Subset lottery probability Most lottery questions are exact, ie N numbers are winning, and you choose N numbers (chosen subset is as big as the winning subset). But how do you calculate the chance to win when you choose more than N numbers (chosen subset larger than winning subset)?
Say there are 8 numbers and 2 winning ones. You pick four numbers. What is the chance that the two winning numbers are included in your chosen subset?
I calculated the chance manually as 11/56 ~ 0.1964, but can't figure out how to get it via combinatorics.
 A: $$\large\frac {\binom 22\binom 62}{\binom 84}=\small\frac 3{14}$$
A: $$\frac12\times\frac37=\frac3{14}$$
Fix a winning number. The first factor represents the probability that this number is chosen (if $4$ of $8$ numbers are chosen then every fixed number has probability $\frac48=\frac12$ to be one of them). Assume that this occurs. Then the second factor represents the probability that under that condition the other winning number will be chosen as well ($7$ numbers are left and $3$ of them will be chosen).
This answer deliberately avoids binomial coefficients.
A: If you are familiar with how many ways there are to choose a subset of size $K$ from a set of size $M$, i.e. the binomial coefficient ${M \choose K}$, then all you need to know to finish is that the number of ways to choose a subset of size $K$ that contains given $N$ elements is the number of ways to choose a subset of size $K - N$  from a set of size $M - N$, since $N$ of the elements in the subset are prescribed. So the number of ways to win is ${{M-N} \choose {K-N}}$, and the probability of winning is this divided by ${M \choose K}$.
A: $b=$event that $4$ out of $8$ chosen = $8\choose 4$.  
$a=$event that $2$ of $4$ chosen are winners.
$a\cap b$ = ${2\choose 2}{6\choose 2}$.  $2$ chosen are not winners.  the $2$ winners are chosen from the $6$ remaining.
$a$ occurs given $b$ = $a|b={{a\cap b}\over b}={{{2\choose 2}{6\choose 2}}\over{8\choose 4}}$
