An undecided probability problem

126 Views Asked by At

One person ( or n persons for that matter ) equipped with a true random number generator is set to generate the first n digits of π. The probability of doing this on the first trial is of course $(\frac{1}{10})^n\\$.

If the experiment is to be repeated m times ( trials ), the probability of hitting the first n digits of π ( or a given sequence of digits ) on m independent trials is of course:

$\ 1-(1-(\frac{1}{10})^n)^{m}\\$

Now, we are going to set m as a function of n and see what happens as n approaches infinity:

We see that:

$\lim_{_{n \to \infty} } (1-(1-(\frac{1}{10})^n)^{n})=0\\$, for m=n whereas

$\lim_{_{n \to \infty} } (1-(1-(\frac{1}{10})^n)^{10^n})=1-\frac{1}{e}\\$, for $m=10^n\\$

Indeed this probability can take any number from 0 to 1 depending on which type of infinity we use in our m independent trials.

My question is, is this a valid probability question to ask, or is it nonsense and makes no sense to ask mathematically. Thanks.