Lottery odds?

[BIG DAD]

Are the odds of getting the numbers 1,2,3,4,5,6,7 or any sequencial combination in any order on the lotto higher than random numbers coming out i.e 23,1,5,17,11,14,31.

or to put it another way, if i bought 2 lotto tickets with the numbers 1,2,3,4,5,6 on 1 and 11,2,45,31,7,12 on the other, are the odds of winning on the 1st one higher that the second?

[Emu2005]

The odds are exactly the same for each selection. The chances of any 6 numbers coming out are the same as any other group of 6 numbers. (I think the odds are like 1 in 14 million or something).

It may seem like the chances of the scattered numbers coming out are more likely, as say if you got 1as the first number, the chances of getting another number in the 1-10 category is less than getting a number in the tens/twenties/thirties/forties, because they are all remaining and one of the 1-10 has come out already. But the chances of getting one number in particular are the same.

Don't think I explained that very well but hope it makes sense!

[squarebear]

Which just goes to show how hard it is to win the lottery. The odds are exactly the same, even though people would say that the chances of 1 through 7 coming out are zero.

[BIG DAD]

Thanks, we are still arguing this in the office, i know the odds of any ball coming out are the same but surely the odds of the numbers being in a row i.e 22,23,24,25,26,27 are higher?

[Emu2005]

Hmmm now I'm confused!

You're right, the chances of getting a set of numbers in a row is less than getting a selection of numbers. However - that is because there are more combinations of numbers not in a row than there are of numbers in a row. If your mate picked a random selection, and you had numbers in a row - you would each have exactly the same chances of winning.

My dad used to pick only high numbers above 31 - he had the same chance of winning as every other combination, but thought if he did win he'd win more, as most people pick their birthdays so few others would have picked these numbers and he wouldn't have anyone to share it with!

[squarebear]

Which is why one of the first rollovers was won by a single ticket who had numbers in the high 30s and 40s.

[BIG DAD]

ok so you wrote
"You're right, the chances of getting a set of numbers in a row is less than getting a selection of numbers."

so surely the odds for this are higher.

[squarebear]

the odds of 1,2,3,4,5,6 dropping out of the machine in that order is much smaller than the numbers dropping out in any order eg 4,3,5,2,1,6

As there is only 1 combination of 1,2,3,4,5,6 and 720 combinations of the six balls in total.

[BIG DAD]

so are the odds of the numbers dropping out in any order eg 4,3,5,2,1,6 higher than any other random combination where the numbers are consecutive?

[Emu2005]

Hmmm, I don't even really understand the question now!

Like square bear says, if you had 6 consecutive numbers, there is much more chance of them coming out in any random order than coming out 1, 2, 3, 4, 5, 6.


However, when you ask are the odds higher of these numbers dropping out in any order than for any group in consecutive order - i think there are two issues.

1. The numbers themselves - equal chance ofof all selections coming out

2. The order - if you specify the order you want the balls to come out with one set, and you don't with the other set, then the set where you want a specific order is less likely to occur. This doesn't just apply to consecutive numbers, if you wanted the numbers you gave earlier to come out in order there would be equal chance of this happening as there would be 1, 2, 3, 4, 5, 6.

God I don't even understand what I'm saying now!

[jake-the-peg]

For 1,2,3,4,5,6 to come up the first number can be any of 43 ie 43/49 then the next must be one of 5 specific numbers 5/48 then 4/47 then 3/46 then 2/45 then 1/44.

Which I make 1.9 million to 1

There are 43 possible sequences of consecutive numbers so roughly half a million to 1 against one of them coming up.
(some complexities around the end numbers but this is very rough)

So probability of not sequential is half a million to one on.

but the odds of a particular sequential set are the same as those of a randomly chosen set

But you said random that's difficult because you have to define what's random

There's 2,4,6,8,10,12 that could come up or 29, 31, 37, 41, 43, 47 (primes) or 3,4,7,11,18,29 a fibonacci sequence.

So calculating the odds of a consecutive sequence coming up is one thing and the odds are long, calculating the odds of the sequence being random and not displaying a pattern is different and I have no idea how you'd even start that.

[BIG DAD]

Lol Emu, thanks for your answers, i'm getting confused as well, i'll try and make sense of it 1 more time, the lads in the office are arguing that no matter what combination of numbers you put on the lotto the chances of winning are the same.
But i am saying that if you put on 1,2,3,4,5,6 the chances of you winning are less no matter what order they come out in.

Hope this makes more sense.

[squarebear]

Nope the odds are exactly the same of 1,2,3,4,5,6 or 3,6,5,4,1,2 or any other sequence coming out as any other combination.

[kwicky]

On a simpler problem the odds of tossing a head is 1 in 2 obviously. If you kept tossing the coin and you got 5 heads on the trot the next toss of the coin is still 1 in 2.

I queried this with my maths teacher some years ago and still believe that having had a series of heads the next toss should favour a tail. But no!

[4candles]

The lads in the office are right. It might be easier to get your head round it if you forget the numbers on the balls. Let's suppose that the balls have no markings at all. The chances of any six popping out of the machine are exactly the same - right? Now put the numbers, or any marks at all, on the balls. The chances are the same!

[ACB312]

Big Dad, sorry, the lads in your office are correct, the odds are exactly the same. Rather than looking at numbers, just think of the balls as events that can be successful or not. The first event will be successful 1 in 49 times, the second 1 in 48 and so on (Factorial 6 in Factorial 49 = almost 1 in 14000000). There's a book called "Taking Chances", sorry can't remember the author's name, that explains the lottery (and whole lot of other games) and the true odds and how they are calculated. People glibly trot off the 14 million to 1 odds but think of it as if you'll need to put 5 lines on every Wednesday and Saturday for the next 270 years to match 6 events once - assuming lottery is mathematically truly random, which analysis so far bears out.......Cheers, Andy

[reinganum]

This is an old question (probably started in Nov 1994 when the Lottery started).

For simplicity, I am referring only to the classic Lottery where six balls are delivered by the machine. I have not taken the bonus ball into account, because it does not feature in the 'jackpot' prize.

Simply, the answer is in two parts:

(a) The chances of winning i.e. picking the six balls is the same for any combination of numbers, whether in sequence or not, since the machine delivers the balls at random. As everyone knows, the sequence the balls appear is not relevant.

(b) The chances of having to share the prize are much lower if some combinations are avoided, for example sequences, numbers below 31 (birthdays) etc.

[TheWinner]

Here is a link to a site that I have a lot of fun with. I sometimes think of 6 ridiculous numbers and use the site to see how much I would have won if I had played them.


http://lottery.rleeden.no-ip.com/