What is the probability that this problem is interesting?

Long Method: Lets compute the total number of outcomes first.The first number has $100$ choices and the second has $99$ choices. Thus by product rule , total outcomes $=99\times 100 = 9900$ .

Now lets compute the number of $R<1$ . When the first number is $1$ the second number has $99$ choices since all the $99$ numbers are greater than $1$ thus making $R<1$ . Similarly , When the second number is $2$ the second number has $98$ (we must exclude $1$ for $R<1$ ). Similarly When the second number is $3$ the second number has $97$ choices and so on When the second number is $99$ the second number has only $1$ choice that is $100$ . Thus outcomes satisfying $R<1$ are $1+2+3 \dots +98+99 = \dfrac{99\times 100}{2}$

Thus the required probability is $\dfrac{\dfrac{99\times 100}{2}}{9900} = \boxed{\dfrac{1}{2}}$

Shortcut method: If you choose any numbers and compute $R$ , it will be either less than $1$ or greater than $1$ . Thus the probability that $R<1$ is $\dfrac{1}{2}$ no matter how many numbers are in the set.

Notice how the numbers you choose do not matter by symmetry. Let's say you choose $a, b$ given that $a > b$ . There are two fractions you can create, $\frac{a}{b} > 1$ and $\frac{b}{a} < 1$ , therefore there is a $\frac{1}{2}$ probability that the property holds.

Therefore the answer is $1+2 = \boxed{3}$

$R < 1$ translates to "the first number is less than the second number". Consider the two numbers we pick. We pick the smaller one first half of the time, no matter what the two numbers are, so $R < 1$ half of the time as well. Thus $a = 1, b = 2, a+b = \boxed{3}$ .

The simplest solution to this problem goes here: For any 2 distinct integers (say a and b) chosen, they can be written either as a/b or b/a. Obviously only one of these fractions is less than unity. Since this is true for all integer pairs the required probability is 1/2.