Combination maths problem

[Irvin]

Hello all,

I have the following question;

How many combinations are there when you have the following number limits per postion;

pos1= 1 to 13
pos 2= 3 to 22
pos 3= 7 to 29
pos 4= 12 to 35
pos 5= 19 to 38
pos 6= 27 to 40

I seem to be going crazy or brain dead at the moment.

Any help would be appreciated.

Please show all working so I can add to my spreadsheet.




Thank you.

 

[NmbrsDude]

I'm no mathematician but try this on for size...

By using the ranges you gave we obtain that each position can only have the following possible number of outcomes:

P1 13
P2 20
P3 23
P4 24
P5 20
P6 14

Therefore, the odds should be

13 x 20 x 23 x 24 x 20 x 14
6 x 5 x 4 x 3 x 2 x 1

which give us

40185600
720

which gives us odds of 1 in 55,813.3.

Hope this helps,
ND

 

[savagegoose]

yup the way to look at it is for the 1st posistion there are 6 possible results that you will win from. as the draw doesnt need to be ordered.

so thats 6 out of 13 or 13/6

for second pos there are only 5 balls remaining to be drawn so

its x/5

etc.

 

[GillesD]

Possible combinations

Based on the requirements you specified, I calculated all possible combinations meeting those requirements.

I found 2,752,024 combinations from 1, 3, 7, 12, 19 and 27 to 13, 22, 29, 35, 38 and 40. This represents 71.7% of the 3,838,380 possible combinations of a 6/40 lottery.

But this time I am not that 100% sure of my program. I will run it to verify if these calculations are right. In all cases, I assumed that the number in second position is greater than the one in first, that the number in third position is greater than the one in second, and so on.

 

[Irvin]

Thanks for the answers guys.

I originally thought it was around 55000 but have to go with GillesD answer.

If we anchor 1 to the first spot we end up with around 370,000 combinations. And percentage wise this corresponds with the numbers / position drawn to date.

Thanks again.