Any mathematicians or probability analysts?

[tbourner]

Does anyone know the equations for working out probability of Lottery wins?

 

I'm sure I remember it being something like:

....N!

--------

.(N-n)!

 

Or something, it works out to be:

49x48x47x46x45x44

-------------------

....1x2x3x4x5x6

 

Which comes out at 13983816 to 1.

That's for choosing 6 numbers and getting 6 correct.

 

What about for choosing 6 numbers and getting 3 correct?

My calc came out at 152:1 which seems high!! Anyone know the real answer?

 

My question for whoever knows about this is what is the probability of choosing 5 numbers in a 49 ball draw and getting 2 correct? Is it better than getting 3 from 6 choices?

Similarly choosing 5 and getting 3 instead of choosing 6 and getting 4?

 

Some people must know what I'm getting at already!

[Supragal]

AAAAAAAAAAAAAAAAHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

[RobSheffield]

AAAAAAAAAAAAAAAAHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHHH

 

Breathe

[mikeyb10supra]

Hmmm now where did I put my uni notes.....if no one has answered this by tomorrow I will did out my notes and let you know....god how sad am I

[chilli]

1st bit is just nCr = 49C6 (assuming there are 49 numbers and 6 balls drawn) then that gives the value you were after.

 

draw 6 from 49 and get any 3 correct P = (3/49)x(2/48)x(1/47) = 1/57

[tbourner]

hmmmm, ok then, well it seems my plan is flawed then and I won't be rich before next week!

[TrickTT]

http://www.alllotto.com/lottery_odds_calculator.php

[chilli]

ps: the last two calculations are Hypergoemetric I think.

 

i.e. you take R from N without replacement and then r from n without replacement. Probability = (R, r) * (N-R, n-r) / (N, n)

 

so draw 5 from 49 and get 2 = Hypergoemetric [49, 5, 5, 2] = (R, r) * (N-R, n-r) / (N, n) =

 

5C2 * (49 - 5)C(5 - 2) / 49C2 => I'll leave you to work that one out

[tbourner]

It seems to be 49C6 / ( 43C3 * 6C3 ) is the probability of getting 3 correct from 6 balls = 13,983,816 / (20 * 12341) = 57:1

 

So I wanted to know 49C5 / ( 44C2 * 5C2 ) which is 1,906,884 / (946 * 10) which is 202:1 (probability of getting 2 correct from 5 chosen).

[tbourner]

http://www.alllotto.com/lottery_odds_calculator.php

 

Well I was nearly there!!

 

Choose 6:

.Match.|......odds......

.....6....| 1 in 13,983,816

.....5....| 1 in 54,201

.....4....| 1 in 1,032

.....3....| 1 in 57

 

Choose 5:

.Match.|......odds......

.....5....| 1 in 1,906,884

.....4....| 1 in 8,668

.....3....| 1 in 202

.....2....| 1 in 14

 

Lower odds, but that doesn't help!

[black cat]

wtf

[piehound]

glad im builder and i use ft and inches

[Pete]

The stupid thing about probability is that it can't actually predict what's going to happen with random events. Only what's likely.

Those odds might put you off playing the lottery, but if you hadn't seen the odds and you played - your could've won!

[tbourner]

Basically you can reduce your odds at a higher risk - I suppose you can do that simply by buying more tickets!! But this should work out to give you marginally better odds when buying a certain number of tickets.

[Supragal]

are the odds less for getting the numbers 1 2 3 4 5 6 or not?

[andrew7]

You can reduce the odds a little bit by including the following in your choices (if starting with six pick numbers):

 

Two of your numbers should have matching last didgets, i.e. 2,12,22,32,42

 

Two of your numbers should be consecutive, i.e. 2,3 or 18,19 or 30,31 etc...

 

Also, if you eliminate half the numbers, i.e. 1-25 or 26-49 and always use the others to pick from, on the occasions when the draw is one-sided you have a much greater chance of sUccess (but will have a lot of boring weeks when you have no chance of anything).

[chilli]

You can't predict the outcome for truely random, memoryless events no - however you can pick numbers that ensure that if you do win, you are more likely to get a bigger pay out (pick numbers less likely to be picked by others) - that's about it.

 

Yes 1, 2, 3, 4, 5 ,6 is the same probability as _any_ other combination

 

Random events, statistically they should obey chi square - that's about as far as any predictability goes. I.e. over large samples they tend to obey a pattern, but the outcome of any single run can never be predicted of course...

[tbourner]

Law of averages says you're much less likely to get 1, 2, 3, 4, 5 and 6, but obviously the probability is technically the same as any group. Probability is different to likelyhood?

 

I suppose you could say there are only 44 sets of 6 consecutive numbers, but millions of sets of any 6 numbers - so you're more likely to get a different set than a consecutive set.

[chilli]

Law of averages says you're much less likely to get 1, 2, 3, 4, 5 and 6, but obviously the probability is technically the same as any group. Probability is different to likelyhood?

 

I suppose you could say there are only 44 sets of 6 consecutive numbers, but millions of sets of any 6 numbers - so you're more likely to get a different set than a consecutive set.

 

huh? what is the law of 'averages'? I think intuition says it is less likely and this is what most people expect, however it is exactly same same likelyhood, probability or whatever you want to call as any other set of numbers.

 

Where it does differ is the probability of getting 1, 2, 3, 4, 5, 6 out in that order, now that is much less likely, but lottery numbers are not based on that of course...

[bammbamm]

there are three types of mathematicians, those that can add and those that cant

[chilli]

I forgot to mention, the original formula you were looking for is:

 

N! / R! (N-R)! which is odd because you still got the first sum right lol

 

The bottom bit is indeed the R!, the top bit is what is left when you have cancelled all the terms in the N! with the (N-R)! bottom bit, i.e it leaves N-R top terms from N down to (N-R)...

 

this works for all simple cases where the sample size and the number matched is the same.

 

I double checked and the hypergeometric is the right formula and it matches what the online calculator produced (didn't have a calc with nCr handy at the time and was feeling lazy).

 

I think the by hand bit you had was going wrong somewhere, not enough terms in it.

 

anyway that's enough sad geeky math, just thought I'd finish it off because things like this sometimes fascinate me.

 

PS: If you win now, can I have a cut?

[tbourner]

Well the idea was (and it's been used before as a scam) to get 44 lines, choose 5 numbers and have the 6th number as the remaining 44 numbers that you haven't picked - basically that guarantees getting 1 number!!

 

Then if you get 2 numbers right in the 5 you picked you win £30 (because the 3 numbers you picked that weren't drawn would have been in 3 of your rows as the 6th number!).

If you get 3 right out of your 5 though, you get 2 lines with 4 numbers and 42 lines with 3 numbers!!!!

 

Basically I think it increases your odds but is a higher risk.

 

 

Imagine if you got all 5 right!! You'd win the jackpot 6 numbers on 1 line, plus 1 line with 5+bonus ball and 42 lines with 5 balls correct!!!!!

[chilli]

hmm, think there is flawed logic in there somewhere:

 

[49 6 6 3] = .01765040 = 1/56.66

 

[49 5 5 2] = .06953621 = 1/14.38

 

but with the 2nd you've had 44 tries, so pay for 44 in the first and for the same 3 ball match you get 0.01765040 * 44 = 0.7857 = 1/1.27 chance, i.e. nearly almost a dead cert, since it was only a 1/57 chance anyway, hardly suprising, so I think the first option is noteably better, statistically.

 

I think that's right but I could be wrong, just my first thoughts on it and it's been a long day...

[tbourner]

Exactly, but that's a chance of winning £10!!! Not very good if you've spent £44.

 

So a higher risk but you will get £30 back from your £44 - much better return although still not very good.

 

Similarly for getting more numbers correct, slightly higher odds but you return more each time (as in my 5 numbers correct example!).

[SupraHuman]

WTF...banana??