different probabilities)

[@Adrianmk2208]

A park run in which you almost finish, but not quite, is known as a Parker run.

[@charliedobbie8916]

Let me tell you a joke about recursion: two people were sitting at a table, and one turned to the other and said "let me tell you a joke about recursion:"

[@VAXHeadroom]

In one of the early copies of the VRTX operating system documentation there were two entries: Recursion: see Hofstadter, Douglas Hofstadter, Douglas: see Recursion It made the nerd in me laugh out loud...unfortunately nobody else in the room got the joke...

[@Pseudomous]

Pete and repeat were sitting on a bridge. Pete fell off. Who was left?

[@nathankarn5557]

@@Pseudomous Repeat?

[@celestialtree8602]

I was hoping for the third recursion level, but didn't expect you to do it. And I was very pleasantly surprised.

[@robertaries2974]

Geoff Marshall Collab. Gonna be a great video

[@Anonymous-df8it]

I was kinda expecting him to go out a layer as well, and standing up from the table between a giant clock and calendar prop.

[@morscoronam3779]

10:48 Sounds like editing Matt had to edit the right word in. 🤔 Why do I notice these things...

[@anthonydillon2969]

Can anyone read lips to see what he really said?

[@ARKGAMING]

I was waiting for the second prop table Glad you didn't disappoint

[@CR0SBO]

The initial Matt Parker comparison to the props seemed perfectly proportionally sized, but the Matt Parker that we had for the prop set of props was at least an order of magnitude too large, never mind the Matt Parker that was presenting the prop set of prop props!

[@Schlups]

Next challenge: Do the run when a leap second is introduced to tick off the number 60.

[@nathanrcoe1132]

that is possible with an absolute position in time, but never with a duration, I think

[@jurjenbos228]

If the stopwatch is coded by an average programmer, yes.

[@henrym5034]

@@jurjenbos228 but how should the result be displayed for the 61s minute case?

[@jazzabighits4473]

@@henrym5034 61s in minutes and seconds is 1 min 01 seconds, so 01 I guess?

[@henrym5034]

@@jazzabighits4473 I mean it’s definitely correct to say 2017/01/01 00:00:00 is 61 seconds past 2016/12/31 23:59:00. It’s also correct to say it’s 1 minute past that (that minute has 61 seconds). That makes me wonder if it’s okay to say it’s “1 minute and 1 second” though.

[@samp-w7439]

I'm very excited because after Matt stated the problem, I figured out the formula for myself and calculated, got 281, and was very happy when I skipped to the reveal, and he had the same answer!

[@gnfnrf]

All of this was interesting, but I was expecting an entirely different set of math about the odds of completing a 1/n task in n attempts, which is not 50%. If I remember correctly, as n increases, those odds converge on 1-1/e, and its fun to see how the formula to calculate it resembles (one of) the formulae for e.

[@jeffkaylin892]

Yeah, I was pondering this instead of sleeping... If I were to catch a bus, which comes once an hour, my expectation is to not wait for more than half an hour. If the bus were there I'd say it was a miracle. If I had to wait 59 minutes I'd say I was jinxed. But if I waited over an hour I'd say I wasn't paying attention. This probability starts as 1 / 60. So 59 / 60 it wasn't there. Then next minute would be multiplied by 58 / 59, and the next 57 / 58. Hmm... I could multiply that all out... hmm cancel the 59s, then cancel the 58s... so at 30 minutes I have 30 / 60 just as one would expect. BUT, if the bus doesn't come once an hour, but has a 1 / 60 chance of having left the depot, then there is a string of 59 / 60 multiplied together. That would make my expected wait longer. And by "expected" I mean the "life is fair" type of expectation where half the time I'm pleasantly surprised and half the time I'm a little disappointed, and very rarely see miracles or damnations.

[@j.rodolfoprz7713]

this ‘fun activity’ will be in my personal purgatory

[@sorenwestrey4925]

Legendary crossover

[@luca6819]

Recently I saw a rerun on an old TV show where scientists were rating crazy inventions or build made by people (usually using KZbin videos), and you were there! Didn't remembered that, that was a nice surprise!

[@DaTux91]

Matt was like "if I can find them" and I looked at the remaining duration of the video and I was like "he couldn't find them". And that made me sad.

[@okRegan]

that recursion gag is the reason no mater how uninterested i am in the title, i will watch any video you put out, you're awesome!

[@jerry3790]

15:25 “I used to be a runner like you, but then I took an arrow to the knee”

[@jakebradley3998]

Holy crap man you're so close to the big milli! Good Luck!

[@anuzis]

This analysis makes the unsafe assumption of a uniform distribution of finishing times across the seconds. Over long enough distances this this assumption is likely increasingly safe, but consider the distribution of finishing times you'd see for a 50 meter dash: probably a Gaussian distribution (AKA "bell curve"; with Olympic sprinters at one end, couch potatoes at the other, and most of us around the middle). Even at longer distances like 500 meters - 1k meters you likely still don't see a uniform distribution across the seconds. Not to nit-pick: I loved the video and really appreciated seeing the approach taken, just trying to particulate as a supportive KZbin collaborator thinking about other ways to refine the theoretical analysis. Looking forward to future episodes!

[@GabeUnger]

Getting so close to 1 mil Matt! Hope you have a good video idea to celebrate:)

[@tuliosabatino]

The props were definitely helpful to demonstrate your point, Matt. Time well spent indeed

[@ChristianNiederhuber]

encouraged by this video I did a little experiment: I implemented a program to challenge your calculation experimentally ... I did 1.000 instances of this croupon collector´s game from 00 to 59 and my experiment came up with a mean value of 280,946054 tries on average - so this seems to be an experimental confirmation of your calculation ... BUT: the median-value in this sample was only 266 tries (minimum value 147 and maximum value 684) so the distribution of the result values is quite right-skewed, because of some few values pretty far on the long right end of the scale ... now what always fascinates me the most in such cases of right-skewed distributions is, that if you just take any logarithm of the values instead of the original values, then you immediatly get almost perfectly normal distributed log-values ! and if you take the mean value of this transformed log-values and transform it back to the original scale, then you receive a value that is very near to the median of the original distribution ! (in my case 269) how those this "trick" work and where does this relationship exactly come from ? maybe you could also make a video about this kind of transformation once in a while ?

[@Cannongabang]

What about the standard deviation? A naive calculation of mine results in ~83. Let me know when you have time !

[@ChristianNiederhuber]

@@Cannongabang ~74,64

[@seanc6128]

I appreciate the gift of laughter in addition to the gift of knowledge.

[@Cr42yguy]

I was waiting for the prop on a prop table. Thanks for not letting me down, Matt.

[@sbartdbarcelona44]

The miniatures were definitely worth the extra effort. Thx for the fun.

[@findlaysmith6280]

Nice save at 10:48 🤣

[@KarimMaassen]

“diverges”

[@mumblbeebee6546]

Great video as always, but for me the highlight was to see Geoff smile and laugh so much 😎

[@trigonzobob]

Now that's what I call running the numbers.

[@Hakasedess]

My one criticism is about the last bit where you hypothesize that you're going to see two clusters, one for sandbaggers at below 281 somewhere, and one for 'the rest' at roughly 281. I dispute this, and would willingly bet money that the actual average of 'attempts' for people who don't know about the bingo at all is going to be significantly higher than 281. The 281 figure relies on a randomized finishing time, and that's simply not going to be the case for... well, anyone who's going to be running exactly 5km regularly enough to even reach 281 runs in the first place. Their time from run to run is simply going to be too consistent to be compared to a 60sided die.

[@dflosounds]

Fair point, though I don't think you would see THAT much consistency in the seconds value unless you're talking professional or short-distance runners. In my experience of running casual 5 or 10k races (like the one in this video), there is likely to be consistency in minutes, but not so much for seconds. All you need to do is deviate from your average time up to +/- 30 seconds, and any of those 59 values are game. When you consider variables like the weather, how much sleep you got, your general energy level that day, etc, I don't think it's that far-fetched to get (almost) effectively-random seconds values. So while I agree that the value would certainly be higher than 281 (because you're right, it's not like rolling a 60-sided die), I don't think it would be particularly significant.

[@Hakasedess]

@@dflosounds I guess it's possible it won't be a significant deviation, though I still imagine it would be when averaged across a large sample. It'd definitely be very interesting to see data on it though.

[@ps.2]

@@dflosounds Fair enough, but it's not really accurate to say you can deviate by ±30 seconds and expect a flat distribution. It's probably more Gaussian, so you need that 60-second interval to be well inside the meaty part of the bell curve. Not way out at the edges of it.

[@dflosounds]

​@@Hakasedess Would definitely be interesting!

[@eekee6034]

Hakkapeele, it looks like you've missed something, but if we take your objection with the bit it looks like you've missed, we get an interesting question. We're dropping the minutes value of the time; we're only taking the seconds value. Now, I don't know how long a 5km run could be in minutes; I'm deliberately not looking at that. Instead, what I'm thinking is, _if the number of minutes is large enough,_ even the most consistent runner will only be consistent to the nearest minute. Then things get interesting. If a runner has a 1-minute range, would the distribution of times be a bell curve? What if he has a 90-second range of times? I think it gets way more complicated than I can work out in the middle of the night with a headache, at least. :)

[@mustafakalaycioglu9613]

The profound knowledge shared to us by Matt that 60>52. I didnt know that before :) Great video mate!

[@madlad255]

The Parker Run

[@AfonsoCL]

Matt is such a fun guy. Spending an afternoon guinea-pigging for his experiments while listening to his passion for maths would be one of my ideal days.

[@TSutton]

This video is a perfect explanation of predicting fossil collecting in Animal Crossing!

[@Jonny_Marko]

I had the same thought but with collecting all the DIY recipes! My odds are not looking so great to find the one I am endlessly searching for :D

[@gamekiller0123]

You haven't actually proven that the harmonic series diverges. The argument says that as n approaches infinity the number of runs also approaches infinity, but n is not bounded as it approaches infinity. We could have a situation where n(H_n) only diverges because n diverges. EDIT: diverges, not converges.

[@entropie-3622]

At least we know it does not converge to 0 faster than 1/n sooo that is something XD (especially for a sequence with all positive terms)

[@tassiehandyman3090]

All hail, the Amazing Mark - he who digs Matt Parker out of a hole of his own making, by simply being a Thoroughly Decent Chap. Thank you, Mark - you're a good egg!

[@user-bl9of5qe7h]

Absolutely love how this is your typical intro-to-probability problem but solved completely using intuition. So stripped down from bulky theory and just beautiful

[@oddysee3030]

For the record, I really appreciated the recursion bit :)

[@ronnytm]

Call me shallow, but the lighting, colours, and exposure of the video look really good for a cloudy day in a park. Props to the cinematographer. I'm sure the content of the video is great too.

[@pyglik2296]

The average time to get k out of n "coupons" is a harmonic series which can be approximated by logarithms and inverting the question to "What's the average number of unique coupons after time t?" gives us k = n(1-e^(-t/n)) which fits nicely to the graph.

[@hebl47]

You had me there for a second. I was starting to worry you didn't make a second recursive miniature table.

[@WDCallahan]

60 is a bigger number than 52 😲 You just never know what you're going to learn about math when you watch this channel!

[@NPDGX]

To add to this from a computer science graduate: The harmonic series, in asymptotic analysis, is Θ(log n). Because of that, the cleanest way to write the solution is as E[X] = Θ(n log n), where X is our random variable for collecting n coupons. Just some food for thought :)

[@MrTurboTash]

4:50 You had me going. Good on ya Matt, did not disappoint. :P 11:24 ... Assuming a random distribution. 13:58 Of course he was :D

[@BradleyGordon42]

That recursion joke. That's the kind of quality joke I love you for.

[@lambdasaturn]

I think this is not a rigorous proof of the divergence of harmonic series, because if it converged to some constant, n approaches infinity, and time also approaches infinity, therefore, in order for your proof to work, you need to first prove that time grows faster than n.

[@Smithers888]

14:45 [Matt expects a bimodal distribution with the "blissrully unaware" peak at 281] The peak wouldn't be at 281; you calculated the _mean_ time to be 281, but the peak is at the _mode_ which would be earlier. To evidence my point without having to run the stats for n=60, consider n=2. The probabilities are: P(2 tries) = 1/2, P(3 tries = 1/4), p(4 tries) = 1/8, etc. with an expected # of tries of 2(1/1 + 1/2) = 3, while the peak of this distribution clearly at 2.

[@stephenbenner4353]

This may be my favorite Matt Parker recursion.

[@EER0000]

This morning volunteered at my local park run, this evening watched a math video about Park run, a very recursive Saturday))

[@Vares65]

4:57 LOL - I DID know! I was literally sitting there waiting for it.

[@MazerTime]

i really love the recursion joke, definatly worth the extra effort

[@antoninmalon7225]

I don't understand (10:33): "the formula will always increase and n is finite" - how does that imply that the harmonic series diverges? N is ever-growing, so the formula would grow in size even if the harmonic series converged, no?

[@BizVlogs]

He made a mistake. You are correct. On the one hand he calls n finite but then uses that to rationalize about an infinite series inside the parentheses? That’s wrong. The fact that lim n → ∞ n*(1/1+…+1/n) goes to infinity does not imply the harmonic series goes to infinity. It does diverge obviously, but you can’t prove it from that.

[@geeshta]

12:23 Saying "Have a cup of tea or something" in a high pitched voice when upset is the most British thing

[@Wordsnwood]

appropriately majestic closing music.

[@Srearis1]

great video as always. love the props

[@grapesofwraith1066]

I love British KZbinr crossovers and recursion jokes!

[@jacob416]

I gotta give you props, that recursion joke was great.

[@happyestus6688]

Your were correct: that recursion bit was 100% worth the effort. 10/10

[@nicholasvinen]

Actually I'd say it was 50% + 25% + 12.5% ... worth the effort.

[@bobd2659]

As soon as the small table came out, I said to myself..."Wait for it...WAIT for it!" You did not disappoint!

[@K-o-R]

3:47 Whoa whoa whoa, slow down there, braniac! This is a lot of information to process!

[@grahams5871]

If you do ever find that 'oblivious' peak and 'sandbagger' peak, then you can work out the average targeting ability of the sandbaggers. the 'oblivious' runners will have a uniform distribution over 0-59 and the 'sandbaggers' will have a normal distribution around their target. (the problem is muddled a bit because probably not all sandbaggers know the optimal strategy of 'aim for the distribution in your bingo card that is least filled')

[@stevebollinger3463]

I feel like some park runs could ‘help out’ people on their last few coupons by tacitly guaranteeing times which are even or odd. Or in the upper half of decades or lower. People in large enough areas trying to finish up could select between the two local park runs each weekend, one of which produces evens and one odds. In this way the final coupon time would be cut on average by 6 months.

[@tylerm8128]

Great video Matt! I was just solving this problem myself, the other day. I'm trying to collect one of every pokemon card in the latest set. I calculated it to be a LOT more packs of random cards than I'm willing to buy, so I'll just buy my remaining cards individually ;)

[@Marronii]

I was actually just waiting for the third table

[@ohnonomorenames]

Every single revile in this video is more fulfilling than the last!

[@jnaoe]

i would have been disappointed if there was not a second miniature table. Thank you :D

[@svibhavm]

the props were DEFINITELY worth the extra effort. Totally agreeed Matt

[@ahobby]

I liked the video because of the extra recursion effort. That second level in pushed did it for me.

[@davidalderson]

parkrun and maths my favourite 2 things. How do you work out how long it should take to collect every number twice? I completed parkrun bingo on my 270th parkrun (a bit below average), but just 4 runs later on my 274th I had double parkrun bingo. My instinct is that 274 to get every number twice would be well below average, but have never been able to calculate this.

[@jeremydavis3631]

Unless I'm mistaken, the math is exactly the same. Just use 120 instead of 60, since there are 120 coupons you're trying to collect (2 for each second). In other words, it should take around 644 times on average if they're all picked up randomly. However, as others have noted, a uniform distribution is likely to overestimate the true average: even if you weren't trying to, I bet you got a lot of similar times on consecutive runs. A more accurate average would require a lot of data.

[@davidalderson]

@@jeremydavis3631 Getting 60 numbers twice is different to getting 120 numbers once. After you have completed your first set, you already have many numbers 2 (or more) times so the second set should take less than 281 from the point you complete your first set.

[@daky1445]

​@@jeremydavis3631 This does not feel right to me at all. Try thinking about what would happen if you had already collected the 59/60 twice. The maths would suggest there's a 1/119 chance for the next one, but in reality it's still 1/60. And what about attempt number 2 you'd do? There wouldn't be a 119/120 chance to get what you need. it would be 1/1 to get it because you don't care if it is a repeat or not I have not got an idea of how to calculate it, and I will not pursue it. The calculation in the video only works when it's a certain number of UNIQUE items collected

[@jordanlinus6178]

A simple reflection is this: Let us say hn is the average number of runs it takes to collect all n coupons. Among those hn runs, hn-n are doubles. So the question is, among those hn-n, how many different are there? Well, hn-n is the average time it takes to collect n-1 coupons, so by the time you have n coupons, you already have n-1 doubles, and the added time is just the time needed for the final coupons, n runs. This is a simplification, since the doubles are clearly not independent from the firsts, and you don't necessarily n-1 double by the time you collect all n coupons, but I think it is not that wrong, and it would not take much longer than n additional runs to collect all coupons twice.

[@bob-hp1lr]

@@jeremydavis3631 One key detail that really throws off your math: there is no difference between collecting a time the first time and the second time in terms of probability (once you have collected all but one time, your odds of getting it are 1/60 not 1/120). Can't think of a clean way to calculate the second question but a quick simulation says ~403 runs. That matches with the logic that it should take less than twice as long (since you are essentially going for bingo twice in a row but you get a "head start" on the second attempt)

[@2birdbrained4u]

At 10:44, you can clearly hear the audio diverging from the video ;)

[@bijova]

The recursion joke is why i liked.