$\begingroup$Say I pick a number from 1 to 100, and repeat this process 99 more times. What is the probability that no numbers will repeat?
Is it correct to say the answer is 1/100! or is there more to it?
$\endgroup$3 Answers
$\begingroup$Any number $1$ to $100$ can be the first, which occurs with probability $p_1 = 1$.
The second number must be any but the first, $p_2 = 99/100$
The third any but the first two, $p_3 = 98/100$
and so on.
Hence the desired probability is
$$\prod_{n=1}^{100} p_n = \frac{99!}{100^{99}} \approx 10^{-42}$$
which we can also write more pleasingly as $\displaystyle\frac{100!}{100^{100}}$.
By the way, this is about $116$ orders of magnitude larger than $1/100!$.
$\endgroup$$\begingroup$No, you are picking with replacement. There are $100^{100}$ ways to pick the numbers. Of those $100!$ have no repeats, as you have $100$ choices for the first one, then $99$ choices that do not duplicate for the second, and so on. The probability is then $\frac {100!}{100^{100}}$
$\endgroup$1$\begingroup$There are $100^{100}$ different ways to a between $1$ and $100$ a hundred times, but only $100!$ ways if you do not allow repeats, so the probability is $100! \over 100^{100}$
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