# The national lottery!

1. Aug 15, 2013

### MathJakob

For those of you in the UK you'll know the lotto consists of 6 balls and 1 bonus ball, to win the jackpot you need only the 6 balls. The jackpot typically ranges from 3m to 6m. The chances of winning the jackpot. $\frac{49!}{(6!(49-6)!)} = 1$ in $13,983,816$

The thing that bugs me is even with those odds being as low as they are, people win the jackpot almost every week, sometimes multiple people. I want to play the lottery as it's going to be my only ever chance to get rich haha but I just can't help thinking the odds are too low.

Other people may say well someone has to win, and someone does win, that could easily be you. The thing is I've run a program on a website before that picks numbers at random and you just leave it running until it picks the exact same 6 numbers that you picked. I've left it running for hours minimized while playing a game or something and after coming back to it an couple hours later it's played over 5million tickets, I've spent over 5million pound, won about 80,000 back and I because I selected 2 tickets per week that means I've been playing for 2.5million weeks or $\approx48,000$ years!

I know 1 in 14million means for every 14million tickets, chances are 1 will be a winner, but it's strange how odds on this scale still follow that rule. I remember reading an article that said even though some things have odds and they are possible, they may well be classed as impossible, in other words it will never happen no matter how much time you have, unless you have infinite time.

For example, it is possible for a coin to land on heads 1million times in a row, but the chances are so small that it's safe to say it will NEVER happen, even within $googolplex^{googolplex^{googolplex}}$ years... it is only likely to happen with infinite time, even though it's perfectly possible.

Bottom line is do I play the lottery or not? :P it's only £1 a ticket and someone does win! I just can't help to think it will never be me haha

2. Aug 15, 2013

### Jorriss

What kind of answer do you want?

You probably won't win, but it's kind of fun regardless.

3. Aug 15, 2013

### MathJakob

Well I'd like an opinion on the part I mentioned about something never happening no matter how much time you have, unless infinite, even though it is totally possible.

and I kinda wanna hear from people with some wisdom on what I should do :P

4. Aug 15, 2013

### leroyjenkens

Every winner has said the same thing.

5. Aug 15, 2013

### lisab

Staff Emeritus
I live in Washington State, which has a state lotto with a brilliant slogan designed to assuage the doubts of even the most mathematically inclined: "You can't win if you don't play!"

Never, ever play more money than you can afford to burn. Seriously, I mean burn. Imagine yourself burning a pound note to ashes. How does that make you feel? How about £5? £10? (Btw I get queasy after a couple dollars!)

When the jackpot here gets large, I go in with some coworkers and buy a few tickets, just $2 each to buy in. The reason I do this is purely social: we enjoy imagining what we'd do with that much money. 6. Aug 16, 2013 ### Greg Bernhardt ### Staff: Admin I play scratch offs a few times a year for almost a decade and still haven't won anything greater than$5.

7. Aug 16, 2013

Staff Emeritus
To six decimal places, your odds of winning are the same whether or not you buy a ticket.

8. Aug 16, 2013

### HallsofIvy

More importantly, so has every loser!

(And I suspect that was what you meant to say.)

9. Aug 17, 2013

### StevieTNZ

I remember checking my lotto ticket back in 2006. I checked it against the wrong numbers, and had those numbers been drawn I would have won fourth division - about $26. So I checked the ticket using the -correct numbers- and had matched again, winning fourth division. I go to cash it, and it turns out I had actually missed a number and had won$560ish.

10. Aug 17, 2013

### StevieTNZ

With that, you could buy a ticket with every combination, praying your the only winner.

11. Aug 17, 2013

### Devils

Stanford statistics PhD. wins lottery 4 times, won't reveal her secret.

http://www.forbes.com/sites/kiriblakeley/2011/07/21/meet-the-luckiest-woman-in-the-world/

-------------------
First, she won $5.4 million; then a decade later, she won$2 million; then two years later $3 million; and finally, in the spring of 2008, she hit a$10 million jackpot.
-------------
Before you pack up and move to Bishop to camp outside of Ginther’s golden mini-mart, there’s more. Lots more. Would anyone be surprised if I now told you that Ginther also happens to be a former math professor with a Ph. D. from Stanford University who just happened to specialize in—arts and crafts?

12. Aug 17, 2013

### MathJakob

There is no formula for determining the numbers, there simply is nothing you can do that can increase your odds. You can use probability to your advantage though. Say the number 8 ball hasn't made an appearence in over 100 games, it was due 49 games ago and so you'd expect the 8 to make an appereance very soon. Although you simply can't know.

This doesn't work. For a start the jackpot is only 2million - 6million and secondly I doubt you could buy every combination, simply because if you do the lucky dip, the machine picks the numbers for you, if you wrote every combination and took it to a shop, you'd be there for days printing off tickets, and you'd also need the cash ect.

It just isn't possible :P

13. Aug 17, 2013

### dilletante

You can't expect the 8 to show up because it is "due", that is a fallacy at best. It certainly isn't using probability to your advantage. Assuming the game is honest, meaning all balls are weighted the same and have an equal chance of coming up, the 8 has the same chance of coming up that it did in the last 100 games, whether it came up in any of those games or not.

I can think of no logic for expecting a greater chance than random of the 8 coming up due to the fact that it hasn't been seen in a while. The opposite is more likely, that there may be some physical reason it hasn't shown, such as it may be slightly larger or smaller than the other balls, or weigh more. So given the situation, I would be more inclined to bet against it showing up the next game rather than picking it as one of my numbers.

14. Aug 17, 2013

### MathJakob

That makes no sense though, the entire point of probability is knowing the frequency of which something "should" occur.

So if you have a die and roll it 6 times, you'd expect the number 5 to appear once. Roll the die 12 times and you'd expect the number 5 to appear twice ect. So we can say that the chances of the number 5 appearing is 1in6.

If 50 rolls have gone by and the 6 still hasn't shown then it's well over due and based on probability, it "should" occur any moment. Of course with probability it's just an average and you can't ever know for sure when a certain number will or will not appear.

15. Aug 17, 2013

### Jorriss

This is a very out dated view of probability.

If you toss a coin 10000 times and get 10000 heads on an unbiased coin, the probability of getting tails and heads respectively the next flip is still 50-50. Each flip is an independent event and is not influenced by prior coin flips. The same reasoning applies to the lottery.

16. Aug 17, 2013

### MathJakob

Yeh I understand that fully. But isn't there some sort of law of averages or something?

17. Aug 17, 2013

### Jorriss

Yes, (I assume you are referring to the law's of large numbers) but these laws refers to limiting outcomes.

18. Aug 17, 2013

### Aero51

If you have a bunch of lotto balls bouncing around in a container, couldn't you use the discreetized boltzmann statistics to determine the probability that a given ball will be selected? Then again that would require a lot of info about the ball container. Or if it was one of those suction kinds you could run a CFD model!

19. Aug 17, 2013

### micromass

Sure. If you have $n$ balls in a container then your probability of selecting a given ball is $1/n$.

20. Aug 17, 2013

### Aero51

I may have misspoke. I was referencing the Boltzmann velocity distribution. In other words, we can say that each ball that enters that "catching" chamber must satisfy a a certain transnational velocity criterion.

For example, well I cant post the example because apparently I cannot access LATEX.
See wikipedia: http://en.wikipedia.org/wiki/Maxwell–Boltzmann_statistics

where the energy in the exponent represents the root of the sum of the x,y,z kinetic energies
Obviously we would need some analogous variables for mu, T,and k, but the fundamental concept is the same.

Once the probability of the acceptable velocity components are obtained, we can then apply Bayesian statistics weighted by the position, IE the probability of observing [vx,vy,vz] given [x,y,z].If we assume they are independent we simply just multiply the two probabilities.

And that should be the likelihood a number is selected!

21. Aug 17, 2013

### micromass

But I already know the likelihood a number is selected. It's obvious that it's $1/n$. I see no single physical or mathematical reason why it should be different.

22. Aug 17, 2013

### Jorriss

How does this situation follow a maxwell-boltzmann distribution at all? A maxwell-boltzmann distribution gives the velocity distribution of thermally equilibrated particles, which is not the system we have here at all.

I dont even see in your analysis where the numbers on the balls come up.

23. Aug 18, 2013

### Aero51

I have a small proof but for some reason I cannot access to latex when posting a thread if someone can help me with that I will gladly post one

24. Aug 18, 2013

### Jorriss

Does your proof give the final result that the probability of selecting any given ball is 1/n? If not, it's wrong.

25. Aug 18, 2013

### Aero51

It does for a special case