Given three integers in $\{0,\ldots,100\}$ which sum up to $100$. What is the probabilty that two of them are the same? We pick $3$ numbers (one by one) from set $\{0,1,...,100\}$.
What is probabilty that two numbers are the same if sum of those $3$ numbers is $100$?
My solution:
Which two are the same we can pick in $\binom {3}{2}$ ways. Suggest $x_2=x_3$- we need to find compositon $x_1+x_2+x_2=100 \implies x_1+2x_2=100$ which implies that $x_1$ is even so we can divide this by $2$. Now we get $y_1+y_2=50$ , and using formula there are $$\binom{50+2-1}{2-1}=51$$ compositions. So, probability is $$\frac{51*3}{\binom{100+3-1}{3-1}}$$
Is this right answer?
P.S.$\binom{100+3-1}{3-1}$ is number of compositions of 100 into 3 parts (allowing $0$)

First, we count the triples $(a,b,c)$ such that $a + b + c = 100$.
For the choice of $a,b$ we need $a + b \leq 100$. Then $c = 100 - a - b$ is uniquely determined.
For $a + b \leq 100$, the number of possibilities is
$$101 + 100 + 99 + \ldots + 2 + 1 = \frac{102\cdot 101}{2} = 5151.$$
Now we count the triples with the additional condition that two numbers are the same. Note that $a = b = c$ is not possible (otherwise, $100 = a + b + c = 3a$ would be a multiple of $3$).
So there are $3$ possibilities to choose the identical pair of numbers.
For a single such choice, say $a = b$, we have the $51$ possibilities $a\in\{0,\ldots,50\}$ s.t. $a + b = 2a \leq 100$. Again, $c$ is uniquely determined.
So in total, there are $3 \cdot 51 = 153$ triples $(a,b,c)$ such that two numbers are the same and $a + b + c = 100$.
This gives the resulting probability of
$$\frac{153}{5151} = \frac{3}{101} \approx 3.0\%.
$$
