What is the probability of two people choosing the same number if person 1 picked the smallest and person 2 picked the largest? We are given that person 1 and person 2 each have 3 distinct numbers from a set of m numbers (1 to m).
Person 1 keeps the lowest number from their set while person 2 keeps the largest number from their set.
What is the probability that these two numbers match?
So if m = 10 and person 1 chose 5,7,9 and person 2 chose 1,2,5 then this results in a success since person 1's lowest number = 5 = person 2's largest number.
I think my issue is not knowing how to deal with each person getting rid of the two other numbers in each set. If each person picks only 1 number then it should be just P(matching number) $= m * 1/m * 1/m = 1/m$.
 A: The players must have selected $5$ distinct numbers between them, the middle one of which is the common (overlapping) number – player 1 selected the $3$ highest numbers and player 2 the $3$ lowest. Thus there are $\binom m5$ ways the players could have selected their numbers out of $\binom m3^2$ ways in all, so the probability is
$$\frac{\binom m5}{\binom m3^2}$$
A: You should break it up into cases based on what the matching number is, i.e., for each $1 \leq k \leq m$ calculate
$P(A_k)$ where $A_k=\{\text{the two numbers match and are equal to $k$}\}$. 
So for instance, if $m=10$ and $k=5$ like in your example, we need to count how many ways the first person can choose three numbers where the lowest is 5, and how many ways the second person can choose three numbers where the highest is 5. For the first person, their other two numbers must come from $\{6,7,8,9,10\}$ and be distinct, so there are $\binom{5}{2}$ choices. For the second, their other two numers must come from $\{1,2,3,4\}$ and be distinct, so there are $\binom{4}{2}$ choices. Overall there are $\binom{10}{3}$ ways for each person to choose their three numbers from $\{1,2,...,10\}$, so $P(A_5)=\binom{5}{2}\binom{4}{2}/\binom{10}{3}^2$.
Do the same thing for each value of $k$. Then if $A=\{\text{the numbers match}\}=\bigcup_{k=1}^m A_k$, we have $P(A)=\sum_{k=1}^m P(A_k)$ since the $A_k$'s are disjoint.
A: There are $\binom{m}{3}$ ways for each player to choose 3 numbers, so combined there are $\binom{m}{3}^2$ possibilities.
Suppose that we are in the situation where the highest of person 2's numbers equal to the lowest of person 1's numbers. In that case their combined set of numbers are 5 of the $m$ numbers, with the middle one being the one they have in common. There are $\binom{m}{5}$ ways to choose five numbers. Conversely, for any such choice, giving the lowest three numbers to person 2 and the highest three numbers to person 1 is only valid way to get an arrangement satisfying the condition. Therefore there is a one-to-one correspondence between choosing 5 out of $m$ numbers, and the number of valid arrangements.
The probability is therefore $$\frac{\binom{m}{5}}{\binom{m}{3}^2}$$
