Red and black on roulette table (without green "0"). The outcome is independent of prior outcomes, right? It does not matter how many times in a row black (or red) comes up prior, the outcome is still random and 50/50. What is the formula that proves this?
Old question, but still unclear and would like to have an answer.
On
And AGAIN there is no 90% probability of a switch or the number of times in a row of any series like the one here comes up 1 ... n times in a row, as it is NOT CONDITIONAL on the earlier outcome(s) .... Einstein did provide the proof here mathematically but I can't find it. Anyone knows where I can find it ?
Right. But there is no "formula" to prove it. That's the definition of a fair wheel - it has no memory of what came before. Each spin is a fresh start.
The same is true for flipping a fair coin. There's no difference between flipping one coin ten times or ten coins all at once. In each case it's 50/50 for each toss independent of the others.
It is true that if you flip a fair coin over and over again in the long run you see half heads and half tails on average. But that's not because the coin thinks it has to catch up after a run of one kind or the other.
There's some serious mathematics that tells you how many flips you should make in order to be within some interval around 50/50 with, say, $90\%$ probability.