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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.

V.Mach
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  • It's a correct calculation of something related to the game, but maybe not of what you understood it to be. What are his precise words he uses to describe what he's calculating? – kimchi lover Nov 25 '18 at 13:46
  • Red Dwarf is a sitcom, and Arnold J. Rimmer is an idiot. It would be a miracle is he got anything right. – Angina Seng Nov 25 '18 at 14:02
  • @kimchilover The game itself is irrelevant. The whole question really just boils down to the fundamental idea - the probability of the outcome of a random event which has been repeated number of times in a sequence. – V.Mach Nov 25 '18 at 18:23

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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 $\frac{1}{13},\,\frac{1}{17},\,\frac{1}{25},\,\frac{1}{49}$).

J.G.
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  • Thank you for your answer. It still isn't 100% clear though. According to this answer: https://math.stackexchange.com/a/829918/619483 past outcomes do not influence future outcomes. Citing: "The mistake that previous outcomes affect future outcomes is commonly called the gambler's fallacy."

    This sounds like Rimmer is doing exactly that - thinking that his past unwanted outcomes should increase the probability of his wanted outcome in the next throw.

     – V.Mach Nov 25 '18 at 18:19
  • @V.Mach I don't have a transcript in front of me, but I doubt Rimmer made completely explicit whether he was stating $P(\forall i\le n X_i=(2,,1))=(1/18)^n$ (which is right) or $P(X_n=(2,,1)|\forall i<n X_i=(2,,1))=(1/18)^n$ (which is wrong). Even in a comedy setting, I find the latter less likely (since it's not an example of any common fallacy), and the former implies he doesn't think the past influences the future. – J.G. Nov 25 '18 at 18:24
  • The point is he throws up to 5 or 6 times and is gradually more and more surprised with each throw by the repeating unwanted outcomes. My question is why. If in every throw the probability of getting {2,1} is $\frac{1}{18}$, then the measure of his surprise shouldn't rise but rather be constant. – V.Mach Nov 25 '18 at 18:34