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

Question

$100$ 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)?

Answer 1

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

Answer 2

Another approach is to use an approximation based on the Poisson distribution which is easier to calculate.

There are $\binom{100}{2}$ pairs of people. If we assume each one independently picks an integer from 1 to 4900 with each number equally likely to be chosen, then any given pair will match with probability $1/4900$; so the expected number of matches is $$\lambda = \binom{100}{2} \frac{1}{4900} = 1.0102$$ The matches are not independent, but they are "almost independent", so we can expect that the total number of matches has a distribution which is approximately Poisson with mean $\lambda$. Then the probability of zero matches is about $e^{-\lambda} = 0.36414$, which is not bad, compared with the exact value of $0.36164$.