Lottery Probability

[Iron4ever]

In the national lottery (UK version) you pick 6 numbers between 1-49. If you get all six you win the jackpot. What are the chances of these 6 consecutive numbers 1 2 3 4 5 and 6 coming out? Is there an equal probability of these consecutive numbers being drawn as there is a random set of numbers like 3 10 11 23 44 and 49 ?

[Boerseun]

Originally Posted by Iron4ever

In the national lottery (UK version) you pick 6 numbers between 1-49. If you get all six you win the jackpot. What are the chances of these 6 consecutive numbers 1 2 3 4 5 and 6 coming out? Is there an equal probability of these consecutive numbers being drawn as there is a random set of numbers like 3 10 11 23 44 and 49 ?

The chances of picking 1-2-3-4-5-6 is exactly the same as picking any other set of numbers. And the chances suck. Consider:

Your chances of getting the right combination, is 49x48x47x46x45x44. And that equals one out of 10,068,347,520. One out of more than ten billion.

Those odds kinda suck.

No wonder they say the lottery is the tax people pay who can't do math!

[CraigD]

Originally Posted by Iron4ever

In the national lottery (UK version) you pick 6 numbers between 1-49. If you get all six you win the jackpot. What are the chances of these 6 consecutive numbers 1 2 3 4 5 and 6 coming out? Is there an equal probability of these consecutive numbers being drawn as there is a random set of numbers like 3 10 11 23 44 and 49 ?

Yes.

You occasionally find articles, books, and software that claims that some special combinations of numbers in this kind of random drawing are mathematically more likely to occur. These claims, no matter how impressive the analysis behind them, are bogus. “1 2 3 4 5 6” is as good a pick-6 guess as any other (and a darned sight easier to remember!)

[Iron4ever]

Originally Posted by Boerseun

Your chances of getting the right combination, is 49x48x47x46x45x44. And that equals one out of 10,068,347,520. One out of more than ten billion.

I thought the odds were 49 x 48 x 47 x 46 x 45 x 44 / 6 x 5 x 4 x 3 x 2 = 13,983,81 ??????

[Boerseun]

Originally Posted by Iron4ever

I thought the odds were 49 x 48 x 47 x 46 x 45 x 44 / 6 x 5 x 4 x 3 x 2 = 13,983,81 ??????

Why would that be?

[Iron4ever]

Originally Posted by Boerseun

Why would that be?

There are 49 possible choices for the first ball
For each of these 49 choices of the first ball,
there are 48 possible choices for the second ball
(because one ball has already been taken out)
meaning there are 49 x 48 ways of choosing the first two balls.
However, because the order doesn't matter, this has to be divided by 2
(because the second ball can come before or after the first ball)

So the number of different possible choices for two balls can be written as 49 x 48 /2

There are then 47 possible choices for the third ball,
but since it doesn't matter which position the third ball goes in we divide by 3, giving 49 x 48 x 47 / 3 x 2

This sequence continues for all six balls which are chosen hence my calculation above.

[Boerseun]

Interesting point, but I can't see the order actually having anything to do with the calculation?

Fact is - there's 49 balls to begin with. Only one of them will fall on the first selection. Now this can be any one of your six numbers. Let's say your numbers are 1-2-3-4-5-6, and the first ball to come out is six. The fact that it was indeed one of your numbers, doesn't change the fact that there's now 48 balls left. So now you're down to 49x48. And so on. I can honestly not see why the order of the selection should have anything to do with your odds? I mean - 6-5-4-3-2-1 is valid for your selection, but the odds are exactly the same as for 4-3-5-2-6-1, which is also valid.

Maybe I'm just slow on the uptake, but I don't get it.

[C1ay]

Originally Posted by Iron4ever

There are 49 possible choices for the first ball
For each of these 49 choices of the first ball,
there are 48 possible choices for the second ball
(because one ball has already been taken out)
meaning there are 49 x 48 ways of choosing the first two balls.
However, because the order doesn't matter, this has to be divided by 2
(because the second ball can come before or after the first ball)

So the number of different possible choices for two balls can be written as 49 x 48 /2

There are then 47 possible choices for the third ball,
but since it doesn't matter which position the third ball goes in we divide by 3, giving 49 x 48 x 47 / 3 x 2

This sequence continues for all six balls which are chosen hence my calculation above.

Correct, so the odds that any six numbers in any order will come up are 13,983,816:1 but, you asked, "What are the chances of these 6 consecutive numbers 1 2 3 4 5 and 6 coming out?" which implies that you are also specifying the order of the draw which would be 10,068,347,520:1.

[C1ay]

Originally Posted by Boerseun

Interesting point, but I can't see the order actually having anything to do with the calculation?

As far as the lottery is concerned the combinations 1,2,3,4,5,6;1,3,2,4,5,6;1,2,4,3,5,6, etc for all 720 combinations of 1,2,3,4,5,6 are all the same draw.

[Bo]

I would think:
there are 49*48*47*46*45*44 possible sixnumber combinations (basicly this is 49!/(49-6)! );
however with this calculation 123456 is a different combination from 654321, so you have to devide by the number of ways in which you could arange 6 given numbers; this indeed is 6*5*4*3*2*1.
or: P=N! /(x!(N-x)!)
(where N is the total number of things you can take, and x is the number of things you actually take)
these are the so called binomial coefficients.

Bo