Q:

$100$ persons think of a number between $1$ and $4900$. Probabilty: Thinking of same number vs. $1$ number chosen by multiple people

$100$ persons think independently of a number between $1$ and $4900$. 
What is more likely?

All people have thought of different numbers $(1)$
A number has been chosen by more persons $(2)$

It reminds me of the Birthday problem.
I would say 
P$(1) =(\frac{1}{4900})^{100} \times (4900 \times 4899 ... \times 4800)$
P$(2) = 1 - P(1)$
I think that this is wrong.
Can someone tell me how its done (by hand)?

A:

Your reasoning is correct other than you should have
$$\mathsf P(1) =\left(\frac{1}{4900}\right)^{100} \cdot (4900 \cdot 4899 \cdots 4801)$$
Most software that I'm familiar with won't give a numerical answer to $$\frac{\frac{4900!}{4800!}}{4900^{100}}$$
One work around would be to implement a for loop. In R statistical software:
prod = 1
for(i in seq(4801,4900,1)){prod=prod*(i/4900)}
prod

> 0.3616378

1-prod

> 0.6383622

