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) x (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) x (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