Ticket lottery question A hundred tickets are marked $1,2,3,...100$, and they are arranged at random. 
Four tickets are picked from these and given to four persons A,B,C,D.
What is the probability that A gets the ticket with the largest value among the four of them and D gets the ticket with the smallest value?
My answer is $\displaystyle\frac{1}{12}$.
Is it correct?
Please help. 
 A: The number of lottery tickets is not important, cause at the end you choose $4$ random numbers and the choice of those numbers doesn't affect the probability of $D$ getting the lowest value and $A$ getting the largest. Therefore let's say we chose numbers $4,3,2,1$. We can distribute them into $4! = 24$ ways. That is our sample space. But out of all distributions we have only $2$ which are "good". ($4,3,2,1$ and $4,2,3,1$ in particular). Therefore the probability is:
$$p(\text{good configuration}) = \frac{2}{24} = \frac{1}{12}$$
A: Edit: I mixed up which of A and D was the highest and which was the lowest but I'm sure you appreciate that it is the same problem rephrased.
First of all, the total number of combinations for the tickets picked is $100 \times 99 \times 98 \times 97$
Suppose that A picks the ticket with number $a \leq 97$
D must pick a number $d$ between $a+3$ and $100$
B and C pick two numbers between $a+1$ and $d-1$. The size of the range for B and C is $(d-1)-(a+1)+1=d-a-1$. When B is given their ticket they have $(d-a-1)$ options, then when C is given their ticket they have $(d-a-2)$ options. Therefore there are $(d-a-1)(d-a-2)$ options for B and C's tickets.
Considering the ranges of $a$ and $d$ you can sum up the number of combinations for all values of $d$ and $a$.
$\sum_{a=1}^{97} \sum_{d=a+3}^{100}(d-a-1)(d-a-2) $
This gives the total number of combinations where A has the lowest and D has the highest ticket.
