Game of Chance

WLOG, say the numbers picked were 1, 2, 3, and 4. Then, we must find the probability that John picked 1 and 4, which is $\dfrac{1}{\dbinom{4}{2}}=\dfrac{1}{6}$ The answer is $1+6=\boxed{7}$ .

The reason we can use WLOG is because the only thing important about the number is their order.

Once we have picked numbers from the bag, we have 4 monotonic integers. How many ways are there to split these integers so that John has the biggest and the smallest of the 4? There is only one way, of course. How many ways are there in total to split the 4 monotonic integers? We have 4 and we choose 2, so there are ${4 \choose 2} = 6$ ways.

Thus, the probability that John will win this game is $1/6 \approx 16.7\%$ , and $1/6$ cannot be reduced, so the value of $a + b$ is $\fbox{7}$

Note that it doesn't matter what the bag contains as long as all the numbers are distinct and there are more than 4 numbers, since by these conditions, there will always be a largest and a smallest number in the set that John and Mark pick.

Let the four numbers John and Mark pick, collectively, be $a,b,c,d$ , with $a < b < c < d$ . The scenarios where John wins are when John has $a,d$ in any order. The total number of ways for John to select his two out of the four integers is $\binom{4}{2} = 6$ , so John has a $\frac{1}{6}$ probability of winning, and the desired answer is $\boxed{7}$ .