Wait, why? Why would you watch this if you do not enjoy math? And if you do, why wouldn't you want to have endless math facts at hand? Alternatively, why must there be a moral?
You shouldn't be the first as the mathematician would reject you, using the same reasoning explained in the video. Being the first, you'd have 0% chance of being after the "kth" date. :)
Well technically if you're going to be the mathematicians only date then there's a 100% chance he will choose you. Depends what his N value is. But yeah I like the moral anyway lol, works for most scenarios I'm sure.
As a mathematician how can you even consider that to be a relevant point to that? If mathematicians and random people can mix it doesn't mean that mathematicians and mathematicians do.
Or in other words: Your first love will never be the one you end up marrying. It's more complicated than that bc there is cheating and all kinds of shenanigans with people divorcing and remarrying each other but kind of sad to have such a rational, unromantic outcome demonstrated here.
How will you know that the person you're seeing is in the 37% spot in your list of dating partners and if you settle down and remain monogamous, is that a paradox?
This leads to two arkward situations: while dating:"So you really want to date me?" - "Yeah I need reference values" and while married "Honey back then, in your womanizing days, why did you choose me amongst all these women?" - "Because you were the next best after I dated 37% of them"
Depending on how many women you dated, the second one is actually a strong compliment. It could be much worse, though: "I chose you, because the best one was in the first 37% and you were the last option".
These must be the best brown papers ever. Her handwriting and graphs are so good they might as well have been printed! Be honest, Brady: is she a T-888 that the professors of NU managed to reprogram and repurpose as a professor? She can't be human ... can she? (I hope she's not looking for John Connor in those music festivals.)
-xln(x) is also a famous problem in Physics. The entropy according to Boltzman can be written as S=-k*p*ln(p) (K:= Bolzman-Constant). Very nice video by the way!
This only covers how to find the single best one. It is not fine-tuned to optimize the average outcome (although I can imagine that a 37% best-case scenario helps the average a lot). What method do I use to get the best average outcome? And how does this generalize if I would be willing to settle for a top-2 toilet/partner ? Or top x%, for lim x -> 0.
I tried it with 2/N as the probability since 2 toilets are good ones out of N, and it lead to the same answer, at the stage -lnx=-1 I got -2lnx=-2 and they cancel to give the same answer. I may have gone wrong somewhere.
vdeave when you plug it back into the probability function though, I assume the success rate must go up. I believe the stopping point is the same. But since the probably function changed plugging in the same stopping point probably yields a different probability
This is still the best. If that best toilet is in the first 37% you’ll still pick the SECOND best if it’s not, and so on. You can only pick the worst if the best 37% are the first 37% AND the worst is the last toilet.
This method works for bimodal outcomes (in this case, either I chose the best candidate, or I didn't). The problem with trying to optimize the average outcome is that it implies you know something about the distribution of your candidates. (If, say, you're rating your candidates from 0-100, then you have to make an assumption like they're distributed normally on this scale). In the way the original question to the secretary problem was posed, you only know that you have some sort of method of of scoring your candidates, not the distribution from which they are drawn. But you could apply the same method in the case of bimodal outcomes (I want to pick one of the top two toilets, or not).
Not quite. If you set a total limit, a maximum of how many you are willing to date in a 3 year span, then the principle can be applied. Remember, you choose N.
Waaaaaaaaaaait... You can probably say that youd only choose women.. and only women which are in the age between 18 and 25 ... so yeah idk but thatd be kinda smaller than 0,37*8 billion:D
I'd love to see how this maths changes if you want to try to select *one of the best two* toilets, or any toilet that's in the top ten percent of toilets. You'd no longer be looking at a point on the curve but a range.
I would prefer to optimize expected quality (assuming quality was evenly distributed over a known range) instead of picking only *the best* toilet with a known probability.
The answer would be the same, as long as you want the range around the point where d/dx = 0 (or the optimal solution) the answer still holds because that point is the best point for getting a toilet in that range.
This way is still the best. If that best toilet is in the first 37% you’ll still pick the SECOND best if it’s not, and so on. You can only pick the worst if the best 37% are the first 37% AND the worst is the last toilet.
Yeah, so this is called the (1,2)-secretary problem or more generally (1,K)-secretary problem where you are happy with anything in "top K". There is a related but different strategy of how you should do things here. For K=2, this new strategy can succeed with ~60% chance and for even larger values of K your chances improve even more (as you may expect).
Is there a mistake in the video? I'm trying to understand the maths. At 5:31 there's an illustration. Shouldn't the chance of being picked at K+1 be 1 instead of K? Because this is what is explained in the video. So perhaps I don't understand it but I think the image is wrong. Where it says K on the toilet door, it should be 1.
Was thinking the same the illustration is obviously wrong. Regarding the calculus, I won't pretend it's wrong but it's at the very least glossed over which is a shame it's really the interesting part of the demonstration.
DB Fuller So you go to the toilet just after you enter the festival? I think you would be one of the few in this world who does this. So I believe this would not work.
4:02 I do hate it when I'm at a music festival and I miss out on the toilet with the chandelier. Come to think of it, I always seem to. How unlucky am I?
I think the argument being given at 4:44 about the probability is irrelevant. The probability is 1/k+1 because the probability of choosing the (k+1)th toilette is the probability that the (k+1)th toilette is the best of the first (k+1) toilettes. Am I missing something ?
I think that's the correct analysis! I was also struggling with what she said. What you said made a lot more sense and can be easily generalized to deal with the general case of the best toilet being instead in the (k+p)th position in a row of N toilets, where p is an integer >= 1. Thanks for the clarity! xd
Oh thank goodness. I couldn't get past 5:00 in the video because of this! I kept going back and re-watching the previous 30 seconds, hoping it would make more sense on the k+1 th time through. Thank you so much for your comment.
I had to think about this as well. So the probability that toilet k+2 is selected when it is the best toilet it is 1 - the probability that toilet k+1 is not selected = 1 - 1/(k+1) = k/(k+1). The probability toilet k+3 is selected when it is the best toilet is 1 - the probability that toilet k+1 or toilet k+2 is selected before it. The latter is the probability that the best value among all of the initial k+2 elements is not in the initial k elements = 2/(k+2). So prob toilet k+3 is selected when it is the best toilet = 1 - 2/(k+2) = k/(k+2) and so on.
Honestly, while the video wasn’t bad, it was way too short and glanced over many things that should have been given more explanation, like why k/n is the probability after k bathroom trials, the bathroom you choose will be the right one, or why that series shown can be approximated to the Integral of 1/x
Hmm I'm probably wrong here but at around 5:30 the probability for choosing k + 3 should be (k + 1)/(k+2) as its 1 - 1/(k+2) right? Or am I missing something?
no since the probability of a tree being better than the first K in K+2 trees is 2/(K+2) because it can be in position K+1 or K+2.Therefore it is 1-2/(K+2) =K/(K+2)
Brady the video is brilliant as well all your channels are. Just something to mention. In animation at 5:30 min there are displayed some probabilities. I suppose probability for toilet k+1 is not k but 1. Check this out please. Thanks so much for such a wonderful job.
Why didn't you surround that probability between two aproximations ? A bound from above and a bound from below with integrals ? It's not that complicated, it won't have made that video less clear, but it could have shown how accurate that approximation was for high values of N, and how our strategy behaves for low values of N. A video on further strategies would be appreciated : what if we are ready to accept one of the 10% best toilet ? Not necessarily the best one, but one of the best ? And what if on the contrary we really really don't want the worst 10% ones, but don't really care if it's a little dirty ? What would be the best strategy then ?
I'm not sure, but I think that the explanation around 4:40 should be clearer: the probability of having chosen k+1 because best compared to previous toilets is 1/(k+1)
This was a very interesting problem. But it assumes that you have information about the size of N, or the number of options available to you. What if we don't know how large N is. Or what if the dates, the potential secretaries (or the toilets) "arrive" with a certain probability, let's say with a Poisson distribution. How many do you have to reject first before you take the next best one? Does a solution even exist? I feel that the secretary problem should have applications in behavioral ecology. It seems like a nice rational solution. Sorry about the long post. I just found the video very interesting. Thank you Brady!!
It's weird that in this video she's still rushed, even though this is the video that people were warned it would be longer, and heavier on the maths. x3
So if I believe for example that only 5% of women on this planet is fit to be my life partner. That is 1/20 and I should therefore find 20 random women, reject the first 7 and then start picking the first one that is a better match than the first 7?
Id recommend looking up on the secretary problem in other websites. Reason being is that even for someone with a good grasp of maths, I doubt they would be able to fully accept the jumps in conclusion in this video.
There is just one huge problem i have with all these calculations. It clearly assumes that each toilet has an equal chance of being good/bad, while in reality toilets close to the entrance definitely have a higher chance of being worse. Also this only covers the propability of choosing the single best toilet of all these toilets, instead of just saying you could be content with a certain point of being clean. Also when i have to go to a toilet, i do have an issue with time, which means i really dont want to check a large amount of toilets, because afterwards i won´t be needing one anymore. Considering these objections, i´d rather choose the first toilet, that goes into my standard instead of trying to find the best toilet ;D
The use of the toilets was just to be cutesy. In reality, this problem has huge implications in decision theory when the given assumptions (draw from uniform distribution, see each draw w/out replacement, etc.) are met. You can generalize this problem (i.e. don't assume each toilet is equally likely to be good or bad), but it would be too complicated to go through in such a short time (and I assume that it wouldn't have such a nice closed form solution as this problem had).
The number on the (K+1)th toilet at 5:38 should be K/K. Dr Symonds did say at 3:35 it was 1, and it would make the generalization of the probabilities of subsequent toilets being selected sensible.
Just wondering, say you didn't end up picking the best toilet, on average, how good is the toilet you will end up choosing. Top 50% of toilets? 40, 30, 20%?
That was some surprisingly intense math for a problem about dirty toilets XD But seriously, I can see these results being useful in all sorts of situations. The only draw back being that you have to know exactly how many things you have to look through before hand.
Wasn't one of your original problems that you didn't want to look at every toilet? You are going to automatically look at 37% percent +1 of the toilets and then there is a 37% chance that the best toilet was in the original 37% so you would end up looking at every toilet
WRONG. you look for the next toilet that is better than the first 37%, which means you don't look through them all. It might end up like that, but it's not likely
+ismartroman gamer If the best toilet was in the original 37%, then after K, another better toilet doesn't exist, therefore you'd have to check every one.
+MetaKnight68 I said that you might end up looking through them all. I never said that you never have to. Listen before you reply. And also, as stated in the video, there is only a 37% chance of that haplening
+ismartroman gamer My apologies, although I don't see how your statement refutes the original comment. I suppose that I should have directly stated that, rather than trying to explain something I thought you didn't understand (though I see now that you clearly do). You say that there is a 37% chance of the best toilet being in the original (as it is stated by the video);that is what the original commenter said, therefore Shaqtapus is not incorrect. I am curious as to why you said, "WRONG." I agree that I didn't state my intent for the reply, and I apologize for that. But your comment doesn't make sense. Also, please check your spelling before commenting.
11:10 I think this function should have X in the domain [0,1] on the horizontal axis, since that is the parameter of probability function P (or k/N since that is equivalent). In any case not just 'k'.
N is a constant, so it kind of doesn't matter. Picking the right K (the one that maximizes the probability) automatically results in picking the right K/N (the one that maximizes the probability) so if you have K/N on the x-axis you end up plotting the same function on a diff. scale.
I believe this is just a case of notation. She is graphing the function P(k) if you look on the vertical axis. It is confusing to once again use k as a generic value on the horizontal axis but remember that x=k/N so as long as k (being any value on the horizontal axis) is between [0,N] your x is between [0,1]
As you pointed out, x is equivalent to k/n. This makes x a percentage of n, dependent on k. In this way, x converts the number of rejected toilets into a percentage of the total number of toilets. The new value is now in a form that the probability function can use. Speaking in terms of the graph, if you were to increase the value of n, you would stretch the graph horizontally, because n is inversely proportional to the function value. As the total number of toilets gets grater, the probability of any one toilet being the best choice gets smaller. But, since there are also more toilets, there will still be a point at which the 37% can be obtained, even if it's further down the line, or X-axis.
The most annoying part of this video is (I'm sorry to say) Brady's impatience. Though he has access to all the viewer statistics, which I can guess induces such behaviour.
i prefer the method "to definitely not have the worst" to the method "to have the highest chance to have the best". But the Mathematics behind it is nice and *well explained*!!
So for a rule of thumb you can remember anywhere anytime: If you can only check a toilet once, check a third, and then pick any that is better than those. And i would also argue; settle if you find one that is past the point of "good enough" to save time if the selection is large and time of the essence.
I have a questions about this: 1.Why is differentiation allowed at 11:50? The function is obviously discontinuous-is it defined only at certain points. Just because a function can be integrated (by an approximation in this case) does not mean the converse is true. Let us not even forget that some function like the Weierstass function is continuous and does not have a derivative.
+RyanCreatesThings To be more realistic, the dating thing is in relation to your age. Assuming you start dating at 15 and end at 35 if you don't find anyone, reject those from the first 37% of that time frame (which is about 7 or 8 years, so when you're 22 or 23, which makes sense) and then start comparing your current date to the rest.
+arcuesfanatic Which still begs the question, when should a person start commiting? We need estimate a person's dating potential, taking into account age and dating pools. Then apply the optimal stopping theory, and finally translate this back to an age. Or keep using K, maybe both? Which one would be optimal? Should we commit when either condition is met, or wait for both of them? The period between where one is met and both are met sounds like a "sweet spot", where if your partner does not commit, you still have the highest probability of getting the best choice. Actually, the problem itself should consider the other party not commiting, but how can we calculate the probability for this? +Numberphile2 +Numberphile answer in a new video maybe?
Brady, could we get a derivation for the case where we want the best AVERAGE result, instead of the best chance at getting the highest-ranked result? I think the problem was solved by Bearden (2006).
You realize that now more people know this and chose toilets more in the middle, which completely changes the complexion of the question. If now more people use the 37% the toilet, then it becomes dirtier, meaning one should use the therefore less used earlier or later toilets
But not everyone comes to the toilets in the same order so you wouldn't necessarily see higher use around the "k"th toilet as it would be a different one for each person.
The original assumptions of the question were a bit loose to begin with. It is highly unlikely that any given toilet has an equal probability of being the dirtiest or the cleanest or anything in-between. It is more likely that as you get closer and into more high traffic areas that the toilet condition will worsen. A uniform distribution may not be the best assumption to start with.
Not necessarily. You have to figure that people we either a) not know about this solution, or b) be in too big of a hurry to care, and use one of the early ones anyway. Combine that with the fact that more use most likely means worse conditions, and you'll just wind up choosing more towards the end as the middle ones become used more and more, but the first will always be used more because of the most important factor: desperation to use the toilet.
4:40: The one thing I don't understand is why we still consider 1 thru k as being possibly selected. Isn't the probability of selecting those toilets 0 automatically?
sanchit goel Because when we approximate the discrete fraction series as a continuous integral each term must become a bar that extends from its original position to the next position (otherwise the integral would be 0). So 1/k becomes a bar of height 1/k that covers the distance from k to k+1. In exactly the same way 1/(N-1) must be a bar of height 1/(N-1) that covers the distance from N-1 up to N. As such our integral must end at N not N-1 otherwise 1/(N-1) would have no distance and so evaluate as 0 in the integral.
Programmers way: Premature optimization is the root of all evil. Unless you require the optimal choice, consider taking the first acceptable alternative. If the toilet is acceptable but not optimal in group K, choose that toilet.
This math works out for an arbitrary example where you have no context of what is acceptable and what isn't, but there's two flaws with the specific example of toilets. 1) There's a threshold of instant acceptance. You know you're at a music festival and if you get toilets 3 or 4, you'll probably take it, even if it's before k. 2) You can't quantitatively compare toilets fast enough that you won't piss yourself.
Great explanation, thank you.( I did wonder if this might be an audition for Numbers - I could easily see you saying: "We can easily find the murderer with a little calculus.") I'm going to use your calculation to select the best time server from a pool.
what if you wanted to have the best average state of toilet? Let's say each toilet has a random state of hygiene between 0 and 1, what would you need to do to pick - not the best one - but the one that's as close to 1 as possible? because with this method you have a 37% chance of getting the 1, and 63% chance of getting anything else, resulting in an average state of 0.37x1 + 0.63x0.5 = 0.68 average state of hygiene. Is there any better way of optimizing this?
With the given algorithm the odds are better. The odds are 37% to find the toilet with hygiene 1. The odds are 37% you skip the hygiene 1 toilet, and end with a random toilet with hygiene < 1. Let's say this average hygiene is 1/2, but in fact it is a little bit smaller, as you have 0 chance of getting hygiene 1 this way. The odds are 26% you pick a toilet but will miss the hygiene 1 toilet. In this case the hygiene of your toilet still will be very high, as it beats the first 37% you checked. Hence your score will be 0.37 + (1/2 * 0.37) + (nearly 1 * 0.26), an expected value between 0.555 and 0.815 that is close to 0.815. I still believe it can be better, but the estimate of 0.68 is too low.
It still works if you put the toilets into a random order first. Otherwise it drifts into game theory (choosing your strategy depending on what you expect others to do)
If you know the festival is going to be full of mathematicians who know this, you pick one of the first toilets because they're all going to be unused.
I don't know about you, but when I gotta go, I don't care about getting "the best", but "good enough". What's the answer if I only care about getting a toilet that's in the top 10% or 20%
Yes, and that last one might even be the worst! 37% of the times, the best one will be in the first K that you skip; so your chance of ending up in the worst should be P=0.37*(1/[n-1]); 1/[n-1] times the chance of picking the best one!
Yes. You are unlucky then, but it is still more safe to do as said. if you go to 100 music festivals, and do this every time, you will have best toilet more often than you would have bad toilet.
Actually, according to this algorithm, as long as the best toilet is in one of the first 37%, you will have to pick the last toilet,... so there is a 0.37/N chance of picking the worst toilet. The probability of picking then 2nd best to 2nd worst toilet would be somewhere between 0.37 and 0.37/N
Yes. So this algorithm optimizes your chances of picking the best, but it doesn't necessarily give you the best chances of optimizing your choice. Say, for example, you died every time you picked anything but the best. Then this strategy will keep you alive about 37% of the time, and that's all you care about. But chances are, you don't mind settling for a "nice" toilet, if it means you won't get stuck with the worst one. For that, a more complex, more subjective "minimizing regret" algorithm is called for. tldr: This cannot be considered the best way to pick a toilet until you precisely define your objective.
I'm glad you made this. This problem is in my Martin Gardner book but I couldn't follow his explanation. This video cleared things up for me. k = N/e is certainly the formula to remember, but it does yield some wrong answers for small numbers of toilets, 7 and 10 toilets for example. I tried improving on it a bit by using other approximations than the left hand rule shown at 8:23. Using the midpoint rule we set the bounds of integration from k-1/2 to N-1/2, but then later it becomes impossible to solve for k. Using the trapezoid rule we set the bounds of integration from k to N-1 and then add (1/k + 1/(N-1))/2 to the result. This yields the more accurate formula k = (N-1) / e^(1-1/(2N))
The derivation just shown relies heavily on the assumption that you value all toilets execpt the best one at zero. Meaning the *second best one* and the *worst one* have the *same value* to you... This is a quite unlikely assumption in the real world ;)
This is used in finance to price American options. It's easier when you learn dynamic programming (done in a different video by Numberphile: the one with the prisoner on a chessboard)
I'm concerned about the phrase "choose the next toilet that is better than all those we have seen." I understand this selection rule correctly, it means that if the best of all the toilets is among the first 37% of the available toilets, we will be forced to take the last toilet as the best toilet was passed before we have decided to accept a toilet. Or am I missing something?
So since there are 3.5 billion women on this planet, I need to go out on 1.3 billion dates and THEN start looking for my soul mate? That sounds.....time consuming.
Am I supposed to know this mathmatical way of solving as a 14 year old? It's when it comes to k, it starts getting messy for me. I also get the ending, just not the way finding the results.
I think at 14 it's unlikely you have seen Calculus yet, so the integration and derivation that she does near the end are not things you would know how to do yet. And probably you also haven't seen the approximation of the exponential function (the graph at 8:27 that gets smaller and smaller as x increases) but you may have seen exponential functions in general and their opposite: log functions. (e.g. log(10^3) = 3) In Canada, we did "Pre-Calculus" (which covered all sorts of functions, approximations, series, etc) in grade 10 and 11 and Calculus in grade 12 (last year of high school), but many people do not see Calculus until university. Other countries have a more accelerated math program and they get to learn more earlier on. It depends. I wouldn't worry about not understanding all of it right now, you will soon enough. It's very good you had a look and followed along. :)
Oh ok thanks for making that lift from my shoulder :D Also I'm from Denmark, so I also have to translate in my head all the time, that gives another challenge. But yeah I really want to be something along scientist/astronomer, I've had the wish to change the world, and extend the humanities living. I usually watch alot of videos on TH-cam about all the things that is related to Physhic, Math and Chemistry. Because I feel like to be the best, you need all what you can get. ;)
They teach this (differentiation) in high school (or first year in university), so you are not expected to know it yet. You can look into it, however, if you're interested.
No. Calculus isn't even high school stuff in most places, so unless you will take math in college you'll never be taught the basics required to understand this. On the other hand, this isn't *that* complex for calculus, so you probably should be able to grasp it given some time, but don't be disappointed if you don't right away.
Striped Marlin Gaming If that reply was meant as some kind of insult, then you did a poor job. If not, then ... I agree? I don't go to festivals for other reasons. For example, last time I was sexually assaulted and no one cared to help. If applying maths and probability theory to decide which toilet to go to is the biggest problem you have with festivals, then you are either oblivious or delusional. My original comment was merely a quip.
it was kind of a combiation of an insult and a joke, seeing as how i am the most hermit-like internet person i know, and would probably never leave my house if i didn't have to.
Haha If You were on a festival of matematicians, where each of them would pick their toilet with that principal, the would all use the same (last) toilet (if they all check them in the same order)
+BabylonicaMan I was coming with an explanation of why you're wrong when I realized you're right. So everyone passes by the first 37 toilets and then starts looking for a better one. But since all the toilets are in the same unused condition they'll never find a better one, so they keep going and going until the end and will have to use the most disgusting possible toilet. Awesome.
So if everybody has to check the toilets in the same order (for some reason), that would be an example of a game (competing for the best toilet) where, if everybody picks the best strategy ignoring others' strategies, everybody would get the worst possible outcome
On 5:29, shouldn't the probability of k+3 being chosen be (k+2)/(k+3) and so on? It would be 1-(1/(k+3)) which would develop to (k+2)/(k+3)... Am I wrong?
I think there is an error in the graphic shown starting at 5:30. Under toilet K+1 it should be K/K not K. So the sequence should be K/K, K/(K+1),K/(K+2).
I disagree this is the best way to find a toilet, secretary, or soul mate. If the toilets are placed randomly, which this video seems to work with, then there's not only a 37% chance of finding the best toilet, but there's also a 37% chance of skipping the best toilet. That's because, when you have 100 toilets, there's a 1 in 100 chance for each of those toilets to be the best toilet, since it's spread randomly. Skipping the first 37 of those means you have a 37 in 100 chance (=37%) to have skipped the best toilet. What I'd do is simply check what your own standards are, and settle for the first toilet which you think is good enough for you. You got to set a standard for yourself in advance, and if you find a toilet, secretary, or date that meets that standard, you settle for it. If it disappoints you after you've tried it, you move on (i.e. you pick another toilet next time, or you find another secretary, or you break up). I'm assuming nobody would actually dump their first 37% of dates, because then the chance of finding the best partner is biggest. Nobody would want to have a 37% chance of having met your soul mate, but having dumped them just because maths said so.
This assumes you don't know what the best toilet looks like. Also, the problem is mathematical, so doesn't involve 'standards' or else you'd factor in stuff like 'I might miss the concert'
***** Yes I know this is mathematic, but that's basically my point. Not everything is best dealt with by maths. In this case, you should just do it without maths.
This is assuming that no scale or maximal/minimal/optimal solution is known or assumed. Of course in more specific situations, we have the luxury of having standards (i.e. scales or metrics of some sort) or optimal solutions (to which we can approximate or assume nothing we have could be closer) and make a decision based on those preconceived assumptions. However, if you don't have those preconceptions, then the method given in the video is optimal to making a decision (based on the assumptions the video makes).
So if you're on a festival with a lot of mathematicians, go for the first toilet and you'll be the first one to use it.
That's assuming everyone will go and look at the toilets in the same order, which I don't think you can.
Can we not just skip past the "festival of mathematicians" idea here??!
***** I was just about to say that ;)
But many people might have thought of it so go for the second
Wesley Teh - And they would’ve also thought of that so #3 is the better option.
So the moral is: Don't be a mathematician's first ever date.
Wait, why?
Why would you watch this if you do not enjoy math?
And if you do, why wouldn't you want to have endless math facts at hand?
Alternatively, why must there be a moral?
You shouldn't be the first as the mathematician would reject you, using the same reasoning explained in the video. Being the first, you'd have 0% chance of being after the "kth" date. :)
Well technically if you're going to be the mathematicians only date then there's a 100% chance he will choose you. Depends what his N value is. But yeah I like the moral anyway lol, works for most scenarios I'm sure.
True. Both I and my first date were a mathematician, and also each others first date. It didn't end well. ;)
As a mathematician how can you even consider that to be a relevant point to that?
If mathematicians and random people can mix it doesn't mean that mathematicians and mathematicians do.
And in the end, I learned that dating is essentially the same as choosing between nasty toilets.
Or in other words: Your first love will never be the one you end up marrying. It's more complicated than that bc there is cheating and all kinds of shenanigans with people divorcing and remarrying each other but kind of sad to have such a rational, unromantic outcome demonstrated here.
essentially you're right!
How will you know that the person you're seeing is in the 37% spot in your list of dating partners and if you settle down and remain monogamous, is that a paradox?
This leads to two arkward situations: while dating:"So you really want to date me?" - "Yeah I need reference values" and while married "Honey back then, in your womanizing days, why did you choose me amongst all these women?" - "Because you were the next best after I dated 37% of them"
Depending on how many women you dated, the second one is actually a strong compliment.
It could be much worse, though: "I chose you, because the best one was in the first 37% and you were the last option".
Artwork and brown papers from this video - bit.ly/brownpapers - proceeds support the animator!
These must be the best brown papers ever. Her handwriting and graphs are so good they might as well have been printed! Be honest, Brady: is she a T-888 that the professors of NU managed to reprogram and repurpose as a professor? She can't be human ... can she? (I hope she's not looking for John Connor in those music festivals.)
This is brilliant, though is about festival toilets!
Άλκης Δ. It's HAL!!!
Άλκης Δ. She could be a replicant... ;-)
Brown papers sounds wrong...
She's got some amazing handwriting o___o
I just went to the comments to search for someone talking about this
If you were at a concert full of mathematicians the first 37% of toilets would be the cleanest cause no one would of used them
-xln(x) is also a famous problem in Physics. The entropy according to Boltzman can be written as S=-k*p*ln(p) (K:= Bolzman-Constant). Very nice video by the way!
This only covers how to find the single best one. It is not fine-tuned to optimize the average outcome (although I can imagine that a 37% best-case scenario helps the average a lot). What method do I use to get the best average outcome? And how does this generalize if I would be willing to settle for a top-2 toilet/partner ? Or top x%, for lim x -> 0.
I tried it with 2/N as the probability since 2 toilets are good ones out of N, and it lead to the same answer, at the stage -lnx=-1 I got -2lnx=-2 and they cancel to give the same answer. I may have gone wrong somewhere.
vdeave when you plug it back into the probability function though, I assume the success rate must go up. I believe the stopping point is the same. But since the probably function changed plugging in the same stopping point probably yields a different probability
This is still the best. If that best toilet is in the first 37% you’ll still pick the SECOND best if it’s not, and so on. You can only pick the worst if the best 37% are the first 37% AND the worst is the last toilet.
Additionally, 37% of the time, you will have skipped over the best option, and will be forced to use the last option.
This method works for bimodal outcomes (in this case, either I chose the best candidate, or I didn't). The problem with trying to optimize the average outcome is that it implies you know something about the distribution of your candidates. (If, say, you're rating your candidates from 0-100, then you have to make an assumption like they're distributed normally on this scale). In the way the original question to the secretary problem was posed, you only know that you have some sort of method of of scoring your candidates, not the distribution from which they are drawn. But you could apply the same method in the case of bimodal outcomes (I want to pick one of the top two toilets, or not).
Today I learned that you can choose a spouse the same way you choose a toilet.
LOL
Terrific explanation. Things like these must be included in the curriculum to make Math more relatable, practical and fun!
As a math student i really love how you're explaining so clearly !
You're an amazing teacher ! :D
Thanks for the awesome content !
Wait, I have to date 0,37*8billion people before I can start to choose?
If you are bisexual then yes.
Not quite. If you set a total limit, a maximum of how many you are willing to date in a 3 year span, then the principle can be applied.
Remember, you choose N.
Waaaaaaaaaaait... You can probably say that youd only choose women.. and only women which are in the age between 18 and 25 ... so yeah idk but thatd be kinda smaller than 0,37*8 billion:D
Ou wouldnt it be interessing to calculate how much time itd take? :O
.. or how much it cost.
Haha... I'm loving the "prof j grime" scribble on one of the toilet walls!
I love how the first toilette you visit at a music festival is a LSD trip... :D
it is actually radioactive.
When your geiger conter starts buzzing that's definitely not a good toilet!
I'd love to see how this maths changes if you want to try to select *one of the best two* toilets, or any toilet that's in the top ten percent of toilets. You'd no longer be looking at a point on the curve but a range.
I would prefer to optimize expected quality (assuming quality was evenly distributed over a known range) instead of picking only *the best* toilet with a known probability.
I asked myself the same. Also what if you have friends and you want to select good ones for them as well?
The answer would be the same, as long as you want the range around the point where d/dx = 0 (or the optimal solution) the answer still holds because that point is the best point for getting a toilet in that range.
This way is still the best. If that best toilet is in the first 37% you’ll still pick the SECOND best if it’s not, and so on. You can only pick the worst if the best 37% are the first 37% AND the worst is the last toilet.
Yeah, so this is called the (1,2)-secretary problem or more generally (1,K)-secretary problem where you are happy with anything in "top K". There is a related but different strategy of how you should do things here. For K=2, this new strategy can succeed with ~60% chance and for even larger values of K your chances improve even more (as you may expect).
If you tel everyone this everyone's going to leave the first K toilets so they're going to be the most hygenic...
that's assuming everyone checks all the toilets in the same order
If they take enough K they probably can't walk to a toilet
I shat myself after 12 minutes of watching.
Is there a mistake in the video? I'm trying to understand the maths. At 5:31 there's an illustration. Shouldn't the chance of being picked at K+1 be 1 instead of K? Because this is what is explained in the video. So perhaps I don't understand it but I think the image is wrong. Where it says K on the toilet door, it should be 1.
Was thinking the same the illustration is obviously wrong.
Regarding the calculus, I won't pretend it's wrong but it's at the very least glossed over which is a shame it's really the interesting part of the demonstration.
I hope everybody learns this and uses this, because I'll then always pick one from the first 37% that nobody is using. ¦¬Þ
DB Fuller And how you gonna find the first 37% of rejected ones?
Christian Boudjaoui They'll surely be those closest to the entrance. People are sequential like that.
DB Fuller So you go to the toilet just after you enter the festival? I think you would be one of the few in this world who does this. So I believe this would not work.
Christian Boudjaoui what I meant was the closest of the entrance to the bathroom area.
+Christian Boudjaoui Being one of the few who do so was exactly his intention behind it though. :P
I gotta get a date first for this mathematical model to work..
4:02 I do hate it when I'm at a music festival and I miss out on the toilet with the chandelier. Come to think of it, I always seem to. How unlucky am I?
the answer: Toilets with chandeliers don't exist.
Kudos to the artist who painted all the various toilets. The radioactive one at 6:34 got me.
I think the argument being given at 4:44 about the probability is irrelevant. The probability is 1/k+1 because the probability of choosing the (k+1)th toilette is the probability that the (k+1)th toilette is the best of the first (k+1) toilettes. Am I missing something ?
I think that's the correct analysis!
I was also struggling with what she said. What you said made a lot more sense and can be easily generalized to deal with the general case of the best toilet being instead in the (k+p)th position in a row of N toilets, where p is an integer >= 1.
Thanks for the clarity! xd
You are welcome. At least I know that I wasn't the only one confused by that argument.
Oh thank goodness. I couldn't get past 5:00 in the video because of this! I kept going back and re-watching the previous 30 seconds, hoping it would make more sense on the k+1 th time through. Thank you so much for your comment.
I had to think about this as well. So the probability that toilet k+2 is selected when it is the best toilet it is 1 - the probability that toilet k+1 is not selected = 1 - 1/(k+1) = k/(k+1).
The probability toilet k+3 is selected when it is the best toilet is 1 - the probability that toilet k+1 or toilet k+2 is selected before it. The latter is the probability that the best value among all of the initial k+2 elements is not in the initial k elements = 2/(k+2). So prob toilet k+3 is selected when it is the best toilet = 1 - 2/(k+2) = k/(k+2) and so on.
@@timbrown5503 Thank you so much Tim. I really couldn’t let go to kept watching without your explanation.
Honestly, while the video wasn’t bad, it was way too short and glanced over many things that should have been given more explanation, like why k/n is the probability after k bathroom trials, the bathroom you choose will be the right one, or why that series shown can be approximated to the Integral of 1/x
Hmm I'm probably wrong here but at around 5:30 the probability for choosing k + 3 should be (k + 1)/(k+2) as its 1 - 1/(k+2) right? Or am I missing something?
no since the probability of a tree being better than the first K in K+2 trees is 2/(K+2) because it can be in position K+1 or K+2.Therefore it is 1-2/(K+2) =K/(K+2)
I really like how they went through all the maths in this video. Makes it all much easier to understand.
well.. I guess you could say that's a pretty.. shitty video.
... I Love it!
Don't be a potty mouth!
+Noel Goetowski Kids these days are all about the toilet humor.
James Mangum Such a waste...
+James Mangum These days? Kids have always been like that.
You could say it's a pretty helpful video.
Brady the video is brilliant as well all your channels are. Just something to mention. In animation at 5:30 min there are displayed some probabilities. I suppose probability for toilet k+1 is not k but 1. Check this out please. Thanks so much for such a wonderful job.
Why didn't you surround that probability between two aproximations ? A bound from above and a bound from below with integrals ? It's not that complicated, it won't have made that video less clear, but it could have shown how accurate that approximation was for high values of N, and how our strategy behaves for low values of N.
A video on further strategies would be appreciated : what if we are ready to accept one of the 10% best toilet ? Not necessarily the best one, but one of the best ? And what if on the contrary we really really don't want the worst 10% ones, but don't really care if it's a little dirty ? What would be the best strategy then ?
I'm not sure, but I think that the explanation around 4:40 should be clearer: the probability of having chosen k+1 because best compared to previous toilets is 1/(k+1)
I liked this better when this was the Monty Hall problem.
This was a very interesting problem. But it assumes that you have information about the size of N, or the number of options available to you. What if we don't know how large N is. Or what if the dates, the potential secretaries (or the toilets) "arrive" with a certain probability, let's say with a Poisson distribution. How many do you have to reject first before you take the next best one? Does a solution even exist?
I feel that the secretary problem should have applications in behavioral ecology. It seems like a nice rational solution.
Sorry about the long post. I just found the video very interesting. Thank you Brady!!
It's weird that in this video she's still rushed, even though this is the video that people were warned it would be longer, and heavier on the maths. x3
7:40 what is the name of this approximation? Or how can i find out about it on the internet?
conclusion: mathematicians always take the first toilet - they are in a hurry because they spend all the time calculating the optimal choice
I love how at the end she compares the toilets to relationships and being a boss
So if I believe for example that only 5% of women on this planet is fit to be my life partner. That is 1/20 and I should therefore find 20 random women, reject the first 7 and then start picking the first one that is a better match than the first 7?
Id recommend looking up on the secretary problem in other websites. Reason being is that even for someone with a good grasp of maths, I doubt they would be able to fully accept the jumps in conclusion in this video.
finally some good maths! thank you for this video :)
I am so glad I picked the long explanation and not the short conclusion, this was so much better!! Thanks for the tip ;)
There is just one huge problem i have with all these calculations.
It clearly assumes that each toilet has an equal chance of being good/bad, while in reality toilets close to the entrance definitely have a higher chance of being worse.
Also this only covers the propability of choosing the single best toilet of all these toilets, instead of just saying you could be content with a certain point of being clean.
Also when i have to go to a toilet, i do have an issue with time, which means i really dont want to check a large amount of toilets, because afterwards i won´t be needing one anymore.
Considering these objections, i´d rather choose the first toilet, that goes into my standard instead of trying to find the best toilet ;D
Or until your bladder chooses the one you're closest to.
I think your 'toilet closest to the entrance' objection expresses the difference between knowledge and wisdom. That made me think. Thanks!
The problem isn't really about toilets. They just explained it in a way to make it more digestible.
The use of the toilets was just to be cutesy. In reality, this problem has huge implications in decision theory when the given assumptions (draw from uniform distribution, see each draw w/out replacement, etc.) are met.
You can generalize this problem (i.e. don't assume each toilet is equally likely to be good or bad), but it would be too complicated to go through in such a short time (and I assume that it wouldn't have such a nice closed form solution as this problem had).
The number on the (K+1)th toilet at 5:38 should be K/K. Dr Symonds did say at 3:35 it was 1, and it would make the generalization of the probabilities of subsequent toilets being selected sensible.
Just wondering, say you didn't end up picking the best toilet, on average, how good is the toilet you will end up choosing. Top 50% of toilets? 40, 30, 20%?
That was some surprisingly intense math for a problem about dirty toilets XD
But seriously, I can see these results being useful in all sorts of situations. The only draw back being that you have to know exactly how many things you have to look through before hand.
I think I am in love.
this was so fun!! I'm liking these math explanations with calculus, it really gives you an idea of how powerful calculus is!!
Wasn't one of your original problems that you didn't want to look at every toilet? You are going to automatically look at 37% percent +1 of the toilets and then there is a 37% chance that the best toilet was in the original 37% so you would end up looking at every toilet
WRONG. you look for the next toilet that is better than the first 37%, which means you don't look through them all. It might end up like that, but it's not likely
+ismartroman gamer If the best toilet was in the original 37%, then after K, another better toilet doesn't exist, therefore you'd have to check every one.
+MetaKnight68 I said that you might end up looking through them all. I never said that you never have to. Listen before you reply. And also, as stated in the video, there is only a 37% chance of that haplening
+ismartroman gamer My apologies, although I don't see how your statement refutes the original comment. I suppose that I should have directly stated that, rather than trying to explain something I thought you didn't understand (though I see now that you clearly do). You say that there is a 37% chance of the best toilet being in the original (as it is stated by the video);that is what the original commenter said, therefore Shaqtapus is not incorrect. I am curious as to why you said, "WRONG."
I agree that I didn't state my intent for the reply, and I apologize for that. But your comment doesn't make sense.
Also, please check your spelling before commenting.
@@chronos3783 only 37%? That is a very high chance!
11:10 I think this function should have X in the domain [0,1] on the horizontal axis, since that is the parameter of probability function P (or k/N since that is equivalent). In any case not just 'k'.
N is a constant, so it kind of doesn't matter.
Picking the right K (the one that maximizes the probability) automatically results in picking the right K/N (the one that maximizes the probability) so if you have K/N on the x-axis you end up plotting the same function on a diff. scale.
I believe this is just a case of notation. She is graphing the function P(k) if you look on the vertical axis. It is confusing to once again use k as a generic value on the horizontal axis but remember that x=k/N so as long as k (being any value on the horizontal axis) is between [0,N] your x is between [0,1]
As you pointed out, x is equivalent to k/n. This makes x a percentage of n, dependent on k. In this way, x converts the number of rejected toilets into a percentage of the total number of toilets. The new value is now in a form that the probability function can use.
Speaking in terms of the graph, if you were to increase the value of n, you would stretch the graph horizontally, because n is inversely proportional to the function value. As the total number of toilets gets grater, the probability of any one toilet being the best choice gets smaller. But, since there are also more toilets, there will still be a point at which the 37% can be obtained, even if it's further down the line, or X-axis.
The most annoying part of this video is (I'm sorry to say) Brady's impatience. Though he has access to all the viewer statistics, which I can guess induces such behaviour.
RyanCreatesThings Some people are simply unable to get social cues. What can you do.
What?
i prefer the method "to definitely not have the worst" to the method "to have the highest chance to have the best".
But the Mathematics behind it is nice and *well explained*!!
What if this was a concert attended by mathematicians all of which follow this method?
Then everyone would be choosing the last toilets which would be the worst as the first 37% are unused and so are the best and so are unbeateable
@@shadow-leo6519 only if they open them in the same order
they would all choose the last toilet.
So for a rule of thumb you can remember anywhere anytime: If you can only check a toilet once, check a third, and then pick any that is better than those. And i would also argue; settle if you find one that is past the point of "good enough" to save time if the selection is large and time of the essence.
Ph.D. in Toilet Science.
Sounds legit.
I have a questions about this:
1.Why is differentiation allowed at 11:50? The function is obviously discontinuous-is it defined only at certain points. Just because a function can be integrated (by an approximation in this case) does not mean the converse is true. Let us not even forget that some function like the Weierstass function is continuous and does not have a derivative.
Wait, so you're saying i have to date 37% of the entire population before choosing to settle down and have a family?
RyanCreatesThings No but you can estimate the number of persons you will ever consider a relationship with and choose that as your N.
+RyanCreatesThings To be more realistic, the dating thing is in relation to your age. Assuming you start dating at 15 and end at 35 if you don't find anyone, reject those from the first 37% of that time frame (which is about 7 or 8 years, so when you're 22 or 23, which makes sense) and then start comparing your current date to the rest.
Pawcce Fawce But still possible.
+arcuesfanatic I'll just settle for the first one that doesn't run away.
+arcuesfanatic Which still begs the question, when should a person start commiting? We need estimate a person's dating potential, taking into account age and dating pools. Then apply the optimal stopping theory, and finally translate this back to an age. Or keep using K, maybe both? Which one would be optimal? Should we commit when either condition is met, or wait for both of them? The period between where one is met and both are met sounds like a "sweet spot", where if your partner does not commit, you still have the highest probability of getting the best choice. Actually, the problem itself should consider the other party not commiting, but how can we calculate the probability for this? +Numberphile2 +Numberphile answer in a new video maybe?
Brady, could we get a derivation for the case where we want the best AVERAGE result, instead of the best chance at getting the highest-ranked result? I think the problem was solved by Bearden (2006).
You realize that now more people know this and chose toilets more in the middle, which completely changes the complexion of the question. If now more people use the 37% the toilet, then it becomes dirtier, meaning one should use the therefore less used earlier or later toilets
But not everyone comes to the toilets in the same order so you wouldn't necessarily see higher use around the "k"th toilet as it would be a different one for each person.
The original assumptions of the question were a bit loose to begin with. It is highly unlikely that any given toilet has an equal probability of being the dirtiest or the cleanest or anything in-between. It is more likely that as you get closer and into more high traffic areas that the toilet condition will worsen. A uniform distribution may not be the best assumption to start with.
Not necessarily. You have to figure that people we either a) not know about this solution, or b) be in too big of a hurry to care, and use one of the early ones anyway. Combine that with the fact that more use most likely means worse conditions, and you'll just wind up choosing more towards the end as the middle ones become used more and more, but the first will always be used more because of the most important factor: desperation to use the toilet.
4:40:
The one thing I don't understand is why we still consider 1 thru k as being possibly selected. Isn't the probability of selecting those toilets 0 automatically?
Worst relationship advice ever D:
Maths says your wrong. :)
at 9:03 why do you find the area of the graph from K to N (integrate 1/x from K to N) , aren't we finding the sum from K to N-1?
why are we not integrating 1/x from k to (n-1), as (n-1) is the last term of that (1/k) series!!??? why??
sanchit goel
Because when we approximate the discrete fraction series as a continuous integral each term must become a bar that extends from its original position to the next position (otherwise the integral would be 0).
So 1/k becomes a bar of height 1/k that covers the distance from k to k+1. In exactly the same way 1/(N-1) must be a bar of height 1/(N-1) that covers the distance from N-1 up to N. As such our integral must end at N not N-1 otherwise 1/(N-1) would have no distance and so evaluate as 0 in the integral.
Another application is if you're at a store with a lot of lines to check out. You walk by a few and then hop in the best line you see after that.
Programmers way:
Premature optimization is the root of all evil. Unless you require the optimal choice, consider taking the first acceptable alternative. If the toilet is acceptable but not optimal in group K, choose that toilet.
really nice and clear handwriting. I love it.
This math works out for an arbitrary example where you have no context of what is acceptable and what isn't, but there's two flaws with the specific example of toilets.
1) There's a threshold of instant acceptance. You know you're at a music festival and if you get toilets 3 or 4, you'll probably take it, even if it's before k.
2) You can't quantitatively compare toilets fast enough that you won't piss yourself.
Very interesting! It's fun to see someone who isn't afraid of actually doing the calculus :D
somehow the audio is very quiet in this video :/
That was a great video. I'm happy to have a bit more of the mathematical side here :)
There's only so many times I can hear Dr. Ria Symonds say "toilet". Of all the framing devices you could've used… T_T
Great explanation, thank you.( I did wonder if this might be an audition for Numbers - I could easily see you saying: "We can easily find the murderer with a little calculus.") I'm going to use your calculation to select the best time server from a pool.
what if you wanted to have the best average state of toilet? Let's say each toilet has a random state of hygiene between 0 and 1, what would you need to do to pick - not the best one - but the one that's as close to 1 as possible? because with this method you have a 37% chance of getting the 1, and 63% chance of getting anything else, resulting in an average state of 0.37x1 + 0.63x0.5 = 0.68 average state of hygiene. Is there any better way of optimizing this?
With the given algorithm the odds are better.
The odds are 37% to find the toilet with hygiene 1.
The odds are 37% you skip the hygiene 1 toilet, and end with a random toilet with hygiene < 1. Let's say this average hygiene is 1/2, but in fact it is a little bit smaller, as you have 0 chance of getting hygiene 1 this way.
The odds are 26% you pick a toilet but will miss the hygiene 1 toilet. In this case the hygiene of your toilet still will be very high, as it beats the first 37% you checked.
Hence your score will be 0.37 + (1/2 * 0.37) + (nearly 1 * 0.26), an expected value between 0.555 and 0.815 that is close to 0.815. I still believe it can be better, but the estimate of 0.68 is too low.
Wow, i commented yesterday about wanting a section of Numberphile with more details, and there it is! Cool :D
What if the music festival has loads of mathematicians visiting, where they take all the best toilets using a similar method?
It still works if you put the toilets into a random order first. Otherwise it drifts into game theory (choosing your strategy depending on what you expect others to do)
i just bring my dog to pick up my poop after me
Dont waste your time with this, i'd just pee in a bush lol.
If you know the festival is going to be full of mathematicians who know this, you pick one of the first toilets because they're all going to be unused.
what's a mathematician doing at a festival?
5:40 - I small an e^-1 come here soon... It always seems to pop up in problems like this.
I'll just hold it in
I wasn't expecting calculus in a video about probability. That's the beauty of maths :)
I don't know about you, but when I gotta go, I don't care about getting "the best", but "good enough". What's the answer if I only care about getting a toilet that's in the top 10% or 20%
Great video Brady. I appreciate the work you put into these!
If the best toilet is in the first 37%, you'll have to choose the last one obviously ?
Yes, and that last one might even be the worst! 37% of the times, the best one will be in the first K that you skip; so your chance of ending up in the worst should be P=0.37*(1/[n-1]); 1/[n-1] times the chance of picking the best one!
Yes. You are unlucky then, but it is still more safe to do as said. if you go to 100 music festivals, and do this every time, you will have best toilet more often than you would have bad toilet.
The worst case is if ALL of the top 37% toilets are in the first 37% AND the worst toilet is the last toilet.
Actually, according to this algorithm, as long as the best toilet is in one of the first 37%, you will have to pick the last toilet,... so there is a 0.37/N chance of picking the worst toilet. The probability of picking then 2nd best to 2nd worst toilet would be somewhere between 0.37 and 0.37/N
Yes. So this algorithm optimizes your chances of picking the best, but it doesn't necessarily give you the best chances of optimizing your choice. Say, for example, you died every time you picked anything but the best. Then this strategy will keep you alive about 37% of the time, and that's all you care about. But chances are, you don't mind settling for a "nice" toilet, if it means you won't get stuck with the worst one. For that, a more complex, more subjective "minimizing regret" algorithm is called for.
tldr: This cannot be considered the best way to pick a toilet until you precisely define your objective.
I'm glad you made this. This problem is in my Martin Gardner book but I couldn't follow his explanation. This video cleared things up for me.
k = N/e is certainly the formula to remember, but it does yield some wrong answers for small numbers of toilets, 7 and 10 toilets for example. I tried improving on it a bit by using other approximations than the left hand rule shown at 8:23. Using the midpoint rule we set the bounds of integration from k-1/2 to N-1/2, but then later it becomes impossible to solve for k. Using the trapezoid rule we set the bounds of integration from k to N-1 and then add (1/k + 1/(N-1))/2 to the result. This yields the more accurate formula k = (N-1) / e^(1-1/(2N))
Anyone else do this in real life?
The derivation just shown relies heavily on the assumption that you value all toilets execpt the best one at zero.
Meaning the *second best one* and the *worst one* have the *same value* to you...
This is a quite unlikely assumption in the real world ;)
DiEvAlDiEvAl You would only need to date 18.5% of the population first because only half are of the gender you're interested in.
DiEvAlDiEvAl I know not every one is, but I'm pretty sure the vast majority is.
This is used in finance to price American options. It's easier when you learn dynamic programming (done in a different video by Numberphile: the one with the prisoner on a chessboard)
I do now.
I'm concerned about the phrase "choose the next toilet that is better than all those we have seen." I understand this selection rule correctly, it means that if the best of all the toilets is among the first 37% of the available toilets, we will be forced to take the last toilet as the best toilet was passed before we have decided to accept a toilet.
Or am I missing something?
So basically, there’s a 37% chance with this method you will screw up
So since there are 3.5 billion women on this planet, I need to go out on 1.3 billion dates and THEN start looking for my soul mate? That sounds.....time consuming.
tell that to Ted Mosby.
No, just date until you're 37% of the way to your death.
Ppaatt so just assume you'll die at 100 and date until you're 37
Greg. Sym. unless you get hit by a car tomorrow.
What becomes the tipping point given a variable of allowing a return to the best of everything already evaluated?Is there a defined formula?
Am I supposed to know this mathmatical way of solving as a 14 year old? It's when it comes to k, it starts getting messy for me. I also get the ending, just not the way finding the results.
I think at 14 it's unlikely you have seen Calculus yet, so the integration and derivation that she does near the end are not things you would know how to do yet. And probably you also haven't seen the approximation of the exponential function (the graph at 8:27 that gets smaller and smaller as x increases) but you may have seen exponential functions in general and their opposite: log functions. (e.g. log(10^3) = 3)
In Canada, we did "Pre-Calculus" (which covered all sorts of functions, approximations, series, etc) in grade 10 and 11 and Calculus in grade 12 (last year of high school), but many people do not see Calculus until university. Other countries have a more accelerated math program and they get to learn more earlier on. It depends.
I wouldn't worry about not understanding all of it right now, you will soon enough. It's very good you had a look and followed along. :)
Oh ok thanks for making that lift from my shoulder :D Also I'm from Denmark, so I also have to translate in my head all the time, that gives another challenge. But yeah I really want to be something along scientist/astronomer, I've had the wish to change the world, and extend the humanities living.
I usually watch alot of videos on TH-cam about all the things that is related to Physhic, Math and Chemistry. Because I feel like to be the best, you need all what you can get. ;)
It's basic calculus.
They teach this (differentiation) in high school (or first year in university), so you are not expected to know it yet. You can look into it, however, if you're interested.
No. Calculus isn't even high school stuff in most places, so unless you will take math in college you'll never be taught the basics required to understand this. On the other hand, this isn't *that* complex for calculus, so you probably should be able to grasp it given some time, but don't be disappointed if you don't right away.
At 7:07, the term 1/N doesn't go inside the parenthesis, shouldn't it be (k/n)(1+1/k+...)?
Solution: do not go to music festivals.
Well, if this was the first solution you came to, then you should have no problem keeping to that, seeing as how you are currently on the Internet
Striped Marlin Gaming If that reply was meant as some kind of insult, then you did a poor job. If not, then ... I agree?
I don't go to festivals for other reasons. For example, last time I was sexually assaulted and no one cared to help. If applying maths and probability theory to decide which toilet to go to is the biggest problem you have with festivals, then you are either oblivious or delusional.
My original comment was merely a quip.
it was kind of a combiation of an insult and a joke, seeing as how i am the most hermit-like internet person i know, and would probably never leave my house if i didn't have to.
Teragauss Cuddle other solution- do not date.
Redrum
Handwriting is super neat!
Haha If You were on a festival of matematicians, where each of them would pick their toilet with that principal, the would all use the same (last) toilet
(if they all check them in the same order)
+BabylonicaMan I was coming with an explanation of why you're wrong when I realized you're right. So everyone passes by the first 37 toilets and then starts looking for a better one. But since all the toilets are in the same unused condition they'll never find a better one, so they keep going and going until the end and will have to use the most disgusting possible toilet. Awesome.
So if everybody has to check the toilets in the same order (for some reason), that would be an example of a game (competing for the best toilet) where, if everybody picks the best strategy ignoring others' strategies, everybody would get the worst possible outcome
To take away any confusion: the K on the fifth door at 5:30 should be K/K (=1).
Only true numberphilers here :)
On 5:29, shouldn't the probability of k+3 being chosen be (k+2)/(k+3) and so on? It would be 1-(1/(k+3)) which would develop to (k+2)/(k+3)... Am I wrong?
Do you want to tell me that I have to reject 1.257 billion women first to get the best chances (36.8%) of finding the perfect soul mate in the world?
i like you
I think there is an error in the graphic shown starting at 5:30. Under toilet K+1 it should be K/K not K. So the sequence should be K/K, K/(K+1),K/(K+2).
I disagree this is the best way to find a toilet, secretary, or soul mate. If the toilets are placed randomly, which this video seems to work with, then there's not only a 37% chance of finding the best toilet, but there's also a 37% chance of skipping the best toilet. That's because, when you have 100 toilets, there's a 1 in 100 chance for each of those toilets to be the best toilet, since it's spread randomly. Skipping the first 37 of those means you have a 37 in 100 chance (=37%) to have skipped the best toilet.
What I'd do is simply check what your own standards are, and settle for the first toilet which you think is good enough for you. You got to set a standard for yourself in advance, and if you find a toilet, secretary, or date that meets that standard, you settle for it. If it disappoints you after you've tried it, you move on (i.e. you pick another toilet next time, or you find another secretary, or you break up).
I'm assuming nobody would actually dump their first 37% of dates, because then the chance of finding the best partner is biggest. Nobody would want to have a 37% chance of having met your soul mate, but having dumped them just because maths said so.
This assumes you don't know what the best toilet looks like. Also, the problem is mathematical, so doesn't involve 'standards' or else you'd factor in stuff like 'I might miss the concert'
***** Yes I know this is mathematic, but that's basically my point. Not everything is best dealt with by maths. In this case, you should just do it without maths.
This is assuming that no scale or maximal/minimal/optimal solution is known or assumed. Of course in more specific situations, we have the luxury of having standards (i.e. scales or metrics of some sort) or optimal solutions (to which we can approximate or assume nothing we have could be closer) and make a decision based on those preconceived assumptions. However, if you don't have those preconceptions, then the method given in the video is optimal to making a decision (based on the assumptions the video makes).
If there is 37% chance of skipping best toilet then there is 63% of not skipping it.
Zvonimir Mlinarić true, but would you, in practice, really use this to find a toilet/secretary/date?
Lol 10:50 Prof J Grime. What do you have against James Grime, Brady?
Hate to be that person, but the total number of toilets is N+1, not N.
+Srinath Kumar ??
Srinath Kumar who even zero-indexes in combinatorics
Dr. Symonds, you're a great presenter, a brilliant mind, and a very cute cartoon character!