I know that we can calculate the possibility of a winning number in a $6/49$ lotto game with this formula:
$$C(n,r)= {n!\over r!(n−r)!}= C(49,6)={49 \choose 6}=\frac{49!}{6!(49−6)!}= 13 983 816$$
I can say that the probability of winning $6$ numbers is $1 / 13983816$ according to the above formula. I would like to calculate the same ratio for $3$, $4$, and $5$ numbers. Here I am seeking a formula to do this.
My question is how can I calculate the total number of combinations of $5,4$ and $3$ winning numbers out of the $6$ winning numbers in all possibilities? Is there a similar formula for this? Thanks.
The number of possible outcomes is $$\binom{49}{6}$$ which is the number of ways to select a subset of six of the $49$ numbers.
The number of ways to select exactly $k$ winning numbers is the number of ways we can select $k$ of the six winning numbers and $6 - k$ of the $43$ other numbers in the draw, which is $$\binom{6}{k}\binom{43}{6 - k}$$ Hence, the probability of selecting exactly $k$ of the winning numbers is $$\frac{\dbinom{6}{k}\dbinom{43}{6 - k}}{\dbinom{49}{6}}$$ In particular, the probability of selecting all six winning numbers is $$\frac{\dbinom{6}{6}\dbinom{43}{0}}{\dbinom{49}{6}} = \frac{1}{\dbinom{49}{6}}$$ which makes sense since there is only one way to correctly select all six numbers.
For $k = 2$, we get $$\frac{\dbinom{6}{2}\dbinom{43}{4}}{\dbinom{49}{6}}$$ I will leave the calculations for $k = 0, 1, 3, 4, 5$ to you.
The events $k = 0, 1, 2, 3, 4, 5, 6$ are mutually exclusive and exhaustive, so their probabilities should add up to $1$.