How to get infinitely many lottery tickets
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- Опубликовано: 12 сен 2024
- In October 2015 the UK lottery changed from 49 balls to 59 balls. Which means the chance of choosing 6 correct numbers went from 1 in 13,983,816 to 1 in 45,057,474.
But the 45,057,474 number is wrong. Because of a new possible prize, not only can you possible win a never-ending supply of tickets, but the real jackpot odds are 1 in 40,665,099.
Read my lottery article in the Guardian:
www.theguardian...
See the official new lottery rules:
www.national-l...
of course he chose only primes
Now that you mention it...
That's like the worst possible Strategie. he certainly is not the only one to have choosen those, so he would have to share the jackpot (if he would have won)
He also chose 31 41 59. Where's 26 53 58 though?
I am curious strategy*
I used to work at a gas station (here in the US, our lotteries are basically the same). I had a problem gambler who bought an ungodly number of lottery tickets every week. I felt bad for her. "You wanna hack the lottery?" I finally asked her. She listened. I ran the numbers, explained the math.
That was the last time I saw her in the store to buy tickets.
I got fired not long after that.
Absolutely true story.
And yet someone has to win. You need a ticket. So just get ONE.
+Hanniffy Dinn no one has to win, there can just be no winners if no one gets all the numbers
Jeffrey Jeffson
wtf? you are missing the whole point of the word LOTTERY. They have raffle numbers on to guarantee winners of some kind.
Bravo Elliot! OK, you got fired, but you tried to help someone. Well done!
+Hanniffy Dinn Nope, there can be lotteries where nobody wins. That's how the jackpot gets really high sometimes- nobody wins for a while, so the money racks up.
There is also a nonzero chance that a billionaire bumps into me on the street and gives me 10 million to apologize and because he likes my face. And that dream is free. So I still don't see the benefit of a lottery ticket.
Lottery tickets are for those of us with less-lovely faces.
standupmaths Well, I think everybody could construct an absurdly improbably scenario for them to become rich for basically no reason if they thought about it a bit. So maybe lottery tickets simply are for people with no imagination ;)
+Penny Lane Or for people with more initiative. ;)
NoriMori Yes, this "initiative" must be what makes people pray when they feel the need to do something, anything, even though in a lucid moment they will explain to you that God already knows best so any advice on their part would be quite hubristic ;)
+Penny Lane Let's not be too quick to assume that the general public don't understand quantum mechanics. They may be thinking that there is inevitably a universe in which they do win. They may all be right. Me, I spend my money on drink.
Excellent as always! My maths professor, who specialized in statistics and probability, always stated that lottery tickets were a tax on those who failed at grasping the fundamentals of probability. He then continued each week to buy a ticket. (!?!?)
+Martin S. Stoller Mine stated almost exactly the same - although my math prof also seemed to be uncapable of understanding the other factors involved in lottery (excitement, promise of a subjectively better tomorrow etc.), unlike yours. so, you could actually consider yourself lucky in this aspect (i.e. having been taught by an actual human)
Yes, the lottery is about a lot more than just expected return.
+Martin S. Stoller I knew somebody who would play the values that would best fit an uniform distribution of the numbers. Maybe your teacher did something like that?
Looks like your teacher thinks he deserve be taxed for no know good enough basic probability.
now consider that if you win a lower price you use it to buy tickets... what are the oddds now?
That is a really good point! Great, now I have to re-do my calculations…
+standupmaths make another video!! :D
+Ivan Gonzalez +standupmaths 3 months have passed. Is it ready yet?
+John Violidakis (johnv) Let me punish you with an ambiguous statement so that you'll become none the wiser:
You should know what to expect after such an answer, because Matt really is a stand-up guy.
+standupmaths When you calculated the odds of getting 2 numbers correct and therefore getting another ticket, did you exclude the cases when you get more than 2 numbers and therefore DON'T get a new ticket? (but money instead, which is sooo much worse =P)
Bigger carrot, longer pole.
+Aadil Shah that's what she said
What?
Yep
high time we got another Matt maths video. more please! I can't get enough of that amazing voice. I'm even re-watching old numberphile videos to get my fix.
I've been meaning to start making videos regularly. But yes: until then Numberphile will keep you going.
+standupmaths Could you please calculate what is the propability of winning jackpot if you bought another tickets for every 3,4 or 5 numbers hit? Maybe the propability would move again ever so slightly :)
How to get infinitely many lottery tickets
1) acquire an infinite amount of money to buy an infinite amount of tickets, in which case you have no reason to buy them became you already have infinite wealth....
easy.
Shivam Sharma imagin the inflation...*shudder*
it wouldn't change your infinite wealth. It would be like inf./2=inf.
and also crash the economy of extreme hyperinflation
So the lottery seems to be mirroring a general trend in society, a lot less people getting money, but those few that do get even more. Cynical me is cynical I guess. :->
This is something that has puzzled my mind for a while now, why do the jackpots need to be so big? Also why some sports stars get paid such an enormous mountains of cash? I mean, some of us would be happy to clean toilet seats with their tongues if it payed 10 million in an hour, thou not many would do it more than that one hour or maybe even less.
mustekkala tax take .
"...why do the jackpots need to be so big.." More people buy tickets with big jackpots. "...why some sports stars get paid such an enormous mountains of cash..." Sports stars are in high demand, but there's a low supply.
More numbers to choose from means more ways to lose. Only someone in marketing would try to spin this as a good thing.
***** Wait, Europe DOESN'T tax lottery earnings?
+Michael Miller Hmmmm.... where I'm from (and yes, it's in Europe) they are most definitely taxed. And something huge, like 40% or so, too. Not sure what +Sheldon Cooper had in mind
The UK doesn't tax any sort of competition winnings I don't think. I know game show winnings are tax free.
In Aus our Saturday lotto only goes up to 45, and we need six balls, with 2 sups drawn.
Our Powerball game used to be 5 from 5 numbers up to 45, then 1 from the Powerball pool, against from 1-45. The problem was, I reckon, too many people were winning by playing the PowerPick, which gave you the same normal numbers for 45 games, but changed the Powerball every game, so you were guaranteed the Powerball.
They changed the game to fix this. The number of possible balls changed for the normal 5, only from 1-40. And the powerball was changed heaps, from 1-45 down to just 1-20. The catch, of course, was that there was now a sixth ball you had to get in the normal lot of numbers! So suddenly a PowerPick (which is now only 20 games with the same set of normal numbers, with the PowerBall changing every game to guarantee the Powerball was yours) is a great option -- just match 6 from 6 balls with a 1-40 pool size, but the cost was $18.60 a go.
Comparing the new Powerball size to the Saturday lotto size, you have an easier time winning the powerball with a PowerPick then you do of winning a Saturday night lotto, BUT you can play far more games of Saturday night for the $18.60 you spent on that single PowerPick ($18.45 will get you 26 games of Saturday lotto. Fill out two coupons, we're Australia so each coupon only goes up to 18 games :/ ).
My personal theory re the Lotto is to play the game that gives you the best INCREASE in chance, for the LEAST amount of money, which happens to be the single lowest price game you can find. Anything above 0% is great, and you'll never find a better increase than from zero to a positive number, so play the cheapest game and never expect to win!
Honestly! The difference between 10 games and 50 games, when you consider that you're playing a 1 in 12 million or whatever chance game, is ridiculous, so putting $3 on 6 games as opposed to $35 on 50 games is the same thing.
Sorry. Rants over. And yes, I work in a newsagency. When the magazines are boring guess what I think about?
It is still a straightforward calculation, and you don't actually need geometric sums. Just consider the function P(n) to be the chance that you win the lottery with n tickets remaining. If I start with 1 ticket, I get
P(1) = (chance of winning the lottery in one drawing) + (chance of not winning the lottery or another ticket)P(0)+ (chance of winning another ticket)P(1)
Thus,
P(1)(1-(chance of winning another ticket))= (chance of winning the lottery in one drawing) + (chance of not winning the lottery or another ticket)P(0) = (chance of winning the lottery in one drawing)
Thus,
P(1) = (chance of winning the lottery in 1 drawing)/(1-(chance of winning another ticket))
You could also view it like this; if you win another ticket, nothing has really changed, and so we can simply remove that possibility from the sample size.
but the problem is if you don't win another ticket it costs (an admittedly small) amount of money to purchase a new one. we are not assuming infinite money so assuming every time you don't win a ticket you buy another one is a bad assumption especially when you consider the more tickets you buy and don't win the less you stand to earn by winning. besides saying this misses the point of the calculation the point was calculating the probability that buying one ticket will eventually let you win the jackpot not calculating the chance that buying new tickets will eventually win you the jackpot. as for buying more tickets winning you the jackpot that is pretty easy to calculate if n is the number of tickets you buy over time and p is the chance of winning as n approaches infinity p approaches 100% if n is infinity assuming infinite time (which has to be the case for infinite tickets to be the case) then you literally have to win eventually.
In Canada we have a lottery called "6/49" in which -the chance of winning the jackpot is 6/49- one chooses six numbers from a possible 49. One also selects an additional _bonus number._ I don't know why they don't just name the lottery "7/49". Perhaps there is a law requiring the name to be expressed in simplified form.
I don't believe for a moment that you normally purchase lottery tickets, Matt. I feel that you will frame and hang this on the wall, a testament to the public's deplorable maths skills (mine included, regrettably). If it happens to be _the_ winning ticket, I recommend posting armed guards until the 180-day claim period has lapsed, or fill in the "void" box. Options!
Indeed, I don't normally play. But I couldn't resist today! I'll get the ticket framed, win or lose.
If I understand the Canadian System correctly, it is not a simple "7/49".
As far as I understand your description, the system consists of two steps: the first one is the usual "6/49", the second one is an additional "1/10".
For the first step, the chances of having all six numbers correct is 1:14 million; for the second step, the chances for having the bonus number correct is 1:10.
If you want to win the jackpot, you have to have all 6 (out of 49) numbers correct AND 1 number (out of 10 different numbers) correct.
This gives you a possibility of 1:140 million.
Please correct me if I'm wrong.
+Thomas Nimmesgern
Forget my previous reply. I mixed up the Canadian System and the German system. :-(
The German system works as described above ("6/49" + additional "1/10"). For some strange reasons, I thought of the German system, I just misunderstood your description. The bonus number you mentioned led me on the wrong path. :-(
However, I can tell you that the German system has a "bonus number" as well, so it actually works like this: "6/49" and "1/(49-6)" and "1/10". Kind of "regular numbers" and "bonus number" and "extra super bonus number".
I guess that step 1 and step 2 are similar to the Canadian system (regular "6/49" and adiitional "1/(49-6)"); step 3 is unique to the German system (additional "1/10").
So what is it good for? Here in Germany, the winning ranks are similar to the following list:
Rank I - all 6 out of "6/49" plus the correct "extra super bonus number" out of "1/10"
Rank II - all 6 out of "6/49" without the correct "extra super bonus number"
Rank III - 5 out of "6/49" plus the correct "bonus number" out of "6/(49-6)"
Rank IV - 5 out of "6/49" without the correct "bonus number"
Rank V - 4 out of "6/49" plus the correct "bonus number" out of "6/(49-6)"
Rank VI - 4 out of "6/49" without the correct "bonus number"
Look at ranks III to VI. You see where the "bonus number" (adiitional "1/(49-6)") is used. That kind of bonus number just gives some extra winning ranks.
I hope this doesn't sound too complicated.
Thomas Nimmesgern Ah, I don't really know how exactly it works. I do not buy lottery tickets myself. People say buy a ticket when the prize is higher, but that is not convincing.
The one thing that ticket buyers can do to help themselves is to reduce the probability of sharing the prize by selecting unpopular numbers. I'm still not buying a ticket though. :)
Q: Did you include suing the value of the smaller prizes fed back in to buying more tickets in the next round in your calculation?
Because if you did, then with the addition of starting with a sufficiently large (but not infinite) combination set of purchased lines on the 1st round it should be possible to raise the probability to 1:1 after a infinite or perhaps less than infinite number of rounds. Which certainly is better than ~40M:1 Vs Infinite rounds
I only have two balls to choose from
The conclusion in the end is genius!
I never bothered myself to waste a thought about the lottery, but this is actually quite interesting.
I'd love to see some lottery with WAY more convoluted rule sets, just for the fun of working out how unlikely it is to win!
Maybe there should be a mathematician lottery where you have to predict the roots of a random-coefficient polynomial!
+standupmaths they stopped the ability to like replies, so I considered reporting this for awesomeness. sadly I can't choose that option.
Dan Dart
Of course you can like replies
I apologize if this has already been discussed, but in the calculation of the total probability, shouldn't there be another factor of (1-P(J)) for each additional ticket since you could win another ticket only if you didn't win the jackpot in the previous round? For example, the probability for winning by the second ticket would be P(J) [won on first ticket] + (1-P(J))*P(T)*P(J) [didn't win on first, won a second ticket, and that ticket won]; by the third ticket would be P(J) + (1-P(J))*P(T)*P(J) + (1-P(J))^2*P(T)^2*P(J) and so on. So the common factor between terms in the sequence is (1-P(J))*P(T), not just P(T). This probably wouldn't affect the actual calculation much because P(J) is so small and (1-P(J)) is so close to 1, but it might answer Jerry Johns question of possible probability of winning being > 1 if P(J) & P(T) are large. With the additional factor, the total probability would be P(J)/(1-((1-P(J))*P(T))) = P(J)/(1-P(T)+P(T)*P(J)) < P(J)/(1-P(T)) [made the denominator smaller]
I love the significance of your numbers.
@5:12
My calculation of the probability of winning the jackpot betting one ticket is
P(J)/(1-(~P(J) and ~P(5noB) and ~P(5B) and ~P(4) and ~P(3) and P(2)))=1/40712796.920761723828
I am assuming that if you win one of the raffle prizes and matches 2, you receive another ticket.
Of course you must have infinite time.
Not necessarily. There's an infinite number of possibilities that result in a jackpot in finite amounts of time. You don't "must have" infinite time, but of course having it wouldn't hurt your odds any.
+Omar Z Infinite time, even at minimum wage would be infinite money. Since most salaries are above this anyway there will be even bigger infinities! Time is money after all so forget the lottery.
+Andrew Crawford He might mean that infinite time is needed to literally get infinitely many tickets.
NoriMori yes
+Omar Z haha! Plus quite a lot of time to check them to see if you've won, cos they won't pay unless you claim.
No need for an infinite series.
The probability p of winning the jackpot from any one ticket is equal to:
p = p_J + p * p_T
where p_J is the probability of having 6 matches and thereby directly winning the lottery, and p_T being the probability of winning another ticket.
Solving for p yields the following equation:
p = p_J / (1 - p_T)
Plugging in 2.2 * 10^(-8) and 0.0975 for p_J and p_T respectively results in a probability of 1 in 40,665,099, which matches Matt's answer.
That's the same formula as to solve the geometric series, but I'd like to see how you got that algebraically.
I think I found a strange problem with this. As Matt stated, the probability of winning the lottery appears to be P(J) + P(T)P(J) + P(T)^2*P(J) + P(T)^3*P(J)...., however, imagine that P(T)=1. Now the probability of winning is P(J)+P(J)+P(J)+P(J)..., which is divergent, implying you have a greater than 100% chance of winning the lottery! That issue, however, can be resolved by saying there are a finite number of lottery combinations, and after the number of tickets exceeds the number of possible combinations, you are guaranteed to have won the lottery.
But, there's still a problem.
You can also imagine that P(T) is some number such that the geometric series with ratio P(T) and initial value P(T) sums to a number N such that N=1/P(J). To be precise, for that specific value of N, P(T)=1/(1+P(J)). Since the probability of winning the lottery with any number N is N*P(J), since the P(J) can be factored out of the series, this implies that the probability of winning the lottery is 100%. Surely, this is impossible, since you can not guarantee that you will win any extra tickets!
Likewise, you can imagine a P(T) such that N>1/P(J), which would mean the sequence sums to 100% in a finite (albeit, massive) number of terms.
Forgive me if I made any mistakes. Happy mathing!
hey, man, great videos, keep up the work; here's a question: what are the chance of winning the lottery in romania considerring the rules are the same as they were in the uk with the 49 numbers, EXCEPT you can put 3 sequences on the same ticket?
+radu nicolae The chance of one of your sequences of 6 numbers matching all 6 (out of 49) winning numbers is 1 in 13983816. So if your three sequences are all different, divide that by 3 to get 1 in 4661272.
Matt, do you plan on more regular uploads, or are you going to stick with the numberphile format and upload videos about cool little stuff like that?
I plan to eventually go for regular uploads, it's just really hard to find enough time. But if I can: I will.
The title of the videos is misleading!
But great video! ;)
+Gustavo Amarante Well in theory u can buy a set amount of lottery tickets which guarantee u to always have 2 right. Dont know the math excatly maybe someone will give it a try ;)
+Peter Scheringer yeah, but the amount of the tickets you have quickly declines. say you beginn with 1,000,000 tickets, after 6 or so rounds you are left with only a hand full of tickets. and no jackpot. probably.
Oh right wrong thinking i guess.
no problem
Brilliant like every time ! I hope i have a professor like you in the near future.
Study maths! But remember that whoever the professor: it's the maths which is the star.
if you got 5 numbers correct this would have been known as the parker ticket.
And he gets £1000, and by extension he can obtain 500 lottery tickets.
na, he would have to have all six, but off by one
In my state, the lottery funds a public university scholarship. It's called a "lottery scholarship," but the recipients are chosen by a combination of merit and need. I won $22,500 in my lottery scholarship. I would call that winning the lottery, and I've never bought a ticket in my life.
So I don't know if the UK changed it against since this video was published, but I am curious how the odds of winning the jackpot would be affected if you said 'Any winnings that are not the jackpot are going directly towards buying more tickets'.
More importantly is there some Self Sustaining Number of tickets you might be able to purchase at the start. That is for instance, say you started with one ticket, he says that you have about a 9% chance of getting a free ticket, but there is also those chances you win some money that isn't the jackpot. All that money you then win then goes towards more tickets. Of course if your 1 single ticket doesn't win your system is dead.
But if you purchase 2 tickets at the start your chances are now a touch better that you win some money to buy more tickets. You might go down to only having 1 ticket. You might go up to having a whole bunch of tickets.
So obviously no matter how many tickets you buy (assuming you don't just purchase every single combination of numbers) there is the off chance that they all loose. Or even if they don't all loose enough of them might not win that your number of tickets is significantly reduced.
So like there being no such thing as a Perpetual motion machine, there truly would be no Self Sustaining Number of Tickets. But we could perhaps at least figure out an expected life span of a system of self buying lottery tickets. Say we were just doing this for fun with no real expectation of actually winning the jackpot. How many tickets would we need to initially buy such that assuming the lottery rules never change again, that our self buying lottery tickets would have an expected life span of an actual person?
is this done by permutations or combinations? is 123 different from 321 etc?
When you compute P(T) are you computing the probability of ONLY matching 2 of the 6 numbers?
Newsagents hate him :D
+standupmaths what's the song you used? and great videos!! I really like them
The positive is that if you happened to have £91 million, you could buy every single combination, be guaranteed to win the jackpot, and will get millions of free tickets while still getting at least most of your money back if not making a profit.
Of course, if you have £91 million going spare, you don't really need to win the lottery, because you already have more money than a single person can reasonably spend in a lifetime (assuming you don't spend it all on hookers and blow).
How do you spend 91 million ANYTHING on hookers?
manudude02 People have actually tried to buy out lotteries before, but it required large companies.
Does this change in the lottery have any effect on the expected value?
That ending was rather Vsauce-esque. Great video!
I noticed that all the numbers on your ticket are Prime. That made me smile. Have you seen Harry Baker's love poem about Prime numbers? It's called 59. The Ted talk in which he delivers it is on RUclips.
Matt! Thanks for the new video first of all. PLEASE keep more videos coming! I would love to hear some proper new maths jokes, my friends aren't annoyed enough of the old ones anymore! :) seriously, those jokes as well as your terrific stand up and the rest of the videos are beyond awesome. PS: Greetings from Germany && i'm gonna be a teacher! Crazy, right? :D
What are the new chances of getting 5 to match? Or 4? Or 3?
I have a question about the 9.748% chance of winning a new ticket.
Does your calculation use "2 correct numbers exclusively" or "at least 2 correct numbers"?
Sadly, my chances are even worse. I live in Germany, where there is an even meaner system.
You choose 6 in 49, but also a so called "supernumber" between 0 and 9.
So your chance to get the jackpot is (49! / (6! * (49-6)!)) * 10. And that is 1 in 139,838,160...
2 correct exclusively you win money for 3 or more and nothing for 1 or 0 so technically speaking you don't win a new ticket for anything but 2 (though with 3 or more you can buy one but that isn't the point).
What edge does the lottery have on you? For example in blackjack the house has like a 1% better chance of winning. You have 49.5% they have 50.5% (that's if you play perfectly)
Barney Laurance in swizerland, the excess money that isnt giing to be paid out gies to a fonds to support arts installations, start ups and other cool shit. How is it handled in th uk?
question: getting 3, 4, 5 or 6 numbers correctly, implies that you also get 2 numbers correctly. Will you be eligible to get a free ticket then as well?
If you buy a lottery ticket every week then your expected return remains the same: £1, because they cost £2 and 50% of the ticket price goes into the prize fund, same as before.
If, on the other hand, your strategy is to buy tickets only on the 5% (e.g.) of weeks which have the biggest jackpots, then your expected returns have gone up: as you mention, rollovers are now more common so the biggest jackpots will be bigger than before.
Well this is quite late, but you don't actually have to sum the geometric series: let's call the probability of winning the lottery (with any number of free tickets in between) W, probability of winning with any single ticket X and probability of winning another ticket Y. Then with some simple conditional probability we see that W must satisfy the equation
W = X + Y * W
Hence W = X / (1 - Y)
Chance of winning anything -> 1/3423
Chance of winning a new ticket -> 1/3481
And I thought Slot Machines were cruel.
Only prime numbers, ofcourse... Any reason to choose those?
of course not
Because prime numbers, duh...
I walked in to the newsagent and when I realised I could choose my own numbers I panicked. Primes are always my go-to. But I'll be super-annoyed if a different set of primes p ≤ 59 win.
+standupmaths You should choose randomly. Other people presumably like prime numbers, so if you win the jackpot with prime numbers, you're more likely to have to share it. If the numbers ever come out as 1,2,3,4,5,6 then there are going to be a lot of disappointed "jackpot" winners.
hey Matt , what's the music that you use in the background?? thanks!
That's a song called "Stand-up Maths Theme". A friend of mine composes all the music for my videos. This guy: tardis.wikia.com/wiki/Howard_Carter_(composer)
I loved that ending. Very nice video, Matt. Thanks a lot!
Thanks! I enjoyed the ending myself, more than I probably should.
Almost finished with your book!
I have been thinking recently, if you spent all your winnings (from 3 balls to 5 ball matches) on more lottery tickets, what are your chances now of winning the jackpot?
I forget how long ago it was, so it might've been before you left Australia, but we changed one of our lotto a couple years ago, too. Not to increase the range of numbers to choose from (it's still 1 to 45) but to increase the number of choices you make (pick 7 numbers, not 6) - but again, the marketing says "More prizes now!"
Also, the sum of the geometric series you mentioned, after taking P(J) common is evidently greater than 1. That didn't immediately bother me, but then I thought about it a little bit more. It's only really not an issue because P(J) itself is already so small that it overshadows the magnification of the sum of the series (let's call the sum S). What happens if you have a lottery tickets for a much more likely to win lottery such that S>(P(J))-¹. Then it would appear that the likelihood of winning this kind of roll over lottery is greater than 1. Which doesn't compute. Or does it? No, it can't. Or can it? Maybe if... No. PLEASE HELP.
+Jerry T. John P(J) factors out so it's P(J) * (1+P(T)+P(T)^2+...) and because P(T)
Henri Williams I know. But the sum is greater than 1. That should ring alarm bells when you talk about probability. In this case, the sum of the series is 1.1083... If P(J) was a likely event that had a probability of say 0.95, then the result has more than a 100% chance of occurring. My question was, what part of the generalised approach to the problem fundamentally limits P(J) so that the product stays less than 1 when multiplied with the sum of the series.
Ok, let P(T) = x. P(J)+P(T)
+Henri Williams I had to read it twice, but it did help. Thanks!
Hey Matt when will you be coming to NYC again ?
No plans at the moment but I'll probably be around NYC some time in 2016. Watch this space!
Part two: How to fit infinitely many lottery tickets into finite size universe.
The following are the probabilities of winning 1st Division (the top prize) in the main Australian lotteries: Saturday, Monday and Wednesday Lotto (6 from 45) 1 in 8,145,060; Oz Lotto (7 from 45) 1 in 45,379,620; Powerball (6 from 40; 1 from 20) 1 in 76,767,600.
what value N does P(T)^N
if n refers to number of tickets pretty sure n is infinite as there is no limit to the number of tickets you can win the chance of that happening or making a difference gets small fast so infinite tickets doesn't equal infinite chance.
What is the expected value of before vs now?
So if you were to buy a single lottery ticket and spend half of your winnings buying more lottery tickets when you match 3-5 numbers, what is the probability that you will eventually win the jackpot?
October 10th!!!
So how many tickets would you need to win back in this series before the odds of getting another ticket are greater then that of winning the jackpot? (like 10%, then 1%, then 0.1% ect.)
+Sean Bevan Quite straightforward calculation with logarithm. Just take 10-base logarithm on both halves from inequality 1/45000000 > 10^(-x) which results into log(4.5*10-7) > -x x > 7.65 and since X is integer and probability only gets lower at half point between 7 and 8 so therefore 8 is the earliest point at when it's more probable that you would have won the lottery than getting another new ticket.
What does 3.2 times less likely mean?
0.5 times less likely seems clear enough; the probability of winning declines by 50%. 0.75 less likely leaves you with only 25% of the original probability to win. Continuing in this fashion, 1.0 times less likely reduces your probability of winning to zero. 3.2 times less likely means that the probability is -2.2 of the original which makes no sense at all. Please explain Matt.
X times less = amount/X has been a common expression for a long time.
@@kleine.kleeblatt Perhaps so, but I expect better from a mathematician,
I want to know now if you can buy tickets with the lower winnings of 1000 and 100 what are the real odds adjusted for those assuming you used ALL THE MONEY to buy just lottery tickets and stopped when you hit the jackpot.
you have to calculate the fact that you'll be dead before you get your infinite amount of tickets!
This video is also pretty great for explaining infinitesimals!Good job.
How would the maths change if you included the chance of choosing three numbers, and buying a bunch of lottery tickets. Or choosing four or five numbers and getting more free tickets off all those free tickets.
Does it make an impact to the 1 in 40 million chance? Or is it negligible?
I've not checked the numbers, but I suspect the probability of three or more matches is so small that despite the multiple tickets you could buy, it will taper off crazy-quick. Will report back with actual results.
what if there are 1 numbers they chose, and 2 to chose from, and then did you win or not?
What's the outro song?
so, P(J) *(integral of (P(T)^x) dx from 0 to infinite).
That's the idea, yes. In this case you only want whole-number values of x so an integral is not quite what you want.
Check this out:
mathworld.wolfram.com/GeometricSeries.html
That's Parker Square of a selection of numbers.
Nice prime numbers :)
If you were to buy a single ticket for each draw for a year (and if you matched two previously that would count as an extra ticket), how much worse off would you be statiatically and how different is this from the old lottery?
That is a really good question! I've not checked the numbers but my guess is that if will be exactly the same: £104 spent, £0 won.
Also, the lottery have kept their ratio of winnings the same, so you still get back in the very long-term around 50% of what you spend.
I'm getting 1 in 40,665,051 instead of 1 in 40,665,099... Which is the right number?
EDIT: never mind, I used the rounded value for P(T). Using the exact value (not gonna tell it here - spoilers!) I did get the right answer.
Rounding! Gets you every time. (Congrats on fixing it and getting the correct result.)
I love it! Can't wait to see the new show in York on Saturday. I made a present for you guys! 😆
How do you calculate the probillity ?
I like how you chose my favourite primes :)
So, is the integral of the probability of winning * dream of winning = constant?
there is a nice segue from this video to talking about your probablity of dying, as in actuarial tables, since you probably wont be around to play forever
0:41 BWAHAHAAAHAAA. Perfect.
So now, did anyone calculate the expected win for the average money distribution, before and after?
Also, the chance of winning alone (given that you won) increases.
I could see you trying your hardest not to make a clickbait joke
if you could play lotto or anti-lotto (where you get 10 $ with a high proberbility or you loose some Millons with a low proberbility) what would you do, and how often?
You said 3.2 times less. Times means more - at least for positive numbers.
*Every drug store fears him...*
ᴊᴜsᴛ ᴡᴀᴛᴄʜ ᴛʜɪs ᴠɪᴅᴇᴏ ᴛᴏ sᴇᴇ ᴡʜʏ!
how many free tickets for it to be reasonaly expected to win ?
A lot
P(T)/(1-P(T))
Glad I could help :)
What happens if you get more than 1 ticket that match exactly 2 numbers? Now that case you will definitely get more than 1 ticket and thus increasing the chance of the jackpot.
In Italy you can choose between 99 numbers...
+standupmaths do you still get your free ticket if you get 3 numbers correct? did you account for that?
you don't you get (according to the video) about 25 pounds. you only get the free ticket for 2 correct numbers 3 and up give cash prizes.
How do you calculate the geometric series?
+JaredTheTwitch For a series of the form r^n * a, where r is a ratio (in this case the chance of winning another ticket, and a is a constant (in this case the chance of winning the jackpot) and n is an integer starting from 0, then, as long as r is between -1 and 1 (it is in this case), it will converge to a / (1 - r) which results in the value matt showed us.
I love that you had chosen all primes.
+TheStole That is a poor strategy because it will result in more sharing of the jackpot, if I understand the rules correctly. Prime numbers are likely to be relatively popular, particularly the last one which surely will appear on a great many tickets.
What you want are sequences of _unpopular_ numbers. While no more likely to be drawn than any other, they are likely to provide a greater return to the winners if chosen. I can only conclude Matt bought this strictly for making the video. I wonder why no one seems to prepare sequences of unpopular numbers for use in lotteries. Cheers. :)
They're my go-to numbers!
After the drawring is done on the telly, do let us know which numbers turned up so we can work out your prizing, Matt!
Hold on a moment. There is a limit to how many you can buy. You can only buy one for each combination. You cannot buy infinitely many lottery tickets because there is only one of each combo. You could buy the same ticket twice but it will not affect your chances. I am not sure I understand the rules of the lottery, but I am about positive that if there is any chance of winning, then there must be a finite number of possibilities, hence you cannot have that chance because you cannot buy infinite lottery tickets. I am sure that it isn't that much of a problem, but your calculations seem to be a bit off.
you can get infinitely many because you if you get 2 correct you get a ticket for NEXT WEEKS lottery not the week you bought the ticket. if you got a free one for the week you bought a ticket you could just give it the winning numbers (which you have to know given you got this ticket after the drawing) and win money immediately. there is no limit to how many you can get with the method discussed in the video because even if you pick the same numbers every week you still have a chance to win because every drawing uses new numbers.
as long as the jackpot is more than 2.9 times bigger then isnt it technically better value?
59 numbers? why not 60?
I can foresee Disney suing your lottery for the logo once they notice. Unless they already settled.
Can you please make a sequel of the video? You could go a bit deeper on how to calculate these odds, what would be the expected value of a single play, if are there buying strategies to get a better expected value, how often should there be a winner, and so on. Maybe a take on the "St. Petersburg Paradox".
Whoops! You already did something! Ignore comment and thanks!
So if you converted any of the other prizes into more tickets this would also increase the chance of eventually winning the jackpot from one original ticket.
Lol I just read the comments people have already considered this.
I occasionally play the California Lotto. When I tell people I do I sometimes get lectured about throwing my money away and especially, what a "bad investment" it is. And what I say to them is, "The investment money is taken care of, this was purchased with entertainment money. You go to the movies, I buy a Lotto ticket. Who has the better chance of being rich? That's pretty entertaining to me."
This video really makes me curious about the nature of probabilities when the sample size is infinitely large.
All your lottery numbers are prime :O
I think that in general, you should buy a ticket if the prize for winning is greater than the odds of not winning.
And, as my mom used to say, it's cheep hope.