Probability Calculation

[Barquentine]

I'm not sure if this question is in the right category but here goes:

Can anyone tell me how you calculate the probability of getting both lucky stars right in the Euromillions lottery for three draws in a row.

Mind you - didn't get any other number so it's done me no good!

It just seems spooky getting the lucky stars 3x in a row. But maybe the odds aren't that big anyway. Oh - and another factor - on each occasion I did 3 rows (does that affect the probability?)

I wish I had understood combinations/permutations and probability during my O-level maths years!

[ChuckFickens]

The probability of getting just the two lucky stars once is 1 in 95

So getting it twice would be 1 in 9025

Three times.. 1 in 857375

(i think :))

[sidkid]

Don't think its as high as 1 in 95. That's assuming you cant have the same number twice (so no 2+2, 3+3 etc) and that there is no zero.....

[sidkid]

oops - rethink - i now see the numbers go up to 11.....

[ChuckFickens]

I cheated on the 1 in 95 and got it from the wikipedia entry for the euro millions :)

I agree though, because if I work it I actually make it (11*10)/2 so 1 in 55

So two times...1 in 3025

3 time... 1 in 166375

Again... I think :)

[ChuckFickens]

Where's factor30 when you need him :)

[sidkid]

I think the chances for getting the two lucky stars once is 1 in 55....

[sidkid]

Too slow!

[Barquentine]

Thanks.
And having three lines on each of the three draws? Do you just divide by three? So 166375/3 = 1 in 55458?

[ChuckFickens]

Just looked again and the wiki entry for the lucky stars probability was for getting no main numbers and just the two lucky stars. so yeah, ignoring the main numbers totally it's 1 in 55

11 possibilities for the first number drawn. 10 possibilities for the second number so 11*10 = 110, then because it doesn't matter what order the number are drawn in and there are only 2 numbers it's 110/(2*1)= 55

Then to get the probability of successive events it just p^n (p is the probability, n is the number of events) so in this case 55^3=166375

I used to hate stats though :)

[ChuckFickens]

"And having three lines on each of the three draws? Do you just divide by three? So 166375/3 = 1 in 55458?"

yes.

[factor-fiction]

I don't know how the Lucky Stars thing works -I'll have to look on the Euromillions Lottery when i get back this afternoon

[ChuckFickens]

two numbers drawn from a pool of 11, factor.

[Barquentine]

Many thanks.

[Berkshirelad]

IMO, the probability of getting the same stars 2 or 3 draws in a row is irrelevant. Past draws do not affect furture draws; the proability remains the same for each draw.

[sidkid]

Berkshirelad - you're right that each draw is independant of the previous and next draw but that wasn't Barquentine's question.......

[mibn2cweus]

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[New Judge]

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