To try everything Brilliant has to offer-free-for a full 30 days, visit brilliant.org/AnotherRoof/ You’ll also get 20% off an annual premium subscription. ⬣ CORRECTIONS, CLARIFICATIONS, AND COMMON COMMENTS ⬣ 1. (a) There are two issues with my analysis of the US Powerball. First was an honest mistake which was the percentage loss for the powerball: It should be 0.84/2 which is a 42% loss. I mistakenly divided the EV by the cost to arrive at ~59%. My bad! This is still a worse percentage loss than the Spanish Christmas Lottery, but not as bad as claimed in the video. Sorry about that. 1. (b) Secondly was an intentional simplification which is that the jackpot often gets split between multiple players. As the jackpot increases, more people buy tickets which increases the likelihood of the jackpot getting divided, thereby lowering the EV. This analysis is pretty complicated so what I did was take all the jackpots awarded since the latest rule change and take the median, which is $246M. The net effect of divided jackpots, however, is that the EV remains at around the -0.85 which is similar to the value quoted in the video. See this article for more info: www.forbes.com/sites/startswithabang/2016/01/13/the-science-of-powerball/ 2. A simpler way to calculate the EV of the Spanish Christmas lottery would be to take the total prize pool and divide it by the tickets sold. For the case studied in the video, we have 2.5B / (100,000*180) = 139 which is what I obtained in the video (barring rounding errors). Boy, is this much easier but hey, if we'd done it that way we would have got to learn about the linearity of expectation!
26:50 I believe that the loss should be around 41-42% rather than 59%. If you take the profit of -0.84 and divide by the amount put in ($2), you get 0.84/2 = 0.42. I think you used the EV to calculate the profit rather than the complement from $2. This seems more in line with 24:00 where the money spent is 20, the EV is 14, and the profit is -6. The loss is calculated as positive of EV / money spent.
Gordo means fat as in overweight. Big fat prize is the idea. About 25 years ago I asked a Spanish mate to feel the back of my hand. Bemused and suspicious he did it and asked why. I told him the guy in the kiosk had said "que te toque el gordo" / lit. "may the fatty touch you" - a proper translation is I hope you win, good luck basically. My best joke ever.
It is ungoogleable. No matter how I phrase my query, google refuses to believe I want anything except dozens and dozens of articles in a row about the largest lottery _jackpots_ ever awarded.
@AntonioLasoGonzalez Yes, but the page doesn't cite any source. Anyway, the idea is to find it without putting the answer in first. Like, if you don't already know which lottery has the biggest payout, how do you find it?
also the neat thing is that taxes are only applied to El Gordo and the second prize, so most of winnings are free of taxes. what a surprise to see this lottery covered in your channel!
Used to be no taxes for any public lottery. But in 2013 they added a tax of 20% to any prize higher to 40.000 euros. You know why? Because Spain is Spain.... It was used as the best money laudry scheme in the world. Winners went to banks/lottery shops for counsel, they contact politicians/big business (But mostly politicians with lots of questionable "support" money), Politicians pays 10% more for the ticket, shares profit with the informants, and the politician just laundred 40k to 400k euros for just 4/40k. Spain is Spain because everyone knew, including the government and taxes agency, that some famous politicians (Fabra,Roca,..) for example had won more than 10 or 20 times some lottery. No one "suspected" anything for years and years. Juan Antonio Roca was an specialist, most lottery shop owners had his phone. He won the lottery 80 times in about 20 years.
@@framegrace1 So if you win 100k now (just to make the math easier) and the tax is 20%, you get an 80k winning that you can sell to a money laundered for 88k ? The tax makes no difference whatsoever on that part of the business. And any kind of gamlbing winnings can be used this way. You might even show up at a casion, buy $1million in chips for cash, gamble 20% of it and go home with $800k in verified winnings.
Because each ticket necessarily has the same EV (symmetric with respect to all numbers), EV for any ticket is simply total prize pool divided by number of possible tickets. Prize pool in thumbnail is correct (I calculated 2,521,476,000 Euros), total number of tickets is 100,000x180, so EV before cost is 140.082 Euros. It doesnt matter how the prizes are distributed or at which probabilities. I kept waiting for this punchline but never got it. This also helps understand intuitively why these games have the same EV with and without number replacement: if total payout is the same and every number has an equal chance at winning, EV for each wont change.
13:00 Ah, the “expected value is linear” section of the video. I once had a teacher who yelled to the whole class, and also opened the door to an adjacent classroom to tell the neighboring class studying something else, that “THE EXPECTATION OF THE SUM IS THE SUM OF THE EXPECTATIONS!”. And even knowing that, and having had homework in a different class to prove that expectation is linear, it can still mess with my head.
Linearity of Expectation is so overpowered. Here is one example I saw the other day: Let S denote the surface of a sphere. Assume S has surface area 1. Let F be a measurable subset of S with area strictly greater than 1/3, then F contains two mutually orthogonal points. To see this, consider three mutually orthogonal vectors v1, v2, v3 (the example you are most familiar with is (1,0,0), (0,1,0), (0,0,1)), and consider rotating the vectors so they are still orthogonal and hit every point of the sphere. Let X denote the random variable "the number of these vectors that are in F". We will show that E(X) > 1, but this is only possible if X >= 2 sometimes occurs, otherwise we always have X 1/3, so E(X) > 1/3 + 1/3 + 1/3 = 1.
@@kaishang6406 It's not that two of the three vectors are always in F, but that there is some orientation where two of the three vectors are both in F at the same time - which means that those two points are a pair of mutually orthogonal points in F (as required).
@@rmsgrey thanks, i misunderstood what the expect value is averaging over. i thought it was averaging over all possible subset F and couldn't see the relation. If it is averaging over all orientation, the relation become clear.
that's a really pretty proof! The linearity "disentangling" the dependence between the orientations of the three vectors relative to each other is really neat
It does, actually! Although it has never happened, 0 would have 1 and 99 999 as neighbours (aproximaciones). Conversely, 99 999's neighbours would be 99 998 and 0. In a sense, this works as the ring Z_100 000.
Great video! I am however a bit confused how the expected loss for powerball at 26:52 can be 59% if the expected value is more than half of the ticket price. For the spanish christmas lottery the expected loss of 30% is calculated as (EV - ticketPrice) / ticketPrice which correctly evaluates to -0.3 , but plugging in the values for powerball gives (1.16 - 2) / 2 = -0.42 or expected loss of 42%. EDIT: Just realized that it is a known error that was just corrected in the pinned comment, so I needed to reload the page to see it
Why are we approaching this by focusing on it from the perspective of individual participants? Why don't you just add up all of the prizes and divide by the number of participants? We don't have to worry about the amount of money being paid out varying as we do in, for instance, the American lottery.
It's worth mentioning that each ticket is divided into ten tenths but also people can and do make their own further subsplits called 'participaciones'. Individuals may share participaciones of small value with family and friends (this is not fashionable anymore), and all kinds of associations sell participaciones in which you may pay e.g. 5€ of which you play 4€ with 1€ going to the seller as a donation. This means one prize very often reaches many hundreds of people, many of them in modest or even unremarkable amounts.
5:40 I actually LOLed. And there is a reason why I never played, nor have any intent to play the lottery. I know how the maths works, even though I am a lousy mathematician. You buy the feeling of possible great reward. Since I do not need to buy that feeling (or expectation), I respectfully refuse to take part in any lottery (or any sort of gambling). The winner is always "the house". I know about certain techniques but 1. I am too lazy to focus that long to get an advantage and 2. I know that I would be immediately suspicious as my "luck" would be noticed. And 3rd: I do not need it.
Now, I wasn't surprised, but I have put the phrase "due to the linearity of expected values" in a paper. Using this we could calculate probability distributions for X by working out E(X), E(X²)... E(X^n), which were quite simple (well, only due to the linearity of expected values) and then just by definition E(X^n)=sum(i=1..m) i^xP(X=i), so you get a system of linear equations you can solve for all P(X=i)s. It's a neat method, which turned something not pactially calculable into a form which allowed for explicit calculation.
Feliz Navidad for you all!!! In Spain there're drawings of this lottery all year long (on Thursdays and Saturdays), but it takes place on a more simplified fashion. There are five cages to draw every single digit of a 5-digit number, and there is a draw for every single prize: first a few 2-digit ends, then a few 3-digit ends, 4-digit ends, then two or three draws of fully fledged 5-digit numbers (the latter is drawn, the bigger the prize), and finally 2 1-digit endings for refunds (the 1st prize ending also qualifies for a refund), obviously balls drawn are returned to cages between drawings. Next drawing to take place on January 6th (known as 'El Niño') uses that system (known as 'Modern System' opposed to the 'Traditional System' used on Christmas drawings). Most people that got refunds from the Christmas drawing do bet again on 'El Niño' as it also costs €20 for a ticket tenth, year-round drawings have a cost ranging from €3 to €12 for a ticket tenth, but there are also much lesser tickets to be sold.
I'm an algebraic thinker, so around 17:00, I paused and plugged in "n" for the number of balls... and yeah, I was shocked that it ended up simplifying to 1/n odds of any combination of first and second draw in *both* the replacement and non-replacement cases. Probability is so unintuitive.
I'm not sure it's that unintuitive. Instead of probabilities, imagine every combination is realized in practice, so we're just talking about fractions of some actual group. Then the "expected value" is just an average over the population. So if we give, say, 1/4 of the group $1 prizes and 1/3 of the group $2 prizes, then clearly the average prize earnings are (1/4)($1) + (1/3)($2) = $ 11/12. And it doesn't matter how much those groups overlap, because we are just interested in the total amount of money doled out. So it is with expectation. Who cares whether the winners are packed together or spread out? It's about average winnings. So just add the value of each prize.
Hello from Spain! Today's gordo has been 72480, which is... too round @~@, even more so than last year's! Both of them have been multiples of 24, with this last one actually being a multiple of 480, and bothe were (very) abundant numbers, with their propper factors summing over twice the number itself for this year's number, and over one and a helf times for last year's. Anyway, good luck next time! (I doubt any of the big winners are reading this xd)
I saw someone make a similar question herr, but mine isn't exactly the same. They asked about the derivation of the independence of games X and Y, if game Y depends on the result of game X. I think their question was about the probabilities of Y1, Y2... etc, but what's been confusing me are the prizes b1, b2... The available prizes are different for game Y depending on the results of game X. Wouldn't that make them dependent? As in, aren't the prizes b1, b2 etc dependent on the result of game X?
hey another roof. i just discovered that the first 1,000 digits of the 7th root of 17 after the significant ones is a perfect de bruijn sequence. pi, phi, and the square root of 2 dont even come close. i think ill call this new property "supernormality". needs more studying. thanks claude 3.5 sonnet for the help! ✌️✌️
Looking at it from an economoc perspective though, the fact such a large proportion of the population plays it means the expected loss can be lower, as the scaling costs of administrating lottery are relatively low, so the profit will be the same as smaller lottery with a lower expected winnings.
My wife asked me literally yesterday to explain to her why there isn't a winning strategy for this (we bought a décimo for the first time this year). I managed, but I'll bookmark this for when she asks again next year lol. Good luck and feliz navidad from Madrid!
After all these years on the internet, I'm finally looking at the guy all the spam emails have been trying to reach: someone who actually bought a lottery ticket in a foreign country.
Interesting video! I ran a similar calculation last year, but accounted the prize of a décimo in the value of each prize, i.e. a décimo for El Gordo would reward 399.980€ net after deducing the 20€ initial investment. My numbers are very similar to yours because it doesn't make much of a difference, but I was wondering: why is the initial investment not considered when discussing EV in gambling? In your roulette example, the actual prize for winning red/black would be 1, not 2. Does it make sense?
@@josenobi3022 If game Y depends on game X, there is still a fixed probability of the event "X1 and Y1". Independent just means that P(X1 and Y1) is equal to P(X1)*P(Y1). Hope that helps!
@@josenobi3022 Doing Y by itself will have its own associated probabilities. So like the earlier example, game Y was the "2nd prize" with its own prize numbers. If we don't replace the ball then Y is dependent on X. The probabilities for Y's events don't change, but it will adjust the probability for the "X and Y" probabilities -- that's where dependence comes into play.
@@josenobi3022 If you completely ignore game X and just focus on game Y, then the outcomes of Y have their own probabilities of occurring. They are just their own events. Dependence and independence only come into play once we examine two games.
I think there is a typo at 9:51 , assuming there are 2 4th prizes, then there would be 198 tickets with the "first three digits of 4th" as winners *. About this, there are just a couple of things that seems strange to me: because of the linearity, I'm assuming it's possible to win multiple prize with the same ticket, as an example it's possible to win "first three digits of 4th" two times if both 4th have the same leading digits (moreover, in this case, it would mean that both 4th prize ticked would get the prize, for having the same leading digits of the other). It just seems strange to me that it's possible to win multiple prize with the same ticket, do you confirm this is the case? (to be exact, I'm asking if the only exception to multiple awards to a single number is "if it's one of the winner's numbers, it doesn't matter it shares digits with itself")
wait, actually I think I disagree with how the statement written at 13:20 interacts with the table at 14:19 , according to the table there are 198 prizes still (both 1st prize and 2nd-similarity-compensation together is an option (ex: 3,2), like 2nd prize and 1st-prize-compensation (ex: (2,3)), and lastly winning both 1st and 2nd compensation together (ex: 2,4))
As I understand it from the video, when one of the 4th prizes is drawn, that ticket number gets the full prize, and the 99 other tickets with the same first three numbers get the lesser prize. The same thing happens to the other 4th prize, regardless of whether its numbers overlap with any other prizes. So for any given prize, any given ticket will either: win that exact prize; win one or more of its near miss prizes; or not win anything from that prize. Your ticket then wins all the prizes it qualifies for.
Thanks @rmsgrey , makes sense, but in this case what's the point of 13:20; how is the likelyhood ever reduced if a single can win multiple prizes? I mean, how getting two consecutive 1st and second prizes shold decrease the likelyhood of being 1 away, or how having the same three digits decrase the number of winning tickets to 98? I'm just confused by this part then. The fact that those 98 wins are double and that even 1st and 2nd extraction win "first-three-digits" prizes didn't even cross my mind. Moreover, you could just appleal to "invariance" of the amount of money given at any round (/by a certain prize extraction, as ex 2nd prize extraction or 4th prize number two extraction) to calculate the total number
Man, all the people talking about gambling, taxes, and so on don't get the point. The Spanish lottery is about sharing, giving decimos as a present, about envy (everyone at work has the company's décimo just in case everyone else becomes rich), about thinking to go to the bar where you bought it if you win. This year I had 110€ in décimos (we share some decimos and pay half each) and I rarely win something, but everyone plays every year for the illusion.
"Vídeo patrocinado por Loterías y Apuestas del Estado". Sonaban mejor los niños mutantes de San Ildefonso cuando cantaban en pesetas. ¿Os acordáis del calvo? Good luck getting all the references with that one.
Your Powerball calculations assumed that each jackpot winner would receive $246 million. But when the jackpot gets that high, many people play, so the probability of sharing the prize becomes substantial. In the Powerball, if multiple people simultaneously win the jackpot, the jackpot is evenly split between them. So really, the expected value of your ticket is less than you stated (particularly if you choose commonly-played numbers). On the other hand, sometimes the jackpot is more than twice that size, so in those rare cases, a ticket could really have a positive expected value . . . before taxes. Because once you realize that state lottery winnings in the US are subject to federal income tax, it becomes practically impossible for such a scenario to occur. In most states, even state taxes can apply, in spite of the fact that it is the state itself paying you your winnings. And it's actually worse than that, because in the US, lottery jackpots are not paid as lump sums but as escalating monthly payments. The present value of the annuity is not really as high as they claim, so you actually get much less than stated. IMO that should be illegal, since it is literally, factually untrue, but that's how it is. (Of course, once you consider a more reasonable logarithmic utility function, it becomes even more stupid to play the lottery, since each dollar you risk losing today is a lot more valuable to you than each dollar you win at the end of your hypothetical large jackpot. However, this utility calculation doesn't apply the same way to lottery pools.)
Interesting. I've never done a lottery but I invest in UK Premium Bonds and have more or less 'won' what I expected and consider it interest payment on savings.. Could you do a video on UK Premium Bonds?
The word "lotto" is actually the origin of the word "lottery." The 15th century Milan numbers game was called "Lotto," and the word was also used for other sorts of gambling in large pools with large prizes in Italy. The French got the word "lotterie" from the Italian "lotteria" from "lotto" + -"-eria," and in turn the English got the word "lottery" from the French (unless we got it directly from Italian; it's hard to tell). I'm not sure why in modern French and English, we use the -eria/-erie/-ery suffix instead of just the original "lotto." At any rate, all versions ultimately derive from "lot," as in the object one draws to decide a random outcome. All English meanings of "lot" ultimately come from that object (e.g. a straw).
My wife is normally reluctant to watch maths videos but she loved watching you even though it’s 4am here in Portugal. Tomorrow we may need to go over the border to get a ticket! 🎟️
It makes sense that the independent game is the same as the one where you don’t replace the balls, because in both cases the game is fair. Each number has the same odds as any other of getting any given prize, and the number of prizes distributed this way remains the same.
The predatory psychology in this lottery is next level. Entire towns without a single winnner. Extremely high ticket prices, but each ticket is just 1/1850th of this ticket, so the prize sum is grossly inflated. At the same time there is a claim that the big prize is hundreds of milltions of Euro, while in reality being 400 000 euro, and that is before a "lottery tax" is applied, so in reality it is not even that.
@@57thorns They don't claim that the prize is hundreds of millions. They clearly advertise €4M as the biggest prize with a 1/100000 chance of winning with a €200 ticket. The €2.5B figure come from the total prize pool. Hope that helps!
@@AnotherRoof The top-line figure they advertise online is the entire prize pool. They advertise it right next to the top prize of the Euro Millions, which is only half its prize pool. Tiny text points out the difference, but I still think that's skeevy. It doesn't help that media refers to it as "el gordo," even though that's actually just the top prize. That said, most ads for the lottery don't seem to emphasize the size of the prizes as much as they do in the US in general. And I doubt anyone is confused enough to think the top prize is 2.7 billion euros (though idk, maybe some do). You also claimed that these tickets are only sold in official lottery shops, which is false. It's sold in many unlicensed shops around the country, and licensed, privately-owned-and-operated shops also sell them. Your numbers are also a year out of date. In 2024, 193 series of tickets were sold. It was 185 in 2023 and 180 in 2022.
There is a theoretically winning strategy for roulette, though, given that you have infinite money, and that is to double your bet every time you lose. If you have enough money to keep going until luck falls your way, you aren't kicked out for doing this, and there is no limit on the size of your bet, you will win by the amount of your first bet (note: this means betting over 1000 pounds to win 1 pound after a 10 round losing streak, assuming you started with 1 pound, which is pretty ridiculous).
Maybe I'm missing something tricky that makes it more subtle than it is, but to me E(X+Y) = E(X) + E(Y) (with X and Y independent) is obvious. My reasoning is to imagine you do only X on a given day and do only Y on different a day. Obviously, each expected value is E(X) and E(Y) isn't it?
@@pierrebaillargeon9531 The fact that it still works this way even when you don't replace the ball is counterintuitive. If you do X on one day, leave the ball out, then Y on the remaining state -- it *feels* like things might be different...
@AnotherRoof My comments was in response of your demonstration of E(X+Y) = E(X) + E(Y), which your prefaced by stating that you generally found was non-obvious. Maybe you were instead talking about the ball replacement case specifically. If that was the case, it was not clear to me that you were. During the proof, don't you assume that event X and Y are independent? When not replacing the ball, the events are no longer independent since drawing ball N in event X prevents it from being drawn in event Y, which the proof does not address? That is, in the proof, removing ball N after event X means that, for event Y, the Yn event can no longer occurs and thus affect the calculation. So I think the fact of removing or not ball N not affecting the outcome is only true when all outcomes have equal probabilities, which I think you don't state as a pre-supposition during your demonstration? In any case, you fail to remove event Yn in the demonstration, which is required when ball N is removed in event X. Am I wrong on this?
@pierrebaillargeon9531 I'm busy at the moment but will reply while I remember: The proof doesn't depend on independence. I personally struggle with the intuition behind linearity of expectation because it doesn't depend on independence. That is, because it works in all cases including when they are dependent, it's counterintuitive to me. Like I say in the video though -- if you're good at probability it may come as no surprise!
another way to think about the replacement thing: let's say there's a really simple game, you pick a number and it's either the winning number or the losing number, and the number is drawn twice without replacement, you get one number, and then the only other choice is the other number. So it's either win then lose, or lose than win. 50/50 with replacement, sure you have 2 chances to get the winning number, but you *also* have 2 chances to get the losing number. Perfect balance, 50/50. Even as you scale this up to 100 losing numbers and 1 winning numbers, you still have a chance of getting the same winning number twice and a chance of getting the same losing number twice.
Yes, that's a consequence of linearity of expectation. Add up all the prizes paid out and divide by the number of participants. It doesn't matter to whom those prizes are distributed (i.e. how many people win multiple prizes). But this approach isn't useful in some lotteries where winning a higher prize precludes winning a lower one, because calculating the total amount paid out is just as hard as calculating the expected value. (In fact, it's basically the same calculation.)
@@EebstertheGreatgot it. It’s just that the total prize fund is given at 9:52 anyway, so two thirds of the video are, potentially, unnecessary. Or maybe is it that we don’t know how many tickets are sold?
Only if your goal is to profit (and even then, you can go into professional poker or stock trading). Lots of people enjoy gambling regardless of the outcome, and I don't see the problem if it's done responsibly.
@@alexmcdonough4973 The problem is that it's a scheme for smart people to exploit stupid people. It doesn't matter to me that the stupid people don't realize they are stupid. For example I had a best friend who was valedictorian of the engineering school of a prestigious US university who was convinced the gamblers fallacy was true. Also note they don't advertise the vig (house advantage) usually about 50%. Note it's ILLEGAL in Nevada to run a game that is half that unfair! And in my state indian casinos can't even have bets that are as unfair as that! If the games were mathematically fair or very close to it, and prevented large bets, such that casual players would break even in the long run regardless of how dumb they were, then I'd agree they were harmless... but then they wouldn't exist. The argument is that players can be educated and understand the risk. But the reality is that the games couldn't exist if players were vigorously tested about their understanding as a condition to play. It's a scam pure and simple. It's just well marketed.
@@alexmcdonough4973 , there are far more entertaining gambling games out there besides the lottery. If you find the act of buying a ticket just to wait on results fun, you might just have an addiction that needs to be addressed.
@@alexmcdonough4973 Exactly. My mom gives us all scratch-off lottery tickets for special occasions, like Christmas. It's a fun communal activity to scratch them off, and occasionally we'll win $50 or $100. Harmless fun, when it's that limited in scope.
Don't hate content creators because you are in top 20% percentile of you tube knowlegable viewership ;) It's better to take it as a compliment and move on.
@@LightPink He does if some of his audience doesn't yet fully understand expected value. Once again, it's not about what you, specifically, already know.
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⬣ CORRECTIONS, CLARIFICATIONS, AND COMMON COMMENTS ⬣
1. (a) There are two issues with my analysis of the US Powerball. First was an honest mistake which was the percentage loss for the powerball: It should be 0.84/2 which is a 42% loss. I mistakenly divided the EV by the cost to arrive at ~59%. My bad! This is still a worse percentage loss than the Spanish Christmas Lottery, but not as bad as claimed in the video. Sorry about that.
1. (b) Secondly was an intentional simplification which is that the jackpot often gets split between multiple players. As the jackpot increases, more people buy tickets which increases the likelihood of the jackpot getting divided, thereby lowering the EV. This analysis is pretty complicated so what I did was take all the jackpots awarded since the latest rule change and take the median, which is $246M. The net effect of divided jackpots, however, is that the EV remains at around the -0.85 which is similar to the value quoted in the video. See this article for more info:
www.forbes.com/sites/startswithabang/2016/01/13/the-science-of-powerball/
2. A simpler way to calculate the EV of the Spanish Christmas lottery would be to take the total prize pool and divide it by the tickets sold. For the case studied in the video, we have 2.5B / (100,000*180) = 139 which is what I obtained in the video (barring rounding errors). Boy, is this much easier but hey, if we'd done it that way we would have got to learn about the linearity of expectation!
26:50 I believe that the loss should be around 41-42% rather than 59%. If you take the profit of -0.84 and divide by the amount put in ($2), you get 0.84/2 = 0.42. I think you used the EV to calculate the profit rather than the complement from $2.
This seems more in line with 24:00 where the money spent is 20, the EV is 14, and the profit is -6. The loss is calculated as positive of EV / money spent.
@@LumberLopper Well spotted. This is a pretty shameful error from me that I managed to miss -- I've updated the pinned comment with this correction!
Gordo means fat as in overweight. Big fat prize is the idea. About 25 years ago I asked a Spanish mate to feel the back of my hand. Bemused and suspicious he did it and asked why. I told him the guy in the kiosk had said "que te toque el gordo" / lit. "may the fatty touch you" - a proper translation is I hope you win, good luck basically.
My best joke ever.
I'm from Spain and I didn't know it was the biggest lottery!!! 😮
Same LOL
It actually worries me
It is ungoogleable. No matter how I phrase my query, google refuses to believe I want anything except dozens and dozens of articles in a row about the largest lottery _jackpots_ ever awarded.
@@EebstertheGreat there is a wikipedia page on the topic
@AntonioLasoGonzalez Yes, but the page doesn't cite any source. Anyway, the idea is to find it without putting the answer in first. Like, if you don't already know which lottery has the biggest payout, how do you find it?
also the neat thing is that taxes are only applied to El Gordo and the second prize, so most of winnings are free of taxes. what a surprise to see this lottery covered in your channel!
So the two big prices are not as large as advertized becuas of "taxes", so what are the real prizes?
Used to be no taxes for any public lottery. But in 2013 they added a tax of 20% to any prize higher to 40.000 euros.
You know why? Because Spain is Spain.... It was used as the best money laudry scheme in the world.
Winners went to banks/lottery shops for counsel, they contact politicians/big business (But mostly politicians with lots of questionable "support" money), Politicians pays 10% more for the ticket, shares profit with the informants, and the politician just laundred 40k to 400k euros for just 4/40k.
Spain is Spain because everyone knew, including the government and taxes agency, that some famous politicians (Fabra,Roca,..) for example had won more than 10 or 20 times some lottery. No one "suspected" anything for years and years.
Juan Antonio Roca was an specialist, most lottery shop owners had his phone. He won the lottery 80 times in about 20 years.
@@framegrace1 So if you win 100k now (just to make the math easier) and the tax is 20%, you get an 80k winning that you can sell to a money laundered for 88k ?
The tax makes no difference whatsoever on that part of the business.
And any kind of gamlbing winnings can be used this way.
You might even show up at a casion, buy $1million in chips for cash, gamble 20% of it and go home with $800k in verified winnings.
@@57thorns from 400k you get about 325k after taxes
Because each ticket necessarily has the same EV (symmetric with respect to all numbers), EV for any ticket is simply total prize pool divided by number of possible tickets. Prize pool in thumbnail is correct (I calculated 2,521,476,000 Euros), total number of tickets is 100,000x180, so EV before cost is 140.082 Euros. It doesnt matter how the prizes are distributed or at which probabilities. I kept waiting for this punchline but never got it.
This also helps understand intuitively why these games have the same EV with and without number replacement: if total payout is the same and every number has an equal chance at winning, EV for each wont change.
13:00 Ah, the “expected value is linear” section of the video. I once had a teacher who yelled to the whole class, and also opened the door to an adjacent classroom to tell the neighboring class studying something else, that “THE EXPECTATION OF THE SUM IS THE SUM OF THE EXPECTATIONS!”. And even knowing that, and having had homework in a different class to prove that expectation is linear, it can still mess with my head.
It helped me to think of expectations as "sums" to begin with
Linearity of Expectation is so overpowered. Here is one example I saw the other day:
Let S denote the surface of a sphere. Assume S has surface area 1. Let F be a measurable subset of S with area strictly greater than 1/3, then F contains two mutually orthogonal points.
To see this, consider three mutually orthogonal vectors v1, v2, v3 (the example you are most familiar with is (1,0,0), (0,1,0), (0,0,1)), and consider rotating the vectors so they are still orthogonal and hit every point of the sphere. Let X denote the random variable "the number of these vectors that are in F". We will show that E(X) > 1, but this is only possible if X >= 2 sometimes occurs, otherwise we always have X 1/3, so E(X) > 1/3 + 1/3 + 1/3 = 1.
...wow. That's a great proof.
I don't get why having "X sometimes >=2" leads to "2 of the three vectors are always in F". wouldn't there also be the case that "X is sometimes
@@kaishang6406 It's not that two of the three vectors are always in F, but that there is some orientation where two of the three vectors are both in F at the same time - which means that those two points are a pair of mutually orthogonal points in F (as required).
@@rmsgrey thanks, i misunderstood what the expect value is averaging over. i thought it was averaging over all possible subset F and couldn't see the relation. If it is averaging over all orientation, the relation become clear.
that's a really pretty proof! The linearity "disentangling" the dependence between the orientations of the three vectors relative to each other is really neat
Oddly enough an Spanish channel mentioned your channel in one of their videos recently
does the ±1 wrap around? otherwise a draw of 1 or 5 (or 99999) only have one neighbour.
It does, actually! Although it has never happened, 0 would have 1 and 99 999 as neighbours (aproximaciones). Conversely, 99 999's neighbours would be 99 998 and 0. In a sense, this works as the ring Z_100 000.
Didn't expect a video about the Christmas lottery on this channel, and even less a mention of Pinos Puente
Great video! I am however a bit confused how the expected loss for powerball at 26:52 can be 59% if the expected value is more than half of the ticket price. For the spanish christmas lottery the expected loss of 30% is calculated as (EV - ticketPrice) / ticketPrice which correctly evaluates to -0.3 , but plugging in the values for powerball gives (1.16 - 2) / 2 = -0.42 or expected loss of 42%.
EDIT: Just realized that it is a known error that was just corrected in the pinned comment, so I needed to reload the page to see it
@bongi6811 You're quite right! Massive mistake on my part but thanks for checking the pinned comment!
Why are we approaching this by focusing on it from the perspective of individual participants? Why don't you just add up all of the prizes and divide by the number of participants?
We don't have to worry about the amount of money being paid out varying as we do in, for instance, the American lottery.
It's worth mentioning that each ticket is divided into ten tenths but also people can and do make their own further subsplits called 'participaciones'. Individuals may share participaciones of small value with family and friends (this is not fashionable anymore), and all kinds of associations sell participaciones in which you may pay e.g. 5€ of which you play 4€ with 1€ going to the seller as a donation.
This means one prize very often reaches many hundreds of people, many of them in modest or even unremarkable amounts.
3:40 "La Pedrea" in this context translates better to "The Cobbles". It's just an old way of saying the little stones
5:40 I actually LOLed. And there is a reason why I never played, nor have any intent to play the lottery. I know how the maths works, even though I am a lousy mathematician. You buy the feeling of possible great reward. Since I do not need to buy that feeling (or expectation), I respectfully refuse to take part in any lottery (or any sort of gambling). The winner is always "the house". I know about certain techniques but 1. I am too lazy to focus that long to get an advantage and 2. I know that I would be immediately suspicious as my "luck" would be noticed. And 3rd: I do not need it.
25:00 .... "and it is NOT a joke"
27:15 I laughed way too hard at "powerfail"
Now, I wasn't surprised, but I have put the phrase "due to the linearity of expected values" in a paper. Using this we could calculate probability distributions for X by working out E(X), E(X²)... E(X^n), which were quite simple (well, only due to the linearity of expected values) and then just by definition E(X^n)=sum(i=1..m) i^xP(X=i), so you get a system of linear equations you can solve for all P(X=i)s. It's a neat method, which turned something not pactially calculable into a form which allowed for explicit calculation.
Linearity is neat.
Yaaay! Now it's a Merry Christmas! ANOTHER ROOF!
Paused at 3:01. Hmm. St. Alfonzos pancake breakfast comes to mind. Man, I have to listen to the whole LP once again. Frank was a genius.
23:30 easy:
E(X + Y + Z) = E(X + Y) + E(Z), because X + Y itself is just another random variable.
Thus, E(X + Y + Z) = E(X) + E(Y) + E(Z). Easy.
7:09 Direct hit. Emotional damage sustained. Abandon friendship.
Feliz Navidad for you all!!! In Spain there're drawings of this lottery all year long (on Thursdays and Saturdays), but it takes place on a more simplified fashion. There are five cages to draw every single digit of a 5-digit number, and there is a draw for every single prize: first a few 2-digit ends, then a few 3-digit ends, 4-digit ends, then two or three draws of fully fledged 5-digit numbers (the latter is drawn, the bigger the prize), and finally 2 1-digit endings for refunds (the 1st prize ending also qualifies for a refund), obviously balls drawn are returned to cages between drawings.
Next drawing to take place on January 6th (known as 'El Niño') uses that system (known as 'Modern System' opposed to the 'Traditional System' used on Christmas drawings).
Most people that got refunds from the Christmas drawing do bet again on 'El Niño' as it also costs €20 for a ticket tenth, year-round drawings have a cost ranging from €3 to €12 for a ticket tenth, but there are also much lesser tickets to be sold.
I'm an algebraic thinker, so around 17:00, I paused and plugged in "n" for the number of balls... and yeah, I was shocked that it ended up simplifying to 1/n odds of any combination of first and second draw in *both* the replacement and non-replacement cases. Probability is so unintuitive.
Combinatorics can be fun. During my uni days it was interesting to see how it is an art of counting things and what conclusions can be drawn.
I'm not sure it's that unintuitive. Instead of probabilities, imagine every combination is realized in practice, so we're just talking about fractions of some actual group. Then the "expected value" is just an average over the population. So if we give, say, 1/4 of the group $1 prizes and 1/3 of the group $2 prizes, then clearly the average prize earnings are (1/4)($1) + (1/3)($2) = $ 11/12. And it doesn't matter how much those groups overlap, because we are just interested in the total amount of money doled out. So it is with expectation. Who cares whether the winners are packed together or spread out? It's about average winnings. So just add the value of each prize.
9:45 this is the Oprah lottery
10:01 - 10:04 Another LOL. Your humor is excellent!
This was really fun! Rigorous but light.
Linearity of expectation makes sense when you realise it's a weighted sum, which is more noticeably linear
Hello from Spain! Today's gordo has been 72480, which is... too round @~@, even more so than last year's! Both of them have been multiples of 24, with this last one actually being a multiple of 480, and bothe were (very) abundant numbers, with their propper factors summing over twice the number itself for this year's number, and over one and a helf times for last year's.
Anyway, good luck next time! (I doubt any of the big winners are reading this xd)
But wait, in Powerball etc. is it not theoretically possible for every player to have the same number so they all divide the prize? xD
I saw someone make a similar question herr, but mine isn't exactly the same. They asked about the derivation of the independence of games X and Y, if game Y depends on the result of game X.
I think their question was about the probabilities of Y1, Y2... etc, but what's been confusing me are the prizes b1, b2...
The available prizes are different for game Y depending on the results of game X. Wouldn't that make them dependent? As in, aren't the prizes b1, b2 etc dependent on the result of game X?
hey another roof. i just discovered that the first 1,000 digits of the 7th root of 17 after the significant ones is a perfect de bruijn sequence. pi, phi, and the square root of 2 dont even come close. i think ill call this new property "supernormality". needs more studying. thanks claude 3.5 sonnet for the help!
✌️✌️
I love christmas too ;)
Good luck tomorrow!!
7:04 Mr popular over here with nine friends
Looking at it from an economoc perspective though, the fact such a large proportion of the population plays it means the expected loss can be lower, as the scaling costs of administrating lottery are relatively low, so the profit will be the same as smaller lottery with a lower expected winnings.
My wife asked me literally yesterday to explain to her why there isn't a winning strategy for this (we bought a décimo for the first time this year). I managed, but I'll bookmark this for when she asks again next year lol.
Good luck and feliz navidad from Madrid!
After all these years on the internet, I'm finally looking at the guy all the spam emails have been trying to reach: someone who actually bought a lottery ticket in a foreign country.
@@FionSus Hahaha this made me laugh.
Interesting video! I ran a similar calculation last year, but accounted the prize of a décimo in the value of each prize, i.e. a décimo for El Gordo would reward 399.980€ net after deducing the 20€ initial investment. My numbers are very similar to yours because it doesn't make much of a difference, but I was wondering: why is the initial investment not considered when discussing EV in gambling? In your roulette example, the actual prize for winning red/black would be 1, not 2. Does it make sense?
20:00 but if game Y depends on game X, how can you write down fixed values for the outcome and probabilities ?
@@josenobi3022 If game Y depends on game X, there is still a fixed probability of the event "X1 and Y1". Independent just means that P(X1 and Y1) is equal to P(X1)*P(Y1). Hope that helps!
@@AnotherRoofyeah I get that, I'm talking specifically about what you're writing down at 20:00
@@josenobi3022 Doing Y by itself will have its own associated probabilities. So like the earlier example, game Y was the "2nd prize" with its own prize numbers. If we don't replace the ball then Y is dependent on X. The probabilities for Y's events don't change, but it will adjust the probability for the "X and Y" probabilities -- that's where dependence comes into play.
@AnotherRoof wdym "the probabilities for Y's events don't change" ? Isn't that true only if it's independant ?
@@josenobi3022 If you completely ignore game X and just focus on game Y, then the outcomes of Y have their own probabilities of occurring. They are just their own events. Dependence and independence only come into play once we examine two games.
Buena suerte 👏
I think there is a typo at 9:51 , assuming there are 2 4th prizes, then there would be 198 tickets with the "first three digits of 4th" as winners *. About this, there are just a couple of things that seems strange to me: because of the linearity, I'm assuming it's possible to win multiple prize with the same ticket, as an example it's possible to win "first three digits of 4th" two times if both 4th have the same leading digits (moreover, in this case, it would mean that both 4th prize ticked would get the prize, for having the same leading digits of the other).
It just seems strange to me that it's possible to win multiple prize with the same ticket, do you confirm this is the case?
(to be exact, I'm asking if the only exception to multiple awards to a single number is "if it's one of the winner's numbers, it doesn't matter it shares digits with itself")
wait, actually I think I disagree with how the statement written at 13:20 interacts with the table at 14:19 , according to the table there are 198 prizes still (both 1st prize and 2nd-similarity-compensation together is an option (ex: 3,2), like 2nd prize and 1st-prize-compensation (ex: (2,3)), and lastly winning both 1st and 2nd compensation together (ex: 2,4))
probably I just misunderstood what you wanted to say, I need sleep... I'll sleep it out
As I understand it from the video, when one of the 4th prizes is drawn, that ticket number gets the full prize, and the 99 other tickets with the same first three numbers get the lesser prize. The same thing happens to the other 4th prize, regardless of whether its numbers overlap with any other prizes.
So for any given prize, any given ticket will either: win that exact prize; win one or more of its near miss prizes; or not win anything from that prize. Your ticket then wins all the prizes it qualifies for.
@@rmsgrey this is right yeah!
Thanks @rmsgrey , makes sense, but in this case what's the point of 13:20; how is the likelyhood ever reduced if a single can win multiple prizes? I mean, how getting two consecutive 1st and second prizes shold decrease the likelyhood of being 1 away, or how having the same three digits decrase the number of winning tickets to 98? I'm just confused by this part then. The fact that those 98 wins are double and that even 1st and 2nd extraction win "first-three-digits" prizes didn't even cross my mind. Moreover, you could just appleal to "invariance" of the amount of money given at any round (/by a certain prize extraction, as ex 2nd prize extraction or 4th prize number two extraction) to calculate the total number
Man, all the people talking about gambling, taxes, and so on don't get the point. The Spanish lottery is about sharing, giving decimos as a present, about envy (everyone at work has the company's décimo just in case everyone else becomes rich), about thinking to go to the bar where you bought it if you win. This year I had 110€ in décimos (we share some decimos and pay half each) and I rarely win something, but everyone plays every year for the illusion.
The Spanish drive fast!
If you win El Gordo and second prize is the number one above yours, do you also win the bonus for that?
Yes, they do. If you won ell gordo, with 40k, in that case you will accumulates around 1.5k more
Does the Thai lottery work anything like this? I've always been absolutely baffled by their hours-long lottery draws.
"The following algorithm is linear."
Parallel computation in hardware: *SLOT MACHINE JACKPOT NOISES*
Nvidia stocks up another 200%
"Vídeo patrocinado por Loterías y Apuestas del Estado". Sonaban mejor los niños mutantes de San Ildefonso cuando cantaban en pesetas. ¿Os acordáis del calvo?
Good luck getting all the references with that one.
Ciento veinticinco miiiiiiiil peeeseeeeetaaaaas
Your Powerball calculations assumed that each jackpot winner would receive $246 million. But when the jackpot gets that high, many people play, so the probability of sharing the prize becomes substantial. In the Powerball, if multiple people simultaneously win the jackpot, the jackpot is evenly split between them. So really, the expected value of your ticket is less than you stated (particularly if you choose commonly-played numbers). On the other hand, sometimes the jackpot is more than twice that size, so in those rare cases, a ticket could really have a positive expected value
. . . before taxes. Because once you realize that state lottery winnings in the US are subject to federal income tax, it becomes practically impossible for such a scenario to occur. In most states, even state taxes can apply, in spite of the fact that it is the state itself paying you your winnings. And it's actually worse than that, because in the US, lottery jackpots are not paid as lump sums but as escalating monthly payments. The present value of the annuity is not really as high as they claim, so you actually get much less than stated. IMO that should be illegal, since it is literally, factually untrue, but that's how it is. (Of course, once you consider a more reasonable logarithmic utility function, it becomes even more stupid to play the lottery, since each dollar you risk losing today is a lot more valuable to you than each dollar you win at the end of your hypothetical large jackpot. However, this utility calculation doesn't apply the same way to lottery pools.)
Thanks for raising this -- I've addressed the splitting of jackpots in my pinned comment.
@@AnotherRoof Thanks, I wasn't sure where the figure came from, but that analysis looks pretty good.
Interesting. I've never done a lottery but I invest in UK Premium Bonds and have more or less 'won' what I expected and consider it interest payment on savings.. Could you do a video on UK Premium Bonds?
I’m putting it all on green
This lottery is a mind controlling game for the spanish population to keep hope and community up
WAIT LOTTO HAS NO RELATION TO THE SPORTS BRAND?!?
The word "lotto" is actually the origin of the word "lottery." The 15th century Milan numbers game was called "Lotto," and the word was also used for other sorts of gambling in large pools with large prizes in Italy. The French got the word "lotterie" from the Italian "lotteria" from "lotto" + -"-eria," and in turn the English got the word "lottery" from the French (unless we got it directly from Italian; it's hard to tell).
I'm not sure why in modern French and English, we use the -eria/-erie/-ery suffix instead of just the original "lotto." At any rate, all versions ultimately derive from "lot," as in the object one draws to decide a random outcome. All English meanings of "lot" ultimately come from that object (e.g. a straw).
My wife is normally reluctant to watch maths videos but she loved watching you even though it’s 4am here in Portugal. Tomorrow we may need to go over the border to get a ticket! 🎟️
It makes sense that the independent game is the same as the one where you don’t replace the balls, because in both cases the game is fair. Each number has the same odds as any other of getting any given prize, and the number of prizes distributed this way remains the same.
The predatory psychology in this lottery is next level.
Entire towns without a single winnner.
Extremely high ticket prices, but each ticket is just 1/1850th of this ticket, so the prize sum is grossly inflated.
At the same time there is a claim that the big prize is hundreds of milltions of Euro, while in reality being 400 000 euro, and that is before a "lottery tax" is applied, so in reality it is not even that.
@@57thorns They don't claim that the prize is hundreds of millions. They clearly advertise €4M as the biggest prize with a 1/100000 chance of winning with a €200 ticket. The €2.5B figure come from the total prize pool. Hope that helps!
Having a decent propbability to win either "your money back" or "a new ticket" is standard fare to make people think they "won something".
@@AnotherRoof The top-line figure they advertise online is the entire prize pool. They advertise it right next to the top prize of the Euro Millions, which is only half its prize pool. Tiny text points out the difference, but I still think that's skeevy. It doesn't help that media refers to it as "el gordo," even though that's actually just the top prize. That said, most ads for the lottery don't seem to emphasize the size of the prizes as much as they do in the US in general. And I doubt anyone is confused enough to think the top prize is 2.7 billion euros (though idk, maybe some do).
You also claimed that these tickets are only sold in official lottery shops, which is false. It's sold in many unlicensed shops around the country, and licensed, privately-owned-and-operated shops also sell them.
Your numbers are also a year out of date. In 2024, 193 series of tickets were sold. It was 185 in 2023 and 180 in 2022.
Sooo.... There is a way?
I’ve never heard of this wtf
Am I going completely nuts or is 4:20 a holocaust joke?
It is absolutely not intended as a holocaust joke.
Haha good ok😅
There is a theoretically winning strategy for roulette, though, given that you have infinite money, and that is to double your bet every time you lose. If you have enough money to keep going until luck falls your way, you aren't kicked out for doing this, and there is no limit on the size of your bet, you will win by the amount of your first bet (note: this means betting over 1000 pounds to win 1 pound after a 10 round losing streak, assuming you started with 1 pound, which is pretty ridiculous).
lol.. 200 euros…
I'd rather burn my money then buy a lottery ticket
Be ware though, spending your money is usually legal while destroying it isn't ;]
So, did you win anything? 😃
Maybe I'm missing something tricky that makes it more subtle than it is, but to me E(X+Y) = E(X) + E(Y) (with X and Y independent) is obvious.
My reasoning is to imagine you do only X on a given day and do only Y on different a day. Obviously, each expected value is E(X) and E(Y) isn't it?
@@pierrebaillargeon9531 The fact that it still works this way even when you don't replace the ball is counterintuitive. If you do X on one day, leave the ball out, then Y on the remaining state -- it *feels* like things might be different...
Does the fact that E(X+Y) = E(X) + E(Y), regardless of whether X and Y are independent, still feel obvious? 🤔
@AnotherRoof My comments was in response of your demonstration of E(X+Y) = E(X) + E(Y), which your prefaced by stating that you generally found was non-obvious. Maybe you were instead talking about the ball replacement case specifically. If that was the case, it was not clear to me that you were. During the proof, don't you assume that event X and Y are independent? When not replacing the ball, the events are no longer independent since drawing ball N in event X prevents it from being drawn in event Y, which the proof does not address?
That is, in the proof, removing ball N after event X means that, for event Y, the Yn event can no longer occurs and thus affect the calculation.
So I think the fact of removing or not ball N not affecting the outcome is only true when all outcomes have equal probabilities, which I think you don't state as a pre-supposition during your demonstration? In any case, you fail to remove event Yn in the demonstration, which is required when ball N is removed in event X. Am I wrong on this?
@pierrebaillargeon9531 I'm busy at the moment but will reply while I remember:
The proof doesn't depend on independence.
I personally struggle with the intuition behind linearity of expectation because it doesn't depend on independence. That is, because it works in all cases including when they are dependent, it's counterintuitive to me. Like I say in the video though -- if you're good at probability it may come as no surprise!
@AnotherRoof What helps me is the fact that probabilities are sums, series, integrals, or a combination, and all of those are linear.
Did you win?
another way to think about the replacement thing:
let's say there's a really simple game, you pick a number and it's either the winning number or the losing number, and the number is drawn twice
without replacement, you get one number, and then the only other choice is the other number. So it's either win then lose, or lose than win. 50/50
with replacement, sure you have 2 chances to get the winning number, but you *also* have 2 chances to get the losing number. Perfect balance, 50/50.
Even as you scale this up to 100 losing numbers and 1 winning numbers, you still have a chance of getting the same winning number twice and a chance of getting the same losing number twice.
Couldn’t we just divide the total lottery fund by the number of tickets sold to get the expected value?
I think that would imply every winning ticket receives the same amount of prize money, wouldn't it?
@@IceMetalPunk No. I dont know why youd expect that to bear any relation?
This is absolutely correct
Yes, that's a consequence of linearity of expectation. Add up all the prizes paid out and divide by the number of participants. It doesn't matter to whom those prizes are distributed (i.e. how many people win multiple prizes).
But this approach isn't useful in some lotteries where winning a higher prize precludes winning a lower one, because calculating the total amount paid out is just as hard as calculating the expected value. (In fact, it's basically the same calculation.)
@@EebstertheGreatgot it. It’s just that the total prize fund is given at 9:52 anyway, so two thirds of the video are, potentially, unnecessary. Or maybe is it that we don’t know how many tickets are sold?
And taxes. Gambling is just as dumb as going out in the cold with not enough clothes on. Didn’t your mother tell you so?
Only if your goal is to profit (and even then, you can go into professional poker or stock trading). Lots of people enjoy gambling regardless of the outcome, and I don't see the problem if it's done responsibly.
@@alexmcdonough4973 The problem is that it's a scheme for smart people to exploit stupid people. It doesn't matter to me that the stupid people don't realize they are stupid. For example I had a best friend who was valedictorian of the engineering school of a prestigious US university who was convinced the gamblers fallacy was true. Also note they don't advertise the vig (house advantage) usually about 50%. Note it's ILLEGAL in Nevada to run a game that is half that unfair! And in my state indian casinos can't even have bets that are as unfair as that!
If the games were mathematically fair or very close to it, and prevented large bets, such that casual players would break even in the long run regardless of how dumb they were, then I'd agree they were harmless... but then they wouldn't exist.
The argument is that players can be educated and understand the risk. But the reality is that the games couldn't exist if players were vigorously tested about their understanding as a condition to play. It's a scam pure and simple. It's just well marketed.
@@alexmcdonough4973 , there are far more entertaining gambling games out there besides the lottery. If you find the act of buying a ticket just to wait on results fun, you might just have an addiction that needs to be addressed.
@@alexmcdonough4973 Exactly. My mom gives us all scratch-off lottery tickets for special occasions, like Christmas. It's a fun communal activity to scratch them off, and occasionally we'll win $50 or $100. Harmless fun, when it's that limited in scope.
@@alexmcdonough4973 Financial number going up. The motive for anything a person ever does is some kind of profit in any case, material or not.
I hate it when math videos spend 80% of their run time going through stuff you already know.
So you hate it when math UA-camrs aren't psychic and focusing their psychic abilities on what you, specifically, already know?
@@IceMetalPunk he doesn't need to spend 2 minutes explaining expected value just like how a fashion UA-camr wouldn't explain what pants are for.
Don't hate content creators because you are in top 20% percentile of you tube knowlegable viewership ;)
It's better to take it as a compliment and move on.
@@LightPink He spent three videos and two hours proving 2 + 3 = 5. He is not assuming a high-level understanding from his audience.
@@LightPink He does if some of his audience doesn't yet fully understand expected value. Once again, it's not about what you, specifically, already know.