What is the probability of not getting any number in a lottery where six numbers are chosen from $1$ to $60$. A specific lottery works this way: Six different numbers are chosen in the set $\{1,2,...,60\}$ and you have to guess the numbers choosing six different numbers in the same set.
The probability of getting all the six numbers right is $\dfrac{1}{C(60,6)}$.
But what is the probability of making a bet of six numbers and not getting any of the sorted numbers in the lottery?
I though that it was $\dfrac{C(54,6)}{C(60,6)}$, but I am not so sure.
I also think it is equal to $1$ minus the probability of getting some number, which looks complicated, because when you get $6$ numbers you also get $5$ numbers right, so there is the intersection, and this answer doesn't look so simple as $\dfrac{C(54,6)}{C(60,6)}$.
So, how can I find the probability of making a bet of six numbers and not getting any of the numbers sorted by the lottery?
 A: As JMoravitz has already confirmed, you're correct that the probability of getting none of the lottery numbers is $\ \frac{C_6^{54}}{C_6^{60}}\ $.  But you're also correct that it's $1$ minus the probability of getting at least one of the lottery numbers, and that the direct method of calculating this latter probability is a little more complicated.  In fact, what you've done is discover for yourself a combinatorial method of proving a result known as Vandermonde's identity,
$$
\sum_{k=0}^rC_k^mC_{r-k}^n=C_r^{m+n}\ ,
$$
for the particular case $\ m=r=6, n=54\ $.
If you and the lottery choose $\ r\ $ numbers from the set $\ \{\,1,2,\dots,$$\,n+r\,\}\ $, then the probability that exactly $\ k\ $ of your numbers appear among those drawn by the lottery is
$$
\frac{C_k^rC_{r-k}^n}{C_r^{n+r}}\ ,
$$
because the number of ways you can choose $\ k\ $ numbers from among those drawn by the lottery is $\ C^r_k\ $, and, for each of those, there are $\ C^n_{r-k}\ $ ways of choosing your other $\ r-k\ $ numbers from among the $\ n\ $ not drawn by the lottery.  Thus, there is a total of $\ C_k^rC_{r-k}^n\ $ of the $\ C^{n+r}_r\ $ equally likely ways of choosing your $\ r\ $ numbers which will have exactly $\ k\ $ numbers in common with the set drawn by the lottery.  Therefore, the probability  you've chosen your numbers in one of those ways is the fraction given above.  Since the number of your selections which appear in the set chosen by the lottery must be one of $\ 0,1,\dots, r\ $, and these possibilities are all mutually exclusive, it must be the case that
$$
\sum_{k=0}^r\frac{C_k^rC_{r-k}^n}{C_r^{n+r}}=1\ ,
$$
which, on multiplying by $\ C_r^{n+r}\ $, gives the Vandermonde identity quoted above for the case $\ m=r\ $.
If we split the above sum as follows,
\begin{align}
1&=\frac{C^r_0C^n_r}{C^{n+r}_r}+\sum_{k=1}^r\frac{C^r_kC^n_{r-k}}{C^{n+r}_r}\\
&=\frac{C^n_r}{C^{n+r}_r}+\sum_{k=1}^r\frac{C^r_kC^n_{r-k}}{C^{n+r}_r}\ ,\\
\end{align}
the first term, $\ \frac{C^n_r}{C^{n+r}_r}\ $, in the sum on the right is the probability that you've chosen none of the numbers drawn by the lottery, and the more complicated second term, $\ \sum_\limits{k=1}^r\frac{C^r_kC^n_{r-k}}{C^{n+r}_r}\ $, is the probability that at least one of the numbers you've chosen appears among those drawn by the lottery.
