Maths question

This is very sad I know but its puzzling me:

How do work out permutations, ie 15m:1 chance of winning the lottery or 4 digits on my briefcase. I presume you start with multiplying the amount of numbers you have, ie if you have six you'd do 6x5x4x3x2x1=720, so with any six numbers you have 720 permutations. After this Im struggling abit, okay alot. My mum says Im not stupid just god blessed me in a different way. Any ideas??

 

I would have thought that a 4 digit briefcase lock would have 10,000 possible combinations from 0000 to 9999 ?
same as a bank card PIN?

i.e 10 x 10 x 10 x 10 or 10^4 ( assuming that your lock has all numbers from 0 to 9)

lottery is different as there are 49, then 48 then 47 etc possible outcomes, so 49 x 48 x 47 x 46 x 45 x 44 would be my guess, but I'm not 100 % sure of that.

 

I tried that, it doesnt workout, there must be a figure to divide it by, its really bugging me this

 

For the lottery you can choose 49x48x46x47x46x45x44 variations, but each equivalent 6 number choice will turn up in that list 6x5x4x3x2x1 times. In other words the order you pick the numbers in doesn't matter. This is the difference between a permutation, order matters, and a combination, order doesn't matter.

So the number of lottery combinations is 49x48x46x47x46x45x44/6x5x4x3x2x1 which I think is about 13983816 but don't quote me.

Paul

 

have a look at http://money.howstuffworks.com/lottery1.htm

 

1. number: 6 out of 49 = 8.166
2. number: 5 out of 48 = 9.600 x 8.166 = 78
3. number: 4 out of 47 = 11.750 x 78.400 = 921
4. number: 3 out of 46 = 15.333 x 921.200 = 14,125
5. number: 2 out of 45 = 22.500 x 14,125.023 = 317,814
6. number: 1 out of 44 = 44.000 x 317,813.91 = 13,983,816

 

I think it goes like this.

You have a population (49 numbered balls) and from the population you select a subset (6 numbered balls).

If the order of the balls is significant, then the number of permutations of 6 numbered balls that you can create is P(49,6) or 10,068,347,520.

If the order of the balls is not significant, then the number of combinations of 6 numbered balls that you can create is C(49,6) or 13,983,816.

P(x,y) = x! / (x-y)! = 49! / 43! = 49.48.47.46.45.44

C(x,y) = x! / ( (x-y)! . y! ) = 49! / ( 43! . 6! ) = 49.48.47.46.45.44 / 6.5.4.3.2.1

Blimey! I haven't done that for 25 years. Surprised I can remember it

 

just thought I would make this question alot more difficult, what are the odds of getting 5/6 numbers?

 

That's easy. The number of possible combinations is 13,983,816 as already shown.

The number of desired combinations is C(6,n) where n is the number of correct balls.

So to win the jackpot, C(6,6) which is 1 so 1 in 13,983,816 chance.

To win on 5 correct numbers, C(6,5) which is 6 so 6 in 13,983,816 chance.

C(6,4) = 15

C(6,3) = 20 (a 1 in 699,190.8 chance of winning a tenner)

Edited to say: this is of course nonsense See below!

 

Imagine all 13983816 possible draws written down starting 1 2 3 4 5 6. What could I draw that would match any 5 or the first draw? 1 2 3 4 5 7-49 would do, but so would 1 2 3 4 7-49 6 and so on. For each possible draw I can match 5 in 6x42 ways. So it should be 6x42=252 times more likely than matching 6. So the odds are 1 in 13983816/252 or 1 in 55491. From this it follows that the odds of matching 5 plus the bonus ball are 1 in 55491x43 or 1 in 2386127 because having matched 5 we now have a 1 in 43 chance of matching the bonus.

I think.

Paul

 

So a combi lock made up of 4 digits using 0-9 means there are 15,120 combinations?

 

I haven't got that right. The odds of winning a tenner are vastly better than that. I have only calculated the number of winning triplets and forgotten that they are accompanied by three other numbers that can have any value out of the remaining 46

So, to get 5 correct balls, there are 6 different fives in the draw as before. However, the number of desired events, i.e all the winning combinations that we can create on our lottery ticket, is 6 different fives combined with a sixth number which can have any of 44 values. So the desired events is C(6,5) x C(44,1) = 6 x 44 = 264

For 4 correct balls, it will be C(6,4) x C(45,2) or 15 x 990 = 14,850

For 3 correct balls, it will be C(6,3) x C(46,3) or 20 x 15180 = 303,600

So the probability of winning a tenner is 303,600 / 13,983,816 = 1 in 46.06

So if you played the lottery once a week for a year, you ought to win a tenner once on average.

 

No, that's a different problem because the value of each digit is not dependent in any way on the other digits.

Permutations and combinations are concerned with having a fixed number of objects and selecting a subset. How many subsets can we select? Permutations allow the same subset but in a different order. Combinations ignore order.

Perhaps it's the numbering of the balls that is confusing you. Forget the numbers. They are irrelevant. We could instead have 49 different fruits and select 6 of them.

 

Just goes to show mathematics actually work. Everyone spending at least £52/year they will be rewarded by £10 at most but mostly nothing to show for this investment as expected.

 

Er, so is it 9999?

 

No, that would still be 10000, 0000-9999. When you pick, say, 8 for the first digit the others could still be 8. If you weren't allowed to repeat digits then you'd have 10x9x8x7 which is 10P4 in permutation terminology. Still not 15120 though...

Technobear has a minor bug, he's counting the jackpot winning combinations, there are 43 alternatives for each ball to still match 5 rather than 44.

There's about 1/57 chance of winning £10. So a 56/57 chance of not winning. If you play once a week for two years the chances of not winning at least once are 3/20.

Paul

 

Surely once you have chosen 5 correct numbers, you still have 44 choices remaining for the sixth number.

 

One of those 44 will match 6 and make you a bit happier.

Paul

 

Duh! Of course it will

So to win on 5 is C(6,5) x 43 = 258 chances in 13,983,816

For the other cases it appears to get horribly complicated using this approach and a ball by ball approach works better.

 

I'm pretty sure that it's 252. There's another 6 to dismiss somewhere. It's probably because you've got 42 other choices for your discard ball. There's the one you've got now, the jackpot and 42 others. Does that make sense?

No longer sure it's 252. It seems I can't count.

Paul