In this lottery $7$ balls are chosen from $1-60$. In order to win the main prize, you must select all $7$ right. I calculate the odds of doing this as:
$$1:386,206,920$$
The odds of getting $3$ correct would be
$$1:38$$
Now suppose for whatever reason you are only allowed to select balls from $1-30$ and cannot pick $31-60$. Does this change your odds of winning? What are your odds of the top prize and what are your odds of getting 3 right?
Thanks
As far as I understand the question, you select $7$-element subset of $\{1, ... , 30\}$ uniformly, you win if they coincide with $7$-element subset selected from $\{1, ... , 60\}$ uniformly, and those selections are independent. Suppose, $X$ and $Y$ are those subsets respectively. Now if $Y$ is not a subset of $\{1, ... , 30\}$, then the probability of them coinciding is $0$. Also $P(Y = X| Y \subset 30) = \frac{1}{C_{30}^7}$, so here we have $P(X = Y) = P(Y = X, Y \subset 30) = P(Y = X| Y \subset 30)P(Y \subset 30) = \frac{1}{C_{30}^7}\frac{C_{30}^7}{C_{60}^7} = \frac{1}{C_{60}^7}$.
It is quite paradoxical, but that happens to be the same number, as if your choice were completely unrestricted. And it would be that way, no matter to what subset of $\{1, ... , 60\}$ is your choice restricted (provided, of course that this subset contains not less than $7$ elements).