# Probability of drawing numbered balls in a ascending order with replacement

I have a bag with 20 numbered balls from 1 to 20. I will draw 3 balls with replacement. What is the probability of drawing 3 balls in an descending order?

To answer this question, I can only think of considering each scenario. Like the probability of drawing number 20 is $$1/20$$. Then I can pick any ball but not number 1, which happens with a probability of $$18/19$$. Then I pick number 1 which happens with a probability of $$1/18$$.

Is there a way to formulate this instead of spending years to add these scenarios?

Take an overview: First, in order to be descending the balls have be be different, which happens with probability $$19(18)/20^2.$$ Conditional on that, the three balls may be drawn in any of $$3! = 6$$ orders, of which only one is descending order, so the probability is $$P(\mathrm{Descend}\cap\mathrm{Distinct}) = P(\mathrm{Distinct})P(\mathrm{Descend}\,|\,\mathrm{Distinct})\\ = \frac{19(18)}{20^2(6)} = 0.1425.$$

In case it is of interest, here is a simulation of $$100\,000$$ three-draw experiments in R. With this many iterations, the probability should be accurate to at least a couple of decimal places.

set.seed(515)
m = 10^6;  desc = logical(m)
for (i in 1:m) {
x = sample(1:20, 3, rep=T)
desc[i] = ( (x[2] < x[1]) & (x[3] < x[2]) )
}
mean(desc)
[1] 0.142528


Notes on R code: The sample function draws three balls with replacement from among 1 through 20, ylielding a 3-vector x. The notation x[k] gives the $$k$$th element of the 3-vector. In R, & stands for intersection. At the end of the run, logical vector desc contails $$m= 100\,000$$ TRUEs and FALSEs. Its mean is the proprotion of its TRUEs.