1. Comparing UK National Lottery Odds To Hours In A Lifetime.

Comparing UK National Lottery Odds To Hours In A Lifetime.

I was explaining to a not very mathematically minded friend the other day the tremendous unlikelihood of winning the British National Lottery. It occurred to me afterwards that a good way of explaining this would be to equate the odds of winning to the number of hours in a lifetime.

This is what I came up with, there are a couple of questions towards the bottom concerning a bit of help I need finishing it off - it's many years since my Computer Science degree and we only touched on probability at degree level, so I'm a bit rusty.

UK Life Expectancy:

Years:
Male: 76.9
Female: 81.3
Mean Avg.: 79.1

Days: 365.25 * 79.1 = 28,891.275
Hours: 28,891.275 * 24 = 693,390.6

Odds of winning British National Lottery:

To win the main jackpot 6 unique numbers in the range 1..49 must be chosen. 6 numbers and a bonus number are drawn by the lottery organizers. A bonus number is also chosen by players but this is used only for lower value wins and is irrelevant to the main jackpot. [Note: The bonus number is chosen after the main jackpot numbers and so does not interfere with the 6 main jackpot numbers drawn.]

Odds of choosing all 6 jackpot numbers in the order drawn:
49 * 48 * 47 * 46 * 45 * 44 = 10,068,347,520 to 1

Possible orders of the 6 drawn numbers:
6 * 5 * 4 * 3 * 2 * 1 = 720

Odds of choosing the 6 jackpot numbers in any order and so winning the main jackpot:
10,068,347,520 / 720 = 13,983,816 to 1
(Correct - as advertised on the National Lottery website).

To have a 50/50 chance of winning the main jackpot, play this
many times:
13,983,816 / 2 = 6,991,908
Is that right?

The average British person currently lives for 693,390.6 hours, so in ten average British lifetimes you live for this number of hours:
693,390.6 * 10 = 6,933,906

6,933,906 approx. equal to 6,991,908
In words: To give yourself an approx. 50/50 chance of winning the main jackpot of the British National Lottery you would need to play the lottery every single hour of your life for 10 lifetimes. Is that right?

If you were to play the lottery the same number of times as there are possible number choices, 13,983,816, it would take you 20.167 lifetimes of playing every hour.
13,983,816 / 693,390.6 = 20.167 lifetimes

Plainly if you played the lottery every hour of your life for 20.167 lifetimes you would not have a 1.0 probability of winning, but presumably your odds of winning would be quite high. I have no idea how to calculate what the probability of winning would be if you did this. Maybe someone could explain how this is calculated to me please.

Many thanks.

2. Let's say the probability of winning is $p$ (so $p={1\over 13983816}$). The probability to win (at least) once over $n$ attemps is equal to the probability not to lose every of the $n$ attemps.

Losing $n$ times in a row has probability $(1-p)^n$, hence the probability to win at least once in $n$ attemps is $1-(1-p)^n$.

For instance, if you want to make this probability larger than 50%, you need to have $n\geq-\frac{\log(2)}{\log(1-p)}$. If $p$ is small, this is very close to $\frac{\log 2}{p}$ (and not $\frac{1}{2p}$ as you wrote). In your case, instead of 13,983,816 / 2 = 6,991,908, it is rather 9,692,843 times someone needs to play lottery to have 1 chance over 2 to be a winner. Nearly 14 lifetimes are necessary for that.

About your last question, someone who (foolishly) would try every combination would have a probability of $1-(1-p)^{13983816}$ , that is 63%, to have won at least once.

This comparaison with lifetime is quite convincing, by the way...

Laurent.

3. Originally Posted by Laurent

About your last question, someone who (foolishly) would try every combination would have a probability of $1-(1-p)^{13983816}$ , that is 63%, to have won at least once.
No, if you buy a ticket for every possible combination of results your probability of winning is 1 (they are not independent wagers)

RonL

4. Originally Posted by CaptainBlack
No, if you buy a ticket for every possible combination of results your probability of winning is 1 (they are not independent wagers)

RonL
Well, probably I wasn't clear enough, but I was answering mattstan's question (last paragraph of his post): the 13,xxx,xxx tickets correspond each to different (supposedly independent) lottery results.

5. Laurent - Thanks for your informative answer. I'm still a little unclear, can I trespass on your patience and ask for some clarification?

You wrote:

Originally Posted by Laurent
For instance, if you want to make this probability larger than 50%, you need to have $n\geq-\frac{\log(2)}{\log(1-p)}$. If $p$ is small, this is very close to $\frac{\log 2}{p}$ (and not $\frac{1}{2p}$ as you wrote). In your case, instead of 13,983,816 / 2 = 6,991,908, it is rather 9,692,843 times someone needs to play lottery to have 1 chance over 2 to be a winner.
Firstly why $\frac{\log 2}{p}$? Does the log 2 relate in some way to 50%, what would it be if I was looking for 55%?

Secondly you say the value of this is 9,692,843. I got a very different result:

0.301029 / (1 / 13983816) = 4209534.146664

0.301029 being log 2 or am I being stupid? Apologies as I said before it's been quite a while since I did any formal maths during my CSAI degree.

So if you play the lottery on 13,983,816 consequitive (weekly) occasions (you'll be 268,919 years old) the odds of winning at least once would be 63% that's still not very good. Would you mind talking me through how to work out how many times you must play to have a 99% chance of winning at least once? You see I can't work it out as I don't understand how the log 2 relates to the 50% chance in your explaination.

Many thanks again. Regards, mattstan

6. Originally Posted by mattstan

Firstly why $\frac{\log 2}{p}$? Does the log 2 relate in some way to 50%, what would it be if I was looking for 55%?
How I got the log 2: I wanted to find $n$ such that $1-(1-p)^n\geq 0.5$ (0.5 being the 50%), or $(1-p)^n\leq 1-0.5=0.5$. Taking logarithms on both sides, this becomes: $n\log(1-p)\leq \log(0.5)$, hence $n\geq\frac{\log(0.5)}{\log(1-p)}$ (the inequality changes from $\geq$ to $\leq$ because $\log(1-p)<0$). And I used the fact that $\log(0.5)=\log(1/2)=-\log(2)$.

Then, if $p$ is small (remember that $p$ is the probability of a jackpot, so it is ${1\over 13,983,816}$), $\log(1-p)\simeq -p$ provided we use natural logarithm (in base $e$), which I could also have written $\ln$ (like on calculators, for instance). Using this approximation, we obtain $\frac{\log 2}{p}$ where $\log 2\simeq 0.693$. The computation therefore is: 0.693 / (1/13,983,816) = 0.693 x 13,983,816.

Instead of 50%, if one wants a 99% chance of winning, then one needs to have $n\geq \frac{\log(1-0.99)}{\log(1-p)}\simeq 64397849$.

I hope this helps. And don't feel stupid asking questions !

Laurent.

7. Many thanks for the further explaination.

RE: $\log 2\simeq 0.693$

When I put log 2 into Windows Calc and into PowerToy Calc, I get the same result, that being: 0.301029

You get 0.693!! That's why I am feeling stupid. Since I'm not a moron, and I've tried several times maybe you can explain what's happening. After all if I can't get that right I can't play around with what you've shown me.

Thanks again.

8. That's the point of what I wrote about "natural logarithm". The "log" on calculators (or in various contexts) is the log in base 10, whereas the log I used is the log in base e=2.718..., which is denoted by "ln" on calculators.

The notation "log", while ambiguous, is still often used in maths for "ln" (the logarithm in base 10 almost never occurs in maths).

By the way, this choice of logarithm doesn't matter in the formula $
n\geq \frac{\log(1-0.99)}{\log(1-p)}\simeq 64397849
$
. You can use the "log" of your calculator here. (or with 0.5 instead of 0.99 of course)

Laurent.

9. Laurent,

Many thanks once again and for all your help, I've got it (finally).

It's been an interesting academic exercise for me, not to mention pleasant work diversion.

Cheers.