Wednesday, October 30, 2013

Lotto 6/49 Probability Calculation

the probability of getting 6, 5, 4, 3, 2, 1 or 0 numbers in a lottery where you choose 6 numbers out of 49.
step by step walk through.

the equation:
(49!/6!)/[(6!/n!)(43!/[6-n!])] = probability

note:
[(6!/n!)(43!/[6-n!])] can also be written as: (6,n) X (43,6-n).
for 6 numbers, use the 1st part of the equation only.

! = factorial (the product of an integer and all the integers below it. eg. 6! = 6*5*4*3*2*1 = 720)
n = numbers in the lottery you have right.
* = multiplication, or "times" as it were.

"factorialize" the numerator only as many times as the denominator number. using n = 4 as an example, we would calculate as follows: (6,4) (43,2) = (6*5*4*3)/(4*3*2*1) [factorialized 4 times] x (43*42)/(2*1) [factorialized 2 times] = 13,545.

n = 6 numbers
remember, for 6 numbers we only have to calculate the first part of the above equation:
(49!/6!) = (49*48*47*46*45*44)/(6*5*4*3*2*1) = 1 in 13,983,816
n = 5 numbers
we know the answer to the first part of the equation is 13,983,816, so now let's calculate the second part:
(6,5) X (43,1) =
(6,5) = (6*5*4*3*2)/(5*4*3*2*1) = 6
(43,1) = 43/1 = 43
6*43 = 258
13,983,816/258 = 1 in 54,200.837
n = 4 numbers
(6,4) X (43,2) =
(6,4) = (6*5*4*3)/(4*3*2*1) = 15
(43,2) = (43*42)/(2*1) = 903
15*903 = 13,545
13,983,816/13,545 = 1 in 1032.397
n = 3 numbers
(6,3) X (43,3) =
(6,3) = (6*5*4)/(3*2*1) = 20
(43,3) = (43*42*41)/(3*2*1) = 12,341
20*12,341 = 246,820
13,983,816/246,820 = 1 in 56.656
n = 2 numbers
(6,2) X (43,4) =
(6,2) = (6*5)/(2*1) = 15
(43,4) = (43*42*41*40)/(4*3*2*1) = 123,410
15*123,410 = 1,851,150
13,983,816/1,851,150 = 1 in 7.554
n = 1 number
(6,1) X (43,5) =
(6,1) = 6/1 = 6
(43,5) = (43*42*41*40*39)/(5*4*3*2*1) = 962,598
6*962,598 = 5,775,588
13,983,816/5,775,588 = 1 in 2.421
n = 0 numbers
(6,0) (43,6) =
(6,0) = 0 or no "factorialization required". it will not figure in our answer.
(43,6) = (43*42*41*40*39*38)/(6*5*4*3*2*1) = 6,096,454
13,983,816/6,096,454 = 1 in 2.294

5 numbers + Bonus
(the following is from http://icarus.mcmaster.ca/fred/Lotto/)
The pick of six must include 5 winning numbers plus the bonus. Since 5 of the six winning numbers must be picked, this means that one of the winning numbers must be excluded. There are six possibilities for the choice of excluded number and hence there are six ways for a pick of six to win the second place prize. The probability is thus 6/13,983,816
= 1 in 2,330,636

1 comments:

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