# Should I play the lottery or not?

• MathJakob
In summary, the UK lotto consists of 6 balls and 1 bonus ball, with a jackpot ranging from 3m to 6m. The chances of winning the jackpot are 1 in 13,983,816. Despite the low odds, people still win almost every week. Some may argue that someone has to win eventually, but even with infinite time, some things may still be considered impossible. Ultimately, the decision to play the lottery is a personal one, with the advice to never spend more money than you can afford to lose. There are also stories of people who have won multiple times, but there is no proven formula for winning.
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

What kind of answer do you want?

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

Jorriss said:
What kind of answer do you want?

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

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 want to hear from people with some wisdom on what I should do :P

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
Every winner has said the same thing.

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. I play scratch offs a few times a year for almost a decade and still haven't won anything greater than$5.

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

it's only £1 a ticket and someone does win! I just can't help to think it will never be me haha

leroyjenkens said:
Every winner has said the same thing.

More importantly, so has every loser!

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

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.

MathJakob said:
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##

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

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?

Devils said:
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?
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.

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

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

MathJakob said:
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.

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.

dilletante said:
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.

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.

MathJakob said:
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.
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.

Jorriss said:
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.

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

MathJakob said:
Yeh I understand that fully. But isn't there some sort of law of averages or something?
Yes, (I assume you are referring to the law's of large numbers) but these laws refers to limiting outcomes.

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!

Aero51 said:
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!

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

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 can't 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!

Aero51 said:
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 can't 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 sum of the squares of the translational velocities.
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!

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.

Aero51 said:
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 can't 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!
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 don't even see in your analysis where the numbers on the balls come up.

1 person
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

Aero51 said:
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
Does your proof give the final result that the probability of selecting any given ball is 1/n? If not, it's wrong.

It does for a special case

Aero51 said:
It does for a special case

It should give ##1/n## in all cases...

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

Of course it is.

What changes each week is the number of people playing and the jackpot. It would make more sense to play with a large jackpot, and this is where the opportunity to make money lies.

Well maybe if I show you, you can correct me where I am wrong. But if you prefer to be dismissive I won't waste my time.

Aero51 said:
Well maybe if I show you, you can correct me where I am wrong. But if you prefer to be dismissive I won't waste my time.

Go ahead and post it. Don't be afraid of some criticism!

I will when I get back from lunch with my friends

Here's my take on the problem. Lunch turned into dinner which turned into drinks which turn into forgetting.

Here is my proof. Its really not mine but is paraphrased from Hills "An introduction to statistical thermodynamics" We will first start out with the assumption that the lotto balls are bouncing around in a plenum with a single opening. A ball is "selected" when it is sucked out from this opening. Let's assume that each ball is its own system, denoted Bj, with a state Ej, corresponding to (in this case) the balls kinetic energy. We will not define the limits of the possible states, however we will say that any system can be in any available potential state. The number of systems with a state by nj.

It follows that the following conditions must be satisfied
$\sum n_{j} = s$ number of systems found in state E1...E2... etc [1]
$\sum n_{j} E_{j} = E_{t}$ total sypersystem energy [2]

The total number of states found in the supersystem, St is computed by the formula:

$S_t = s! / ((\prod_{j} n_{j})!)$ [3]

It follows that the probability of observing a a specific state:
$P_{j} = \overline{n_{j}}/S$ [4]
and therefore
$\overline{E}=\sum P_{j}E_{j}$ [5]

We want to calculate the most probable distribution, or the most likely observed n (n*) for our supersystem. To do this we take the natural log of the distribution St and utilize LeGrange multipliers to calculate the maximum. Although the number of balls B1, B2, ...Bn, is finite, it is safe to assume that their accessible (but not necessary probable) energy levels are close to ∞. This allows us to apply Sterlings Approximation, which yields the following equation constrained by the our first two equations:

$\partial / \partial (n_{j}) [Ln(S_{t}) - \alpha \sum n_{j} - \beta \sum n_{j} E_{j}] = 0$ [6]

which results in:
$n^{*}_j = S_t e^{-\alpha}e^{-\beta E_j}$ [7]

Substituting this equation into our first two leads to the following
$e^{\alpha} = \sum e^{-\beta E_j}$ [8]
$\overline{E}= \sum (E_j e^{-\beta E_j}) / \sum e^{-\beta E_j}$ [9]

This also leads to the following result

$P_j = n^*_j = e^{-\beta E(j)}/(\sum e^{-\beta E(j)})$ [10]

I probably did a bad job paraphrasing this derivation, but again its not mine. It is sound, but my explanation may not be. I could have just skipped to the probability function and written below to get my point across.

This where I come in and mess everything up:

Recalling that the energy of each ball is defined by the basic kinetic energy equation:
$E_j = 1/2 m V_{j}^2 = 1/2 m (V_x^2 +V_y^2+V_z^2)_{j}$ [11]
Which leads to the probability function given a velocity:
$P_{velocity} = e^{-\beta (1/2 m (V_x^2 +V_y^2+V_z^2))}/(\sum e^{-\beta ( 1/2 m (V_x^2 +V_y^2+V_z^2))_j})$ [12]

And recalling that the ball needs to be sucked into the tube to be "selected", leads to the notion that only a certain combination of Vx, Vy Vz for each ball will permit selection (IE a lotto ball velocity parallel to the orifice would not result in its selection). I do not want to sit down and figure this specific number out so I will not.
Further more, with these velocities comes a spatial constriction, (a location that the ball must be in order to be "selected, which is proportional to the ratio of the orifice and ball diameters).
The obvious conclusion is that each ball must have both a certain position and a certain velocity to be selected. Let Pposition denote the probability of being in a certain position.

$P_{velocity}\cap P_{position} = P_{velocity} P_{position}$ [13]
since we'll say velocity and position are independent of each other. This leads to the following expression for selection.

$P_{selection}= P_{velocity} P_{position} = P_{velocity} = e^{-\beta (1/2 m (V_x^2 +V_y^2+V_z^2))}/(\sum e^{-\beta ( 1/2 m (V_x^2 +V_y^2+V_z^2))_j}) P_{position}$ [14]

As was mentioned before there was actually a set of velocities and positions which had to be met in order for the ball to be "selected" This leads to the total probability of being selected:

$P_{total selection}= \sum_{vx,vy,vz}P_{velocity} \sum_{x,y,z}P_{position}$

Where the sums denote the permissible velocities and positions. I can't prove this isn't 1/n, but if you can prove it is show me :)

Jorriss said:
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.

Agreed, But why does playing six sequential numbers or playing the prior draw's winning numbers just seems like a stupid strategy.

What are the odds of flipping a coin and getting heads 10,000 times in a row? What is the effect on the odds of adding one more flip? and then infinitely?

nitsuj said:
Agreed, But why does playing six sequential numbers or playing the prior draw's winning numbers just seems like a stupid strategy.

Because intuition can easily lead you astray when it comes to statistics.

nitsuj said:
What are the odds of flipping a coin and getting heads 10,000 times in a row? What is the effect on the odds of adding one more flip? and then infinitely?

1) What are the odds of flipping a coin and getting heads 10k times in a row? 0.5^10k, or about 2 in 103010
2) What is the effect on the odds of adding one more flip? No impact whatsoever, if the coin and the procedure are fair. Even if the last 10,000 flips were heads, the next flip still has a 50% probability of coming up heads. The same is true of the one after that, and the one after that.

nitsuj said:
Agreed, But why does playing six sequential numbers or playing the prior draw's winning numbers just seems like a stupid strategy.

The fallacy is that you can easily "see" that winning with 6 sequential numbers, or last week's winning numbers, is "impossible", but you can't "see" that any ONE set of 6 randomly chosen numbers is equally "impossible". With hindsight, almost every set of winning numbers looks like SOME set of 6 randomly chosen numbers.

Another interesting "randomness fallacy" is to "guess" what are the odds of a random set of numbers containing at least one pair of consecutive numbers, and then calculate what those odds are. (You might have more than one pair, e.g. 12, 13, 46, 47, or a run of more than two consecutive numbers, e.g. 12, 13, 14.) if you don't want to do the calculation, estimate the probability from the published lists of past winning numbers after you have guessed the answer.

One way that will increase your odds is not to choose numbers that other people might choose (e.g. based on birthdays, etc) so you reduce your chance of sharing a prize if you do win.

FWIW I approve of lotteries. They are a neat idea for taxing stupid people without them realizing it

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AlephZero said:
The fallacy is that you can easily "see" that winning with 6 sequential numbers, or last week's winning numbers, is "impossible", but you can't "see" that any ONE set of 6 randomly chosen numbers is equally "impossible". With hindsight, almost every set of winning numbers looks like SOME set of 6 randomly chosen numbers.

Another interesting "randomness fallacy" is to "guess" what are the odds of a random set of numbers containing at least one pair of consecutive numbers, and then calculate what those odds are. (You might have more than one pair, e.g. 12, 13, 46, 47, or a run of more than two consecutive numbers, e.g. 12, 13, 14.) if you don't want to do the calculation, estimate the probability from the published lists of past winning numbers after you have guessed the answer.

One way that will increase your odds is not to choose numbers that other people might choose (e.g. based on birthdays, etc) so you reduce your chance of sharing a prize if you do win.

FWIW I approve of lotteries. They are a neat idea for taxing stupid people without them realizing it

I think the reason why people don't play the same lottery numbers after winning once, or getting 10 heads in a row and then betting on heads again is due to nothing more than psychology... It's the same reason nobody plays the numbers 1,2,3,4,5,6,7 just psychologically you think they're less likely to appear. After all isn't that why people who gamble don't pick black 18 every single time, they mix it up thinking they're spreding out their chances to win lol

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