Turtle said:
Keep the 1 2 3 4 5 6 & choose 5 10 15 20 25 30 for comparison, or more precisely any 6 number combination where all elements are multiples of 5
…
_______For 1 2 3 4 5 6: Probability is 6/49*5/48*4/47*3/46*2/45*1/44
___For 6 multiples of 5: Probability is 9/49*8/48* 7/47*6/46*5/45*4/44
____Clearly these 2 combinations don't have the same probability.QED
Correct. However…
In general, there are m!/(n!*(m-n)!) ways of choosing n of m elements where order isn’t important (where “!” is the factoral unary operators, eg: 4! = 1*2*3*4) [1].
The probability P of drawing a specific n of m elements is its reciprocal, (n!*(m-n)!)/m! [2].
The probability of choosing any 1 of x different choices of n of m elements is x*P [3].
As you’ve correctly calculated, and per [1], the probability P1 of drawing exactly {1,2,3,4,5,6} is (6!*(49-6)!)/49!,
while the probability P2 of drawing any 6 balls numbered with multiples of 5 is (9!*(49-6)!)/(49!*(9-6)!).
Note that P2/P1 = ((9!*(49-6)!)/(49!*(9-6)!)) / ((6!*(49-6)!)/49!)
= (9!*(49-6)!*49!)/(49!*(9-6)!* (6!*(49-6)!))
= 9!/((9-6)!*(6!))
= 84
However the probability of choosing a
specific 6 of 49 numbered balls (eg:{5,10,15,20,25,30}) is not the same as choosing
any 6 of 49 where each’s number is a multiple of 5 (eg: {5,10,15,20,25,30} or {5,10,15,20,25,35} or {20,25,30,35,40,45}, etc.)
There are 9!/ ((9-6)!*6!) = 84 possible ways to draw 6 of 9 (the number of ) elements. As expected from [3], this is the same as P2/P1.
So, in a fair drawing of 6 of 49 balls, the probability of drawing {1,2,3,4,5,6} is the same as the probability of drawing {5,10,15,20,25,30}, or any other choice of numbers – 1/13983816.