XXXVIII OM - II - Task 1

From an urn containing one ball marked with the number 1, two balls marked with the number 2, ..., $ n $ balls marked with the number $ n $, we draw two balls without replacement. We assume that drawing each ball from the urn is equally probable. Calculate the probability that both drawn balls have the same number.
To make the content of the task meaningful, we need to assume that $ n \geq 2 $.
In total, there are $ N = 1+2+ ... +n = n(n+1)/2 $ balls in the urn. Two balls can be chosen from $ N $ balls in $ q $ ways, where

Let us fix $ k \in \{2, \ldots,n\} $. From the $ k $ balls marked with the number $ k $, we can choose a pair of balls in $ \binom{k}{2} $ ways. The total number of ways to choose a pair of balls with the same number (from all $ N $ balls) is

Here is an inductive proof of formula (2): the formula is correct for $ n = 2 $; assuming its validity for some $ n \geq 2 $, we have for $ n+1 $

Thus, formula (2) is proved.
The sought probability is the quotient of the found number (formula (2)) by $ q $ (formula (1)), i.e., it equals $ \dfrac{4}{3(n+2)} $.