It has been a long time since I last studied probability and statistics but I am trying to solve a tricky (for me) problem.
If I have any two number ranges, what is the calculation for the probability of a random value in one range being higher than the other?
Simple example…
Range 1: 10-19
Range 2: 20-39
Result: 0% probability that a number in Range 1 is greater than Range 2.
Trickier example…
Range 1: 10-20
Range 2: 15-24
What is the probability that a random value in Range 1 is greater than a random value in Range 2?
Best Answer
You can count all possible variations, and the ones that answer the actual question and divide them (good/all) to get the probability.
Now we are dealing with ordered pairs of numbers, call Range 1 $X$ and Range 2 $Y$. We want $P(X>Y)$. Now the good pairs are $$(16,15),\ (17,15),\ .., (20,15);\ (17,16),\ .. (20,16); ...; (20,19).$$ We can count these for example according to the possible $Y$'s: the minimal possible $Y$ such that $X>Y$ is $Y=15$, then $X$ can be $16..20$, that is $5$ possibilities. If $Y=16$, we have $4$ possibilities, and so on. So it is $5+ 4+ 3+ 2+ 1=15$ possibilities altogether.
And the number of all possible pairs is now $11\cdot 10$.