Following this series in 2024 and its still gold♥.
@TristanSimondsenАй бұрын
That's basically 3 different games where the car doesn't move and you pick a different door each game. Any time you simulate as if player's already picked the car or know where the car is, the simulation or explanation is invalid. Player picks a door, that door has a 1/3 probability to contain the car. So if he opens that door, he has a 1/3 probability to get the car. Monty's action doesn't change that. So if he stays, he can open that door and has a 1/3 probability to get the car. If he switches, that other door he can now open also only contains 1/3 probability of him getting the car. Player has a equal probability to get the car no matter which door he opens. This will go down as the biggest hoax in mathematics history. You guys never should have doubted Paul Erdos.
@antonioruotolo6014Ай бұрын
Horrible explanation. Not clear where the probability comes from. Zero becomes 1/8 without showing how
@antonioruotolo6014Ай бұрын
In this lesson very obscure explanation
@ubiquitous19993 ай бұрын
I don't understand the part P (NA, B). Why not A and B we have 1/3?? Not A are only two options, Not A first option is B, Not A second option is door/room 3. Shouldn't it be then 1/2? P(NA, B) = 1/2x0 in the case when the car in the B and 1/2x1(in the case when the car behind the 3rd door, so it is only 1 option), so P(NA,B) = 1/2, I am so confused
@ChristinaWahlquist-h5z3 ай бұрын
Will you please show how to set up both integrals with the limits of integration.
@rajeshbehera53613 ай бұрын
caar..
@damienthorne8613 ай бұрын
Good stuff! I love it !
@emilygrace9423 ай бұрын
very confusing video
@JinXing-j1l3 ай бұрын
Dude, you sure this is your channel?
@lrvogt12574 ай бұрын
Options............ A, B, C = 3/3 Pick...................A = 1/3 ... B, C = 2/3 Goat revealed...B = 0/3 ... C = 2/3
@gulzameenbaloch93394 ай бұрын
Thank you so much
@mustafizurrahman56994 ай бұрын
We are deriving g1 from sample Momen equation I.e sum of Xi/n = uhat then g1 should be zero and so does g2, g3, g4
@quasimodo19144 ай бұрын
Hahaha, spent a whole week scratching my head as to what "conjugate" meant after my professor's roundabout wishy-washy definition, only for it to be easily explained in like 2 seconds
@S2lyaQ4 ай бұрын
Really? Can you explain what conjugate is?
@andreip93784 ай бұрын
These videos are underrated.
@TimLee3564 ай бұрын
wrong
@Araqius4 ай бұрын
Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.
@TimLee3564 ай бұрын
since you can't possibly choose the door that the host reveals which is always a goat, you start with 1/2
@Araqius4 ай бұрын
Assume you stay with your first pick. If your first pick is Goat A, you get Goat A. If your first pick is Goat B, you get Goat B. If your first pick is the car, you get the car. You only win 1 out of 3 games if you stay with your first pick. Switching means the opposite. It's just basic math/logic kids understand. Sadly, it's far too hard for idiots.
@aryamandwivedi12774 ай бұрын
Sorry for waiting for 10 years But your notation for gamma distribution is wrong because you interpreted the order of parameters incorrectly😊
@kaiwan53824 ай бұрын
I didn't understand why the probability of not A is 1/3. If the probability of A is 1/3, shouldn't the probability of not A being 2/3? Could someone help explain this? Thanks!
@kaiwan53824 ай бұрын
I can obtain the same joint probability of not A and B as the video, but using a different approach. Considering the definition of B as the host choose door 2 to reveal the goat explained by @dihan6130. The probability of not A is 2/3, and the probability of B is 1/2 given that there are two doors between door 2 and 3 to reveal the goat, so the join probability of not A and B is 2/3*1/2=1/3, which is the same as the result in the video.
@roberthuff31225 ай бұрын
Great stuff, but why is the impedance of free space (a vacuum) 336.7 Ohms?
@akramnajjar5 ай бұрын
A small note (given the wonderful videos) . . . this should be no 28 . . no?
@LaurentBourgault-Roy5 ай бұрын
For those a bit confused by the problem, I'll rewrite it here with more words - There is two tribe labelled 0 and 1. The 0 and 1 are not probability but just names. - We know with perfect accuracy if the individual are healthy or not. This is not a case were we have an imperfect test and it is possible that an individual has the disease but test negatively. In this universe, either someone is lying on the bed dying, or is running around in the jungle, with no in between. Ox Educ confusingly used the term "asymptomatic", but then say they are healthy. The rest of the problem assume we know they are healthy. - Theta = 1 does not mean that all the members are sick, it just mean "We sampled the individual from tribe 1, the one where some member are sicks" - Our model of the probability of an individual being sick is the function f, and it depend on the tribe we sample from. In tribe 0 no one is sick. so f(0) = 0 . In tribe 1, the infected tribe, half the member are sick, the other half is in good health, so f(1) = 1/2. Remember that 0 and 1 in the function parameter are just the tribe label, and not really a number. - Since all the individual we picked are healthy, we cannot be sure which tribe we picked them from. If any of them was sick, then we would have instantly knew that it was tribe 1, because it is the only on that has sick members. But with our current data, either we sampled from tribe 0 that has only healthy individual, or we sampled from tribe 1, and by chance, only picked from the healthy half of the tribe. So P(data|theta = 1,model) is 1/8, since we picked three individual, and our model say we have 1/2 chance for each individual for them to be healthy if we sample them from tribe 1
@amirzare55815 ай бұрын
thank you for going through a complete example from scratch. helped a lot.
@Garrick6455 ай бұрын
The videos are good information and knowledge wise but kinda boring. I loose my focus every now and then.
@grin2a7156 ай бұрын
i needed this so much! when i started my horribly organized probability theory class i needed simple and clear explanations to all the terms that were thrown in . well i didn't get it then but luckily you exist ! wishing you all the best !
@Andreas-q4k6 ай бұрын
good
@krstev296 ай бұрын
Do you know how complex is the induction proof?
@thevegg32756 ай бұрын
At four minutes and 23 seconds, you started to talk about geometrically defining dual vectors. I got really excited because I was sure you were going to talk about parallel projection versus perpendicular projection which describe contra variant components and covariant components a.k.a. dual basis vectors, respectively. Is there a connection between one forms and the perpendicular projection of vectors components called dual basis vectors ?
@marcobaccichetto72046 ай бұрын
Brutal video
@irawardani13187 ай бұрын
where does the values of 0.25 and 0.4 comes from?
@hdrevolution1236 ай бұрын
Best guess
@MarkPerry-lr4xq7 ай бұрын
This is a triumph of excellent teaching. Thank you.
@olichka16017 ай бұрын
Cool video! I finaly understood it! There are not so much about this theme in the web... thank you!
@ddddsdsdsd7 ай бұрын
I'd like to mention that we can go beyond the Monty Hall problem and touche on a fundamental issue in probability and statistics: the comparability of events with different sample spaces or magnitudes. 1. A first event with a magnitude of 3 (three doors) 2. A second event with a magnitude of 2 (two doors) While a great introduction into probability, the Monty Hall Problem only works if one accepts the comparison of two events of different magnitudes as logical. To dismiss people who cannot agree with this comparison as they do not get it is a problem if you ask me. I see the importance of highlighting both ways of thinking. In many contexts, comparing probabilities from events with different magnitudes or sample spaces is indeed problematic or even meaningless. For example, to provide another example of fundamentally different events: Comparing the probability of rolling a 6 on a die (1/6) with the probability of flipping heads on a coin (1/2) doesn't make much sense in isolation. In more complex scenarios, like comparing stock performance across different markets or time frames, not accounting for differences in magnitude can lead to serious misunderstandings. Additionally, people think if your initial choice was the car (which happens 1/3 of the time), then switching would be the wrong move. In this case, the host's reveal of a goat door doesn't help you at all. You've already won, and switching would make you lose. If one accepts the comparison of two events of different magnitudes, the Monty Hall strategy isn't about "always switch" but rather "switch because it's more likely you initially picked a goat." The host's reveal doesn't create new winning chances. The host doesn't change the fact that you probably (2/3 chance) guessed wrong at first.
@EffectiveMuscle7 ай бұрын
Great explanation. Thanks for this 👏👏👏
@fordland087 ай бұрын
Exactly, you go from picking 1/3 of the doors to 2/3 of the doors. It’s more about choices….
@Structuralmechanic7 ай бұрын
Lovely video Ben! As usual.
@muzaffergurersalan85298 ай бұрын
How could you say that theta is fixed when you integrate over it’s “different” values in 1:06
@mitch61518 ай бұрын
Thank you very much Ox, your class is extremely outstanding and great!
@SWJ-j8m8 ай бұрын
How dare I found this now.. Amazing works !
@Seanz20888 ай бұрын
I am confused by p(a,b)= 1/3 x 1/2. Does this assume A and B are independent? If they are independent, p(a|b)=p(a); no need to go through all the rest of the derivation steps.
@adityagarg2598 ай бұрын
why did you write P(\theta) that way, isn't that the density function of \theta and not the actual probability ?
@joneschilufya8678 ай бұрын
First day on Bayesian statistics. Lets go!
@franklyvulgar18 ай бұрын
This was useful for me in understanding why with entropy and cross entropy with KL Divergence the Cross entropy will always be greater than the entropy (you have to flip the inequality though because the function in question is a -log which is concave
@staggeredextreme82138 ай бұрын
Can't believe I'm learning what people learned 9yrs ago and I'm surprised what are they doing now 😳
@SphereofTime9 ай бұрын
2:26
@SphereofTime9 ай бұрын
1:48
@SphereofTime9 ай бұрын
0:13 0:14 0:14
@pomegranate85939 ай бұрын
fantastic! absolutely fantastic explanation!
@activision417010 ай бұрын
Would this also imply the reverse is true for a concave function? I.e., g(E(x)) > E[g(x)] ?
@hyperduality283810 ай бұрын
Stretching is dual to squeezing -- forces are dual. The Ricci tensor (positive curvature, matter) is dual to the Weyl tensor (negative curvature, vacuum). "Always two there are" -- Yoda. The Ricci tensor is the symmetric projection of the Riemann curvature tensor. The Weyl tensor is the anti-symmetric projection of the Riemann curvature tensor. Symmetry (Bosons) is dual to anti-symmetry (Fermions). Covariant is dual to contravariant -- dual basis.
@Suav5810 ай бұрын
Is an average football fan better educated (more sensitive to other cultures?), than average mathematician: kzbin.info/www/bejne/aaLUpKOLiN2fgJI