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?


1 Answer 1


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.

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]) )
[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.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.