This really shows that if somebody is passionate about educating the world, it really doesnt matter whether they teach physics or math or anything else, they will do a fantastic job. Thanks sir!
Lectures by Walter Lewin. They will make you ♥ Physics. sir can you tell me how can I apply for MIT (being an ex- MIT teacher , you can guide me better)to do PhD. Especially in Cosmology, PS: am an undergrad physics student
Many thanks, Professor. This was a difficult problem for me and took me a good few hours to think through how to calculate the probability of 2 wins (without knowing the factorial formula). But I got there, and enjoyed helping other people towards the solution. I learned a lot, both from solving it myself and from trying to understand where and why other people had gone wrong. Can't wait for the next one.
The first 2 days about 5% had it right. The next 2 days 15%, the last 2 days 20% had it right. People got slowly educated when there various solutions were posted (which meant they were wrong). Few who used computer simulations got it right with 3 digit accuracy (they did not run the simulations long enough).
Probabilities get complicated fast. I didn't have the math for this anymore. In my youth even calculating 1000! would have been a challenge, since calculators didn't handle numbers that big. My estimate was that it would be less than 3rd of the first problem, obviously, but more than 6th. I was a bit optimistic there. Which is really what lottery organizers want you to be.
I enjoy your lectures tremendously. One area of potential confusion is the use of the term “arrangements.” In combinations, order (or sequence or number of arrangements) does not matter, whereas in permutations, order does matter. In the case of this problem, with the assumption that the jackpots are identical, the lottery player wouldn’t care which particular tickets were winning tickets, merely that he or she held at least three winning tickets. Hope this helps!
>>>the lottery player wouldn’t care which particular tickets were winning tickets>>> CORRECT, that's why the probability of winning ONLY 1 ticket is 1000 times higher than winning only a prize on his first ticket.
I was thinking to revise the probablity chapter but after watching ur two videos I can say that, my basic revision is almost complete. Thank you sir. 😊
Hello Professor Lewin, Again I remain amazed how a multiple of the quantity (1/e) shows up in these problems: Probability of zero wins = 1/e; Probability of one win = 1/e accurate to three places; and (now here's the surprise) how the probability of two wins ==> .184 = one-half the quantity (1/e). Never in my time at IIT in Chicago, certainly not the MIT in Cambridge, had I grasped the implications of "e" -- beyond its being a mathematical oddity that happens to "work" in a number of physics equations, including the very famous one of raising e to the imaginary power (i)(theta) which unifies algebra and geometry via the differential equation of an ellipse! Thanks a million!! Joe
Professor, here evidently you used the binomial distribution to obtain the probability. I on the other hand, calculated it using the poisson distribution since this is a case where success is low and trial is high. Luckily enough our answers do match up to 3digits precision. This was fun. I am eagerly awaiting your next problem. :)
Well done. May we also present problems to the public? Well this one a friend gave me while sitting at an afterwork bar. Although it is very easy by the math, it is unintuitive at the first view. Let's have a bottle with a bottletop. Booth together have a height of 11cm. The bottle ist 10cm higher then the top. What is the height of the bottletop? Cheers
Yeah! I was wondering how you would introduce factorials... this introduction was very easy to understand for me! Wonderful job done! Thanks :D I feel lucky to have a great friend :D!
Sir I did it correct(I tried that my probability of solving this problem is +1)Sir thank you for this video because I and many of us will learn something new from this video...keep asking questions(loves maths)...STAY HAPPY : )
I got it!!! I'll admit I did not get it until I starting watching this video. As soon as I saw the equation for M, I paused the video, finished my calculations and got 8%!!! So cool to see it was right, even though I needed the help of the equation to finish it.
Respected Professor, I am sure that there are many ways to solve the problem. I particularly like the phrase "Marry different ways" where you discussed the "Combination of Things". But I would be glad to share the way I reached the solution. First, I want to state how I personally convert a "Sentence Problem" to "An Equation Problem" for probability: It is well known that while describing any EVENT(the objective) 1) When we say "AND", we perform 'MULTIPLICATION' of two probabilities 2) When we say 'OR', we perform 'ADDITION' of two probabilities Now, we come to the !st problem. The probability of getting at leat 1 Lottery Prize. Let, The probability of winning 1 ticket of a Lottery be A = 1/1000 = 0.001, The probability of losing a lottery be B = 999/1000 = 0.999. Now our problem. It may so happen that the 1st Ticket I purchased wins. So The probability is A OR (Addition) I lost the first Lottery AND (Multiplication) won the 2nd one. so the probability is A+B*A OR ( Addition) I lost the 1st And 2nd Lotteries And won the 3rd one. So the probability is A+B*A+B*B*A OR (Addition) and so on Therefore the Total Probability becomes A+B*A+(B^2)A+(B^3)*A+ ... = A*(1+B+B^2+B^3+...) which is a GP (Geometric Progression) series till 999 lotteries are lost and the 1000th is WON Thus the Total Sum becomes By using Formula A*(1-B^999)/(1-B) = 0.001*(1-0.999^999)/(1-0.999) = 0.6319 ~ 63.2% Now we turn to the Harder Problem It may so happen that the 1st 3 Tickets I purchased wins. So The probability is A*A*A (1st AND 2nd AND 3rd) Now let us find no was for winning the 3 tickets out of 4 lotteries B*A*A*A(1st lost and rest won), A*B*A*A, A*A*B*A = 3 ways (The last combination A*A*A*B is not taken because three lotteries are already WON) So here we introduce the Factorial and combination concept: thus we can win 3 out of 4 tickets in 3C1 ways (Lost ticket can be 1st, 2nd or 3rd) NOTE: 3C1 = 3! / (1!*(3-1)!) OR (Addition) I lost 1 our of 4 tickets AND won the rest three AND in 3 different ways. So The probability is A^3+(3C1)*B*A^3 OR(Addition) I lost 2 our of 5 tickets AND won the rest three AND in 4C2 different ways. So The probability is A^3+(3C1)*B*A^3+(4C2)*B^2*A^3 OR (Addition) and so on Therefore the Total Probability becomes A^3+(3C1)*B*A^3+(4C2)*B^2*A^3+ ... = A^3 * (1+(3C1)*B+(4C2)*B^2+ ... ) which is a Complicated binomial series till 997 lotteries are lost and the 3 tickets are WON in 1000C997 ways ((1000C997)*B^997) I used computer program to calculate the sum which came out Just 8.02% Thus it provides the Generalisation and ONE LINE solution of the problem. If we want at least 'n' number of lotteries to win out of 1000 lotteries then the Total probability becomes A^n * (1+(nC1)*B+((n+1)C2)*B^2+ ... + (1000C(1000-n) * B^(1000-n)) Hope I have explained it properly. But I know that the solution you presented is way smarter. Thank you.
Yes, of course, there is always more than one way. To go from Boston to New York City, you can fly straight South (distance about 300 km). You can also fly from Boston North, over the North Pole, then over Russia, India, South America, Cuba, Florida, New York City (about 40,000 km). It's nice to have a choice.
Actually I am one of those suffering, I don't know if my code is flushed or something but I did up to 1 billion iterations all night long computing several times and still got either 8.01, 8.04 or 8.06, never 8.02 :( Here's the code if someone wants to take a look: codepad.remoteinterview.io/NUZDTWBRWF
Isidre, your issue is the rng you are using, it's not random enough, and is biased to low numbers. The I used libsodium's rng for my tests and that produced the right result after ~ 50million iterations. (I did the math way to get the correct answer first, but I thought the simulation may be a fun project) maybe this will help :) #include int draw(){ uint32_t myInt; myInt = randombytes_uniform(1000); if (7 == myInt) return 1; else return 0; }
Thank you very much Simon, I already thought about that but I've been programming only for a two months or so right now (only c++) and I know there are ways to generate very random numbers but didn't know how to implement those easily in c++, I undersand the function you sent me, but I don't know how to install the library and where to download it (I am using DEV c++ as IDE), I will probably look it up after the exams.
Enjoy :) it was a fun problem. also look into multi threading the problem, most modern computers have many processor cores. instant 4x speedup on a quad core.
I got it. 1 - { (0.999)^1000 + (0.999)^999 + 499500 * [ (0.999)^998 ] * (1/1000)^2 } = 0.0802 -- where the coefficient 499500 is derived from the nCr function when n=1000 and r=2 ... i.e. the binomial expansion coefficient (aka "Pascal's Triangle")
I had the correct intuition but verytime I did the some calculation error just to give you the correct answer quickly after my embarrassment in fluids........ :(
hello sir, i had a hypothesis that explains the dark energy. Even my physics teacher approved it but we got no one to contact with. So please help us Professor. You are the only physicist I can contact directly. The theory starts this way: every object with mass has gravity.The gravity always attracts the other objects. And therefore their gravity always does work. But work is energy and since the total energy in the universe is always constant the ability of gravity to do work must slowly end throughout the time. So the gravity of objects must decrease over time. And since less gravity means less mass , every object will lose gravity and so mass. If we were to take the momentum of any object in the universe assuming no forces are applied on that object the momentum is constant. If momentum remains constant, while mass reduces the velocity must increase over time. So the velocities of objects will increase continuously. This is my theory about dark energy. And I can also explain the dark matter as well with this theory.But I am not sure if I am right about dark energy. Thank you sir.
you should contact a world authority on cosmology. You can find their names on the web. There are at least 2 at MIT several in Cambridge UK, Univ of Chicago, Princeton Univ. . . . .
You're not missing anything. In the simple problem, we deducted the probability of getting 0 wins from 1, to find the probability of 1 or more wins. To find the probability of 3 or more wins we deducted the probability of fewer than 3 wins (ie 0 wins, 1 win or 2 wins) from 1. So the first part of the calculation, finding P(0), we had already done in the simple problem.
I was wondering... if we're buying 1000 tickets, anyway, shouldn't we buy all of them from the same lottery, which gives us 100% chance of winning the prize? On the other hand, buying the tickets from different lotteries gives us the possibility of winning multiple prizes, so is that better? Interestingly enough, on average, we'll win exactly one prize. In other words, that's the *expected value* of the number of prizes we get from buying one ticket of each lottery. Nice challange, Professor Lewin!
Oh, I'm sorry, Professor! I hope you're not upset, perhaps my comment wasn't very clear. I didn't mean to present a solution to the problem. I was just thinking about what option would give us better chances for higher "profit": buying 1000 tickets from different lotteries or buying 1000 tickets from the same one. What I found out is that, actually, both options give the same profit on average: one prize. Of course, this reasoning goes beyond the scope of the problem, I'm just extending the discussion.
Hello MR. LEWIN I hope good health and more wonderfully life . i have suggestion for you if you can use writing board it will be better for us beyond you are teacher and we are your student . because with board very good to take the idea more than . thanks
I have considered that but it's NOT a good idea. I can install my camera such that it covers an area of about 1.5 x 1.5 m so that my writing shows up large enough so that the viewers cab read it. As I write on it I would be in between the camera and the board. Thus I would write a few lines and then move to the side and wait a minute and then write more etc etc. NOT good. Keep in Mind that during my lectures at MIT we had always 3 cameras + professionals who could move them if needed.Thus I rarely ever block the view to blackboards. We also had nine HUGE boards over a length of about 10 m. I would have here at most 2.5 m if I buy 3 black boards. I may for fun buy a chalkboard to remind the viewers that I can still draw dotted lines, but that would be its only purpose.
For 3 digit precision there is only 8% and it is in accordance with Your assumption at the begining - 3 digit precision, so I assume this is the right answer too :)
Sobota there may be three digits in 0.08 but that doesn't mean you have three digits of significance. Also it's not exactly that Walter increased the precision at the end. You generally want to not round off numbers in the middle of your calculations, only for your final number are you allowed to round off.
Very difficult to understand professor. First time i didn't get something from you. I am not getting why we are doing 1-(p0+p1+p3).How someone come to know he should use this formula :(
The question was, what's the probability of getting at least 3 wins? That is, 3 wins or more. The only way that you can not get 3 or more wins is if you get 0 wins, just 1 win, or just 2 wins. So, if you deduct these probabilities from 1, you will be left with the probability of getting three wins or more - ie at least three wins. We solved the simple problem with the same approach - we deducted the probability of getting 0 wins from 1 to find the probability of getting 1 or more wins.
You didn't need the factorial formula. I worked it out without it. If my first win is on the first lottery, I have 999 lotteries left for my second win. If my first win is on the second lottery, I have 998 lotteries left for my second win etc. So the number of ways of winning two lotteries is 999 + 998 + 997... +1. I used a spreadsheet to add them, but then realised there was an easy way to do this, since 999+1 = 1000, 998+2 =1000 etc.
+Lectures by Walter Lewin. They will make you ♥ Physics. i meant the probability of getting zero prize one prize and two prize (p0 p1 and p2). i am really very sorry professor from the bottom of my heart but i m got getting how the math is going around. how it is working? the thing that u explained about arrangements of the marbles, i completely grasp that but after that "The probability of getting at least 3 prizes" I didn't get that i mean i am not getting the logic u hav used.......... I don't u r getting me or not !! I am sorry ....
Jatin Your son calls you at work. He tells you that the tickets of all lotteries have been drawn. You say: "I assume we did not win any Prize". You son says, "no no dad", we did better than that. You say: "Really, did we win 1 prize?" "No dad we did even better than that". You answer: "you must be kidding, did we really win 2 prizes?" Dad NOOOOO we did even better than that. You answer: "are you serious, did we really win 3 prizes?". Your son answers: "no dad we even won 4 prizes". In summary. *If you do not win 0 prize and not 1 and not 2 prizes, then you must have won at least 3 prizes!* Do you get it now?
Guess i was a little overconfident .... apparently 25.28 % was wrong. Still don't get why my reasoning was so horribly wrong. I will simulate in a c program one of these days to confirm the result. That is really fast and should have accurate results. Thanks for the problem. I will learn something one way or another now.
Several students used computer simulations. Some told me that 50 million - 100 million simulations were needed to get the required 3 digit precision. Many of them ran enough simulations and sent me 8.02%. Some who did not go far enough came up with 8.13, 8, 8.05, 8.01, 8.03%. Close but not the accuracy that I required.
I could not wait and programmed for 20 minutes ... 1.0E7 iterations gives me 8.02 % ( 8.019 ) . Next time I will simulate before posting. I program a lot faster than figuring it out by math. Anyway, I was hoping to prove you wrong with this :-) ..... not the case. If i buy 2 tickets from each lottery probability to win at least 3 prizes is 32.33 % :-). Once the simulation is up and running i can fiddle with things ..
