@@prawtismA low voice is quiet, not deep. Think "She spoke in a low voice, as not to be heard by the guards" "A deeper voice" is what you're looking for.. English is dumb
Newcombs paradox is a really interesting problem. Here's my solution to it: The way I understand it, the argument for choosing A+B is based on game theory (not the channel lol) and the idea that whatever is in B doesn't matter, because you might as well take an extra 1,000 dollars. The reason that's incorrect, is because you're assuming there's 4 options (0 dollars, 1k dollars, 1M dollars, and 1,001k dollars) when there's only really 2 (1k or 1M) because assuming that the predictor is perfect means you can't make a decision that "tricks" it. Technically, you can't change the past, and what the predictor put in each box. But practically, you sort of can. If you choose B, then you get 1M. Not 1M/0, because that implies that there's random chance in this completely deterministic problem. Likewise, if you choose A+B, there's only 1 possible result you can get, which is 1k. You can't get 1M from the B box, because that would mean the predictor was wrong, which he can't be. I think this kind of resembles the idea of a superposition which is pretty neat. Also the concept of free will being an illusion. I really like this problem. I'm not sure if I managed to explain myself the way it's explained in my head, but this is what I view as the objectively correct answer. If anyone thinks something different I'd be glad to argue in the replies. Edit: ok, I think I found a better way to put it: The flaw with the reasoning of: "well I took the 1M and I got it, but I know for certain that A had an extra 1k, so why didn't I just take that as well?" is the idea that you could have chosen differently and gotten the same result with B because it was predetermined what the box had. It's like the A box has a sensor that automatically removes everything in the B box, even though it's seemingly just a free 1k. It didn't have an actual sensor, but it might as well had had one, because the A box is pretty much a trap. B's contents are what I like to call "yet to be predetermined". You can't actively change the past, but the people in the past act differently based on the future, which is your present. Yeah wow this is really hard to put into words Did anyone understand a single thing I said
The content in box B becomes $0 if and only if you pick both boxes. Since the predictor is said to always be reliable, choosing just box B would always get you $1,000,000.
On the first one, an interesting observation: Take 4+5+6 = 7+8 If you take out the first number, 4, and then split it evenly to the other numbers, 5 and 6, you get whats on the other side, 7+8. Another (barebones) example: 16+17+18+19+20 = 21+22+23+24 16/4 = 4 (17+4)+(18+4)+(19+4)+(20+4) = 21+22+23+24 21+22+23+24 = 21+22+23+24
Dude it's literally the same thing as adding the number.. you are adding 4 + 4 + 4 +4 to each of the other numbers, which is the same as adding 16 to the LHS
The problem with the last question is that "reliable predictor" is vague and not clearly explained. The difference in answers just depends on how someone interprets what "reliable predictor" means to them. This is yet another example of a question being more of a language problem than a math problem
Btw if anyone is wondering what my view is, I think that you can basically ignore the "reliable predictor", because we are told that "people seem to divide almost evenly on the problem", so how can you possibly reliably predict something that is close to a 50/50. Now the problem becomes much easier and we can see that taking both boxes is optimal as it grants us an additional $1000 no matter what the "reliable predictor" does.
@javinleong3433 i don't think i understood the problem. Isn't taking box b just better? Because the predictor will have predicted that you will have taken it. And so there is a million dollars in there ??
@@Smallpriestyes I agree with you. If the predictor can literally see the future, then you choose box B. If the predictor can’t see the future then you pick both boxes. There’s nothing profound about it. Just intentionally vague language which disguises the fact that it’s merely a paradox that only arises due to the use of time travel/seeing into the future. And well… of course there’s going to be paradoxes with time travel. Someone tell me what I’m missing here
My answer: if the predictor is reliable, then the world must be deterministic, and then you don't have any free choice when "making your decision". Just pick what you pick and be content with what you have because there is no alternative in which you could have picked differently. (Or, opposite, if you have a free choice, then indeed a reliable predictor cannot exist; but I don't believe in free will. 😉)
I simply love this video, and I don't think word can describe how much. This is the first time I've seen a math video structured like this before, and it's fascinating. Thank you for this!
3:04 the explanation to this: Sum of first n odd numbers in n², so the fraction will become n²/((2n)²-n²) which is n²/3n² thats why it always results in 1/3
If I open box B and it is empty, I open box A and get at least $1000. Predictor was right that I would open two boxes. If box B contains $1 million, I don't open box A, so the predictor is correct again.
*And then after you see the million in box B, you also take box A because it’s free 1000$* 😀 Be honest. Imagine yourself in front of both boxes, do you really believe that you wouldn’t take box A right after ?
Undulating squares and cubes are easy to completely list. It is not that only 4 have been found and one cube, it is that only 4 EXIST (in base 10) and that is fairly easy to prove/check (modular arithmetic mod 100000 say)
The only proof I could find was David Moews one, and thats only for undulating squares (and not "fairly easy" by my standards :D) I couldn't find any proofs for cubes so are you sure that's right?
The way I see it, Newcomb’s paradox is just a malformed question regarding free will. If free will exists, then there is no such thing as an “accurate predictor”. Maybe a very-nearly-accurate one, but never a perfectly accurate one because someone could just _choose_ not to follow its prediction. In that case we have 4 options (ignoring the “take only A” answer, which is funny but rather obviously never the best) - Predictor guesses taking both and you take both => $1000. - Predictor guessed taking both and you take only B => $0. - Predictor guesses taking only B and you take both => $1001000. - Predictor guesses taking only B and you take only B => $1000000. Notice how out of these 4 cases, it’s always better to take both boxes regardless of what the predictor guessed. Hence why that’s some people’s “obvious” answer. By contrast, let’s take the predictor being 100% accurate as a definite truth, thereby requiring free will to be false. In that case we have only two options (again ignoring “take only A”): - You take only B => $1M - You take both => $1000 We don’t even need to mention the predictor because it’s always right and so only your action is relevant. In this case there’s no good answer to “what do you do?” because if free will is a myth, you don’t know what you do until it happens. The best you can say is “I hope I take only B” because that would mean $1M. So, the paradox is actually the incompatibility between determinism and free will. If you lean to the side of free will being real, you should take both. If you lean to the side of determinism, you should hope you take only B. What would I do in a real situation? Well, depends on context. I don’t believe in free will, but also know that we (humanity) definitely do not have a 100% accurate predictor. So if a human came to me with this situation, I’d take both boxes knowing the predictor isn’t flawless. If, however, I was somehow given this prompt in a context where it’s plausible that there is a 100% accurate predictor, I’d only take B. I don’t expect to ever be in the latter situation, so devoid of context I’d take both. TL;DR: Free will is incompatible with a perfect predictor. IRL we don’t have perfect predictors, so it’s best to just take both unless there’s a good reason to think the predictor involved is perfect (in which case you better hope you take only B!) Edit: fixed ‘a “accurate predictor”’ to ‘an “accurate predictor”’.
Thats very cool, but also easy to explain. This happens because 9 is exactly one less than 10. So if we are adding, the ones digit decreases exactly one and the tens digit increases exactly one. So the sum of the digits stays the same. If we are going to another base this property remains for the digit one less than the base number. In base 8, if you are adding to 7s, the result is 16 and another 7 gives you 25, so the sum of the digits is also always 7.
0:51 another fun fact about 11. When you square a string of 1s, you get a number which goes up, and then goes down, and it goes up to the number equal to the number of 1s you started with (if it goes above your base, you have to carry over the extra bits). So 1111^2=1234321
For undulating numbers, isn’t it exclusive to base 10. If you worked in binary you would have way more and hexadecimal way less and it wouldn’t be the same numbers as in base 10. As such what is the correlation between undulating numbers and their base
I think Newcomb's paradox depends upon one's personal circumstances, within a spectrum. For example, at one end, your well being wholly depends upon your receiving at least a thousand instantly, in which case take both boxes. Somewhat towards the other end an extra thousand means nothing to you, so you might just as well take box B and go for the million. At the far end even a million makes no difference, so walk away - or better, perhaps, flip a coin to choose and donate any gain to charity.
Well, you aren't wrong, but also, I don't think that's what the paradox is about. What you're doing is measuring the resulted happiness out of 2 possible situations: 1. Getting 1,000 dollars guaranteed 2. Maybe getting more, maybe getting nothing. To begin with, in this type of theoretical game, usually when each option results in a certain payout, the money isn't meant to be taken literally, and is just a means of quantifying what options are preferred and to what extent. Clearly, a million dollars is better than a thousand, both of which are better than nothing. The given amounts were probably picked semi-arbitrarily. Secondly, you're missing the whole point of the puzzle, which is the predictor. You're just viewing it as a certain probability. Sure, in the real world, if I knew there was some logic-puzzle-esque giveaway, my answer would likely be highly influenced by the factors you mentioned, regardless of what I think the solution to this riddle might be. But in the highly idealized theoretical world, the question is only meant to test your thinking, and no money is involved.
@@HelloIAmAnExist Great response - I do know that I'm missing the mark, somewhat. However, my difficulty is that I cannot believe in the perfect predictor; to me, therefore, the situation cannot exist and any perceived paradox is illusory. Even the puzzle itself doesn't really seem to believe what it claims, inviting us, as it does, to try to catch the perfect predictor out. Cheers :)
@@s1lv3rfir3 But once you have to pick your box B, the boxes are filled. So why not change your mind only then and also take the extra 1000 from box A? Or phrased differently, once you have taken your million from box B using your strategy, could you then also just grab the 1000 from box A, if allowed to? (That is the counterargument at least.)
Great video. Apart from 8^3 of the very few numbers who have the property that the sum of its cube is equal to it are all next to each other. Why is eight an exception?
It is unclear whether one could make one's decision to open box A just after opening box B. It is also unclear whether the player knows what amounts may be deposited in the boxes and whether you can get the total amount or just the contents of the last box opened. It is also unclear in which order the boxes may be opened. Furthermore, it is unclear whether the player knows the predictor's way of thinking and his strategy for the prediction.
Veritasium used to be my favourite channel for science and math,, until i found this channel.. This is the best channel for math and science! But.... Why is this channel have only 60k subs?? People, subscribe
For Newcomb's Paradox, if the predictor predicts that you will choose Box B, it will contain $1,000,000 dollars, which, if they are reliable, means you get $1,000,000 dollars, almost guaranteed. Whereas if you take both you will be almost guaranteed $1,000. You win if you choose Box B and the predictor is correct; where the predictor makes 'highly accurate decisions.' Your odds are pleasing. However, the keyword is 'can.' The predictor >CAN< make highly accurate decisions, but may not choose to. We do not know the motive of the predictor. In this case I would assume the predictor does not want you to make much profit (kind of like gambling) and so will 'predict' that you will choose both boxes every time. In this case whenever you play the only profit you will get is if you choose both, because Box A always contains $1,000. In these circumstances I would choose Box A. But who really knows the identity and motives of the predictor? And when has lots of money really brought happiness? Both Boxes???
Fun fact: if you turn the area of n gon w side length of one into function (call it f(n) ) and took second derivative of it it (f’’(n) and as n approaches infinity the result approaches 1 rad to 1 revolution (= 0.159154943092 Revolution) My case: f(n)=n/(4*tan(pi/n))
It's a 50/50 chance to get 1000000 or nothing if you choose just box B It's a 50/50 chance to get 1001000 or 1000 if you choose both boxes So statistically speaking choosing both boxes gives you a higher yield
Newcomb's paradox: Box B, since it contains more money if the predictor predicts I will choose Box B, but he can't predict Box A because that doesn't make sense because of the statement above so Box B.
4:16 I feel like this cant be right. even if the odds of there being no 9s gets increasingly closer to 100%, there will always be numbers that contain no 9s, even at numbers with tree(3) digits, or any other large number for that matter. infinity * any percentage, even 0.0000000000000000000000000000000000001% will still be infinity, therefore, this sequence even with all 9s removed, will eventually reach infinity, regardless of the speed at which it gets there.
Why’s the last puzzle so controversial? If I understood it correctly, isn’t it just as simple as this: do you want to take both boxes and guarantee to receive $1000, or do you risk it, and pick box B, which will result in a 50/50 where you either receive $0 or $1000000?
@@landsgevaer how come? the so called predictor who choses the contains of the boxes, does so before we pick which box to chose? so there's definitely chances involved, no?
@@DylanLCutshall No, the perfect predictor supposedly knows the future and knows what you will do. So whatever you chose to do, the prefictor will have acted accordingly. The paradox claims there is no uncertainty or chance. (I didn't make it up; that is the paradox.)
@@landsgevaer if what you've written is true, then I don't think that was presented very well. in the video he says that the predictor has already chosen his prediction, and there's no info about him being able to see the future. but let's say you're right, then what's the point with the puzzle?
I would take box B only. Even if I opened it and found it contained no money at least I wasn't trying to be greedy. Besides, that would be a good dose of karma for the predictor, having misjudged me as out for all I can get, when I'm really not.
The video should be titled "6 Minutes of Fascinating Base 10 Coincidences." This video isn't generally focused on "math"; it's specifically centered around base 10.
With the Newcomb's Paradox, it seems like a 50-50 shot for whether Box B will contain anything, which can't be altered by us in any way, no matter what we chose. Why not always chose both boxes then? Does our selection have to line up with the predictor's prediction for us to be eligible to get any money at all?
No, because the predictor is always right. So if you take both, he predicted it already, so you get 1000. If you take B only, he predicted it and you get 1000000
My argument for choosing both boxes is that a perfect predictor doesn't exist. Just like you can't change the past, you also can't predict the future. Let me put it this way. Lets assume that a perfect predictor does in fact exist and that he can see a snapshot of the future based on current events. Lets also assume that the player is also a perfect predictor that will always make the best possible move. If the player decides to go for only box B the predictor will put 1M dollars in it, which means that now the player should take both boxes, but then the predictor would remove the 1M, so the player changes his choice to box B again. This will go on in an infinite loop. This all arises from the fact that we assumed that a perfect predictor can exist, but he can't. So the most logical choice is to choose both boxes.
The harmonic series one can't be right. If you remove numbers from a series that contains 9s, you still get infinite parts, so the sum of them also have to diverge to infinity.
Big Number Abbreviations for you to use: K = Thousand M = Million B = Billion T = Trillion Q = Quadrillion q = Quintillion S = Sextillion s = Septillion O = Octillion N = Nonillion D = Decillion
I'm curious about the first problem you've shown. How do you formalize this series of sums, and how do you prove your claim? You don't seem to formalize anything or show interesting relationships, which could really help with the entertainment value of this video. Update: I managed to find a proof all by myself. It's not that hard if you know about Euler's summation formula, and it probably wouldn't take over five minutes of runtime.
So you are telling me 343 is the rarest number out there... My new lucky number Edit: I just now realized 7³ has the two luckiest one-digit numbers, 7 and 3
The number of special (positive integer) numbers cannot be finite, because if it were then there would be a smallest non-special number and that in itself would be quite special...
what is a reliable predictor? he's always right? If so i chose to firmly believe i must take box B, that way he anticipates it and put 1000000 dollars in it.
I will choose box-B, of course, the predictor has already accurately predicted me to earn 1M😂 UNLESS instead of a mere 1K, box-A contains schroedinger's cat; in that case, I would smashed it with Thor's hammer (provided I was worthy to wield it, otherwise I lend Hulk's arm instead). Damn! I've just written 3 insteads haven't I?
