I watched the new episodes Red Dwarf series XI. In episode 3 there is a scene where Rimmer and Lister play a game called minopoly. Rimmer is throwing a pair of dice while he wishes the outcome NOT to be "a two and a one".
Before his first throw he correctly calculates that probability of getting that unwanted result is $\frac{1}{18}$as there are 36 possible outcomes, two of them (two and one, one and two) are unwanted which yields $\frac{2}{36} = \frac{1}{18}$.
He then proceeds to throw again several more times getting the unwanted outcome every single time. With each new throw he announces the probability of getting the unwanted outcome in that throw. The probability he calculates is getting smaller and smaller because he multiplies the whole sequence of probabilities together so in $n$-th throw he gets $\frac{2^n}{6^{2n}} = \frac{2^n}{36^n} = (\frac{1}{18})^n $.
I wonder if that is a correct computation. Shouldn't the probability of getting the unwanted result with each new throw be always $\frac{1}{18}$? What am I missing?
Sorry if it is a dumb question but no one was able to explain it to me since high school.
Thank you for your replies.
While $(1/18)^n$ is the probability of everything so far, $1/18$ is the conditional probability of the latest upset given what came before, so both values are right on some reading. The Red Dwarf V episode Quarantine has a similar scene, where they again use cumulative rather than conditional probabilities. The luck virus allows Dave to pick all 4 aces from the pack, with successive probabilities $\frac{1}{13},\,\frac{1}{221},\,\frac{1}{5525},\,\frac{1}{270725}$. Again, the stated values are all correct, if interpreted as cumulative (otherwise Kryten should have said $\frac{1}{13},\,\frac{1}{17},\,\frac{1}{25},\,\frac{1}{49}$).