Euler’s Pi Prime Product and Riemann’s Zeta Function
ฝัง
- เผยแพร่เมื่อ 7 ก.ย. 2017
- NEW (Christmas 2019). Two ways to support Mathologer
Mathologer Patreon: / mathologer
Mathologer PayPal: paypal.me/mathologer
(see the Patreon page for details)
What has pi to do with the prime numbers, how can you calculate pi from the licence plate numbers you encounter on your way to work, and what does all this have to do with Riemann's zeta function and the most important unsolved problem in math? Well, Euler knew most of the answers, long before Riemann was born.
I got this week's pi t-shirt from here: shirt.woot.com/offers/beautif...
As usual thank you very much to Marty and Danil for their feedback on an earlier version of this video and Michael (Franklin) for his help with recording this video..
Here are a few interesting references to check out if you can handle more maths: J.E. Nymann, On the probability that k positive integers are relatively prime, Journal of number theory 4, 469--473 (1972) www.sciencedirect.com/science/... (contains a link to a pdf file of the article).
Enjoy :)
"Euler pushed that infinite sum to the limit.."
You deserve a beating for that joke.
Even better, he said 'to the absolute limit' to make it more complex. Is that good enough for a total beat-up?
From the Inner Mind, to the Outer Limits!
That was a Parker Square of a joke
@@tejnadkarni131 Ah, I feel like you're a traitor, even though I do like Numberphile too
This channel and 3Blue1Brown have fantastic visuals that are very helpful. Both also have excellent, clear presentation. No other mathematics channel I've found come close to your skill in communication, which obviously involves a great deal of work to create all the graphis.
Best channels for insomnia so relaxing
The expressions 6n +/- 1 produce all prime numbers greater than three, and many more composite numbers. If we knew exactly where the composite numbers would appear in these sequences, we could infer the location of all of the prime numbers. Am I understanding this correctly? Of what use would this be to anyone?
My mind is so blown that I might cry a little later at the indescribable beauty of what I just learned. Thank you for existing Mr. Mathologer.
"Take the differences between 1 and the square of reciprocal of every prime number. Multiply them out to each other and take the reciprocal of this product. Multiply this reciprocal by 6 and square root the result. The value thus obtained is the ratio of the distance all around a circle to the distance across it".
@@aaronleperspicace1704 something interesting about this is how mystical pi seems when it is classically defined in terms of circles. No wonder hardcore analysts think of pi as defined based on infinite series and do not appeal to geometry.
14:00 with the primes < 1.000.000 you get 3.14159254713, 6 decimals correct!
(Used Python...)
I love how this shows yet another intuitive explanation on 1is neither a prime nor a non-prime.
By all means it's a non-prime.
@@viharsarok So, it has other factors, beside itself and 1 (which is itself)? I’d like to know more about these mysterious factors. 🤔
Mate, your pressentation is STELLAR! Seriously, the transitions, coloring, everything! Great job :D
This might be my favorite video of yours yet. So much amazing information - especially how the prime product connects to the zeta function. That has always confused me - and I was probably too lazy to look it up - but that it's so simple is just ..beautiful in a weird way. XD
"especially how the prime product connects to the zeta function. That has always confused me - and I was probably too lazy to look it up - but that it's so simple is just ..beautiful in a weird way." Ah, me too, I'm glad that I understand it now
I solved the Riemann Zeta function! I posted the solution, check it out!
I solved and proved the Riemann Zeta function. I just posted the solution check it out!
I was deeply astonished by Euler's geniusity!! Thank you for introducing his research to me
I liked this format... with you floating around the equations. Much better than you presenting into a board, or scaring me with sudden sound effects. Keep it up!
Your graphics make these lectures.
I foresee a time when all math is taught like this.
I can't follow half of it, but I can at least follow.
Thanks for these
Professor, these are perfectly paced. They don't go so fast that I lose the chain, but never so slow that it becomes unexciting. It's high level, so there's room for mulling over what's needed to ground the proof. There's always hints of ways to take it further. And the use of the animated background blackboard to do the steps while you explain the ideas keeps it flowing.
If you offered this type of freewheeling 'class' at your university it would be packed.
And PI is related to the primes?! This is at least as amazing as e^(i*π).
Glad these videos work for you exactly as planned, and thank you very much for saying so :)
12:00 Woah!! That was awesome!!! I just finished taking my Random Processes course as an Electrical Engineer last semester, and I'm so glad I watched this video after having taken that, cuz that was brilliant!
Please tell me how it was relevant! I'm doing math and physics research
You are my favourite math channel. I learn a lot from your videos!!!
Thank you... beautiful video, your explanations demonstrate deep understanding and passion at the same time!
Awesome video! This channel is probably my favourite math subscription :)
Great video, amazing explanation of how Euler arrived at that form involving just primes.
I love the Euler product formula. Glad you did a video on it.
Fiiiinally, now I understand how all the prime numbers got in there! Thanks for this video :)
I really appreciate your effort in doing these videos :)
many thanks for this video. it's the first tim i find the Euler's formula so beautiful and could really understand them
Probably my favorite video! Thank you mathologer!
Thanks for another fun video! It is great that you still make the time to do this work even though the crush of the school term is once again upon you. Looking forward to more fun with zeta in the next instalment!
If only I had a bit more time. It would be great to be able to make more of these videos :)
IMHO your math videos are some of the greatest to be found on TH-cam!! Thx a lot for your work, Mathologer!
And how come you speak German that perfectly!??
Hi Mr. Mathologer(for lack of a better name), while I was doing my math homework today, I remembered a problem my math teacher gave me and my friend in 5th grade after we'd finished the work a little early. The problem was as follows: the king's daughter is to be married, and to choose the prince, the king has decided to seat the knights around a very big round table. Then he will start at the first knight, and say "you live." To the second knight, he says "you die." The last surviving knight is the one who wins. The challenge was to make a formula so that you could find which seat you wanted to sit at, to live(given between 1 and infinite knights). I made a little progress, finding that (obviously) even seats are bad, and that the good seat starts at 1, then counts up odd numbers to 2 higher than before, then resets to zero, but I only had a little time and I'm not good at math. In retrospect, I don't think I've heard of any similar problems, which is why I still find it curious.
+Elie Simsch This problem is called the Josephus problem :)
Really great video man, keep it up :)
An absolutely great video. Keep up the amazing job!
Beautiful, I'm seeing stars and little numbers arround my head, this is the stuff that make me love math!
David Herrera Loma?
I think 32 years ago I was playing about with numbers and I discovered this relation between the prime product and the integer power sum (for real s>1) from a sort of sieve argument. I did not know this was a known result, nor had I ever heard of the Euler prime product or the Riemann Zeta function, but thought it was an interesting result. It was in the days before you could just google things. My father persuaded me to hand over my scriblings, so he could show it to a friended mathematician. Of course I got it back with the flaws in my 'proof' pointed out, without any hint that the result was known. My intuition was right, but I had no idea what a proof required.
brilliant video, as always. Keep up the good work.
your channel and videos are awesome, thanks!
unreal video keep it up man i love this
8:02 wow, that pronunciation was perfect - well done!
Greetings from beautiful New Zealand. If you happen to be in Auckland today as part of the Maths craft festival I'll be giving a talk about the best ways to lace your shoes at the Auckland Museum at 5.15 p.m. Come and say hello :) www.canterbury.ac.nz/news/2017/maths-craft-festival-2017-hits-auckland.html
As usual if you'd like to support the channel please consider contributing subtitles in your native language.
Welcome to Nz, hope you have an enjoyable stay
Please do some videos on fractional calculus!
+Josh I'll put this topic on my list of things to ponder :)
Mathologer I'm not sure whether your proof (with conclusion at 5:16) is complete. Is that really a contradiction? Even if one function is bounded above by another, couldn't they still converge to the same value, as long as the difference between the sums converges to 0?
You should consider doing a video taking infinite products to infinite fractions (which is again due to Euler), especially condiering you already have videos on infinite fractions you can reference.
I have watched many of your videos and I think this was the best one.
Just love your videos!
Hello Mathologer,
thanks you so much for your video, which is the most impressive one I have seen in regards to the relation between Euler's Pi Prime Product and the Riemann Function.
Just one small remark: I think on 6:21 it should say 1/27^z + ... and not 1/28^z + ... as stated in the video, simply because 3*9 = 27.
Nice and neat explanation. I would have wished to have a Math teacher like you when I was at school.
Amazing explanation I have been trying to understand this forever
Best math channel!
Salamo aalaikom..you are posing a very important maths issues ..i want to say that resect you and you have to continue ..respect from morocco
beautiful story like always, love your presentation. strong logic and very valuable historical facts.
Thank you for existing 🎈
Toll erklärt und sehr instruktiv. Ich danke dafür!
one of the best youtube chanel about math
Thanks a lot for this wonderful lesson!
very nice presentation. This is how math should be taught! Herzlichen Dank!
This was a really great video.
For this, you've just got another subscriber!
This is so satisfying!
AWESOME insights ! Expanded my knowledge / understanding INFINITELY ?
Very informative. Thanks!
Awesome video, very instructive and clear. I specially liked the probability connection of the zeta function
Glad you like it :)
Hello there, great video. I used a fast function I found using a boolean sieve to list primes in Python and listed primes up to n=1,000,000,000. There are 50,847,534 primes and the last one is 999,999,937. Using this list estimated pi with Euler product at 3.1415926536126695 accurate up to the 9th decimal. Can't find more than up to n=1,000,000,000 prime, memory error. It would be interesting to graph how the accuracy of pi evolves with the number of primes.
You are amazing sir.
Fantastic video!
I really hoped that you could prove the Riemann Hypothesis.....
😂 No, its ur chance bro.
The number of views of this video compared with, say, the number of views of your Harmonic Series/Gamma video suggests to me that this video is criminally underviewed... which might be why it seems like we're still waiting for that Riemann follow-up video you talked about at the end. I would love to see something like what you hint at, explaining the connection between the Riemann zeta function and the Euler gamma constant.
10:20 The killer instinct. Mathologer isn't afraid to strike at the heart.
Comment on 10:45 in video. The concept of a "random integer" is tricky. When you say, "choose one of these items randomly", if there are N items, then usually each item is chosen with probability 1/N. However, the word "randomly" assumes that you have defined a probability function on the items. 1/N is the the "uniform" probability function, but a different probability function would make it more likely to select certain items than others. *PROBLEM*: The integers can not be assigned a uniform probability function, with all integers equally likely, because there are infinitely many. The sum of the probabilities assigned to all items must equal 1, and if all integers had the same probability, these probabilities would sum to infinity. So, in choosing an integer "randomly", you must specify which probability function has been defined on the integers. There is an unlimited variety of probability functions that may be defined on the integers.
This is a really good channel.
I watch your videos with a lot of wow!s. Great fun!
Great, looks like "mission accomplished" as far as you are concerned :)
This is even better the second time around.
mind blowing excellent sir keep it up
Awesome video!
You should do a full video on the reimann heipothesis
Wonderful work. Congratulations! By the way, what software do you use to make this animations? I'd like to use it to prepare my lectures. Sorry for my wrighting. English is not my mother language.
When z = 1, the manipulations required to get the prime product result require multiplying infinity by integers, and then subtracting one infinity from another, and finally dividing one by zero, just for good measure. The conclusion at around 9:27 that there are infinitely many prime numbers is certainly correct, because we know that to be true, but I'm not comfortable that reasoning there is valid.
It is about time I subscribe .
I remember our math teacher gave this to us in a probability question(Find the probability that 2 randomly chosen natural numbers are coprimes). The answer is 6 by pi square
title seen - and already hyped
I want to make subtitles for your videos and give them to my friends who dont know English well enough. Is it OK with you?
Sure, that's be great :)
Bavarian Blue Wolf Sure thanks. Thats exactly what I'm gonna do
I had a math team practice about a week ago with the derivation of one of the (easier) identities being involved in the answer..
the question was:
Consider a infinite number of concentric squares, with side length 1, 1/2, 1/3, and so on, with the nth square having side length 1/n. Starting with the outermost square, the regions between the square are colored alternating blue then red. Given a dart thrown at the square lands inside the square, what is the probability the dart strikes a red region?
1-π^2/12?
Yep!
@Catharsis it's 1 - the alternating sum for n=2 shown in the video
Great video!
Hi - your video lecture is truly amazing. I am thinking about adding a video component to some physics courses I teach at Boston College in the US. Could you advise as to what software you use to create these amazing lectures. Thanks a ton if you can!
9:37 Even Euler "saw" that one coming :D
6:22 it's supposed to be 1/27^z not 28
כמה משעמם לך אחי
dmagen1994 אותה כמות שמשעמם לכולנו 🙂
lol, made the exact same mistake in a math exam. I also assumed 28 was dividable by 3
Yeah
Like
I want it to be 69 so I am replying
Riemann what a genius was involved in general relativity and prime numbers
amazing video!
I've seen no videos regarding the odd powers of this function. Can you do a video on it?
5:30, if there is strict inequality between 2 sequences a_n < b_n, it implies only non-strict inequality between limits lim a_n
It doesnt?
Assume your statement is true, then:
A = 1/1 + 1/3 + 1/5 +...
B = 1/2 + 1/4 + 1/6 +...
Let C= 1/3 + 1/5+...
D = 1/4 + 1/6 +...
By your statement C>=D.
A -1/1 = C
B -1/2= D
A - 1/1 >= B - 1/2
A >= B + 1/2
A > B
13:16 To approximate pi i used Euler's formula for acceleration of series convergence to the arctan(1)
After acceleration each iteration sets one bit of approximation
(10 iterations gives three correct digits)
Since you're a mathematician and you like Rubik's cubes, have you ever looked into the fewest moves challenge? You have one hour to come up with a solution to a scramble, and the goal is to use as few moves as you can. I recently started to look into this, and there are some wonderful mathematics that have helped people achieve averages of less than 30 moves! There is some really interesting group theory involved. There is a technique called insertions, which uses commutators and conjugates. But the most interesting thing I have yet found is NISS, which is also based on group theory.
I looked at this a while ago, but from your comment it would seem that there have been some new developments. Do you know of some good references? I should really do another Rubik's cubed video sometime soon .... :)
Well I learned most things from this video series on youtube:
th-cam.com/play/PL0yL0AZiHw10Kx5um2l4MdOAICtRqJwac.html
But this video series only explains how to apply some techniques. I came up with explanations for the techniques myself and you probably could too, but I later found this pdf:
fmcsolves.cubing.net/fmc_tutorial_ENG.pdf
This pdf goes more in-depth and has some explanations for the techniques it discusses as well. It also describes even some techniques not discussed in the video series.
I do think the video series is still very good and will probably help more with visualizing than the pdf, even if the pdf has pictures and you can follow what the pdf is talking about by applying examples to one of your own Rubik's cubes. But I also recommend reading the pdf, even if it is pretty long, because this stuff is pretty interesting.
Did you ever make the video you said you were going to make at the end? I am very interested in this
th-cam.com/video/YuIIjLr6vUA/w-d-xo.html (he made this - it addresses some of it)
That was really cool
I've been writing questions in my comments so far, but I want to say also Thank you, Thank, Thank you, for your videos. It's a delight to watch them, especially this one. I hadn't heard it before, because I'm interested in geometry and calculus, not in primes, but I must admit this is a beauty.
Now, to my question - sorry I can't let it be ;-) I would be grateful for an answer by everyone, of course, not only by Burkhard.
@12:13 - it was said: "at least two of the ingredients of this proof need a little bit more justification and, well, can you tell which?" I cannot tell it now. Could anybody tell me, please, which two ingredients that could be?
I like this video a lot.
At @9:46 the video seems a bit unclear of what is required to show for a second time that there are infinitely many primes, using equation:
pi^2 / 6 = 1 / ( (1 - 1 / 2^2) * (1 - 1 / 2^3) * (1 - 1 / 5^2) * (1 - 1 / 7^2) * ... * (1 - 1 / prime(max)^2) ) with an assumed maximum prime number, prime(max).
On the right side is a rational number, and it would follow that pi^2 is rational. But that pi^2 is irrational is a stronger statement than pi being irrational, like the square root of 2 is irrational, but its square, 2, is rational. Therefore, it's not enough to show that pi is irrational, which is difficult enough, but we would have to show that pi^2 is irrational for showing that there are infinitely many prime numbers. I personally find the proof using zeta(1) more beautiful.
8:15 That was perfect. Are you from Germany?
I grew up in Germany but have been living in Australia for the last 20 years :)
I always thought I can hear a slight German accent in you voice :D
Slight? You are being so kind... But it doesn't stand in the way of a great presentation.
german accent is good for math and ph ysics :)
Geniuses here. As if it's a slight accent, every person on Earth recognizes German accent. Breaking news: You're not special.
May I post an extension question? Consider this summation from 1 to infinity: k(s) = sum( n! / (n+s)! ). It converges for s > 2. k(-1) = 1+2+3+4+..... Does k(-1) converge to the same value as z(-1) by analytic continuation?
Hi mathologer, i love your videos! The pi^2/6 is beautiful. Do you know if there is a pretty value (free from infinite sums or products except maybe pi, e or the golden ratio) for zeta (3)?
Euler later did find a fairly simple derivation that gives, for even numbers only, the value of zeta as a rational number times that power of pi. In other words, zeta(102) is an explicit rational number times pi to the 102 power. It was a bit of mathematical good luck that left Euler the possibility to turn up the value for zeta(2). I can't imagine that Euler, and other mathematicians since, haven't done more than ponder whether something like it exists for odd numbers, like 3, but I have never seen any such natural thought expressed in print, pro or con. From that, I take it that whatever they have done shows no indication that it is a rational number times that odd power of pi, or any power of pi. It is possible, though, that professional mathematicians consider such speculations as the realm of crackpots, and do not care to be classed as such, and that explains their reticence. However, sometimes it is mentioned that, not so many years ago, it was proved that zeta(3) is irrational.
The nice formula for even values of zeta can be turned into a formula for odd powers, but it would involve the odd Bernoulli numbers, which are zero (beyond 1), and powers of i, which would make zeta for odd numbers seem to be both imaginary and 0. How could you fix that? It is the kind problem you know Euler would love.
Illuminating video. Have you ever shooted the video on the alternate inverse squares series? Any link? Thanks.
Great lecture. The only problem is with the probability of two random numbers to be relatively prime. You didn't define the probability space, namely the probability p_k of choosing a natural number k. It is an interesting question on its own, can p_k be chosen in such a way that
P({k, 2k, 3k,...}) = 1/k, as is stated in the video.
Love you!
This is amazing
I would really like to know what the Riemann hypothesis would show about the way the primes are distributed
One error, minute 6:24 the 28 should be a 27. Nice video, thanks for the info.
Thank you very much .
Hi there! I really love your videos! I'm curious what video editing software you use for your videos? It looks like you shoot the videos with a projector in the wall and then overlay the projection over it with your video editing software? I'm actually looking to get some new video editing software, and would love to be able to have the accuracy to be able to do a similar thing.
Yes, I superimpose a slideshow onto video of me dancing around in front of a blank wall. I just use Adobe Premier for video editing :)
13:00 Since any sample contains only finite numbers, the product will be cut off at a certain point (primes greater than our numbers will never get the chance to show non-coprimality) so that the measured probability will be a slight overestimation of 6/pi^2, resulting in an underestimation of pi. Could calculate the standard deviation of the estimation for pi using this method, and the effect of the part of the tail effectively cut off. Comparing these two could indicate whether this idea could explain that you actually found an underestimation of pi.
this is a good point, 6/pi^2 is just the limiting behavior
Hi :) I've always had a nagging suspicion about drawing (uniform) random numbers from an infinite set. I'm not convinced it's possible: here's an intuition about why I'm sceptical: it would be nice if someone could explain where I go wrong.
First we need to define:
URD: A "uniform random draw" for a set S, (S = {1, 2, 3, ... } in this case) is a random selection x from S such that every y in S has an equal chance of being picked. This is what we're doing in the video, and it's what we usually mean when we just say "random".
At 11:12, we said that Pr(n even) = 1/2 -- this is what I assume most people would agree with. I disagree that this is true, I think it depends on arbitrary selection rules. Firstly the intuition we normally have in mine: if you represent a natural number as a rounded scaler from 1 to infinity or a finite string of bits defining a binary number (like your computer does), then it's true that Pr(n even) = 1/2 : because it boils-down to whether the last bit is 0 (even) or 1 (odd), and since both 0 and 1 bits are equally likely, it follows that Pr(even) = Pr(not even), which must sum to 1 so Pr(even) = 1/2. You can jiggle this for Pr(n is a multiple of m) to get 1/m. It's easy to see that this way of drawing number satisfies URD. I think most people would just agree that, Pr(n is a multiple of m) = 1/m ... but I'm sceptical.
There is another way to uniquely represent numbers, not just as a line or scale: we can represent n as a product of primes via the fundamental theorem of arithmetic.
Consider n = 2^(a1) * 3^(a2) * 5^(a3) * 7^(a4) * 11^(a5) * .....
If we start by drawing random URD numbers a1, a2, a3, ..., over {0, 1, 2, ....} and then defining n as above, we have drawn a random n. In fact, since every different sequence {a1, a2, a3 ... } defines one, and EXACTLY one, natural number n, it also follows that this method of drawing n should also satisfy URD.
But under this last method, prob(n is even) = prob(n not odd) = 1 - prob(a1 = 0) = 1 - 0 = 1
If you want to be more formal, you can note the following observation: suppose you have N (non-empty) sets S1, S2, ...., SN. We want to choose a random element X from the cartesian product set S = S1 x S2 x S3 .... x SN. Then drawing X is equivalent to drawing N elements si from Si, and then setting X = (s1, s2, ..., sN).
In effect, I've used this result in my argument above that Pr(n even) = 1, because I've quoted the fundamental theorem of arithmetic and defined Si = {powers of the ith prime}, e.g. S2 = {1, 3, 9, 27, 81, ....}.
Since I've shown that Pr(even) = 1/2 and shown that Pr(even) = 1, it follows that it makes no sense to even assign a probability to drawing random numbers from the infinite set of natural numbers.
I assume I'm wrong somewhere, but I don't know where I've gone wrong?
It's definitely a bit tricky to pin down exactly what we mean by "chose a natural number randomly". Having said that the first thing that does come to mind when you think about it for a bit can be made to stick: Choose a natural number n and calculate the probability that two randomly chosen natural numbers less than n are relatively prime. Then this probability will tend to 6/pi squared as n goes to infinity. Ticking all the boxes proofwise if you choose to go this way is a little be tedious but not that hard :)
Thanks :D
There are indeed no uniform probability distributions on N that satisfy the usual axioms of probability distributions. However, there are some that exist by weakening one or more axioms, usually countable additivity. Countable additivity says that the probability of any countable set of disjoint events is equal to the sum of the probabilities of each event on its own. So for instance, in a Poisson distribution on a variable X where P(X=n) = 1/(e·n!), we can say that the probability that X is even is P(X even) = P(X=0) + P(X=2) + P(X=4) + · · · = 1/e (1 + 1/2 + 1/4! + · · · ) = cosh(1). This is sensible and does lead to the conclusion you posted above: that such a distribution over all natural numbers cannot be uniform. After all, if it were uniform, P(n) would be 0 for each n, yet P(n≥0) must be 1, which is definitely not the sum of 0 + 0 + 0 + · · ·.
By weakening the requirement of countable additivity to finite additivity, we can define a generalized probability distribution (sometimes called a "probability charge") that does have most of the properties we want. Specifically, it will still be true that P(X=n) = 0 for all n, but it will no longer be required that the total probability be the sum of 0 + 0 + 0 + · · ·. Instead, the total probability can simply be 1, and the probabilities of other countably infinite sets of outcomes can similarly be larger than expected by conventional rules. Because we still require _finite_ additivity, that means that finite sums work normally. For instance, P(X is even) + P(X is odd) = P(X is even or odd) = 1, and P(X=1 or X=2) = P(X=1) + P(X=2) = 0 + 0 = 0. We can define a "uniform" generalized distribution in a number of ways, but usually we require that the probabilities of subsets are equal to their asymptotic densities. So for instance, we would want P(X is even) = P(X is odd) = 1/2, P(X is divisible by 5) = 1/5, P(X is prime) = 0, etc. This distribution seems pretty reasonable at first, but picking at the edges reveals some very strange behaviors.
For instance, in a paper by Kadane, Schervish, and Seidenfeld titled "Is Ignorance Bliss?" the authors ask us to imagine a gambler who has bet on one of two mutually exclusive and equally likely outcomes, A and B, one of which must be true. If A is true, he will lose $1, and if B is true, he will win $1. Suppose further that a random number will be selected according to one distribution if A is true and according to a different distribution if B is true. Specifically, the gambler knows that if A is true, then the number n will be selected with probability P(n|A) = (1/2)ⁿ, but if B is true, then the number n will be selected according to the uniform distribution described above, meaning P(n|B) = 0, whatever the n. Thus at the moment, before n is selected, both A and B are equally likely, and the gambler has an expected value of $0 for his bet. But after n is selected, no matter what it is, Bayes' Theorem shows that P(B|n) = 0. Therefore, the gambler should actually pay money _not_ to know what number was selected, since if he does find out the number, he is guaranteed to lose.
Ultimate beauty of maths ......
Great. Only that I'm unable to find the next video which was advertised at the end.