TITLE: If the gambler's fallacy is false, how do notions of "expected number" of events work?
QUESTION [19 upvotes]: Imagine there is a fair spinner that could land on any number $1$ through $100$. I understand that the chance of any number appearing on the next spin is $\frac{1}{100}$, and if you spin the spinner $100$ times then the expected number of $5$'s, for example, is $1$.
However, I have the following questions:

If the spinner does not land on $5$ in the first spin then this does not make the chance of getting a $5$ on the second spin any more or less likely (to assume otherwise would imply the gambler's fallacy). This means that we now must spin the spinner a total of $101$ times in order to expect exactly one $5$. Doesn't this cause some kind of infinite regress? If we never expect a $5$ on any given spin, and no $5$'s have come up thus far in the first $n$ spins, then won't it take $n+100$ spins to expect a 5? I know that at some point if we examine a group of spins then we can expect something with low probability like spinning a $5$ to occur, but it seems strange and counter-intuitive to me that though we expect a $5$ to come up in a sufficiently large group of spins, on each individual spin we do not expect a $5$. Furthermore, we do not expect any number to come up on the first spin (in that the probability of any particular number appearing is low), and yet we know for certain that some number will still come up.
For there to be a greater than $50\%$ chance of spinning a $5$, we must spin the spinner $69$ times according to the following calculation:
$$P(\text{Not rolling a 5 in } n \text{ spins})=\left(\frac{99}{100}\right)^n \\ \left(\frac{99}{100}\right)^n < 0.5 \\ \log_{0.99}0.5=68.967... $$ Hence, it must be spun $69$ times for there to be a greater than $50\%$ chance of there being a $5$. Why is this not $50$ spins, as we expect $0.5$ $5$'s to come up in this period? Also, I have the same question that once a $5$ does not come up, don't we have to spin it $70$ times for there to a greater than $50\%$ probability, and can't this cause the same infinite regress as described above?

REPLY [3 votes]: People's answers focused on probability calculation, but the thing that bothers me when I read your question is more semantic: what exactly is meant by "if you spin the spinner 100 times then the expected number of 5's is 1."
What you actually mean by "expected" may be the root of the "problem", the expected number of 5's is in average 1. Not just 1. Could be 0, could be 10. You seems to get the math to be able to calculate the probability for each of these event.
The invariant here is: whatever happened before, for the next 100 spins, the expected number of 5's remains in average 1.