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So, forget the dice.
You're interested in two outcomes: A and B (and there are another 9 outcomes that we're not interested in).
One of them has a probability of 1/36 and the other 1/18.
How many outcomes do you need to observe to work out which is which?
Obviously observing just one outcome is not enough, nor is two. Observing an infinite number will be enough. So the answer is somewhere between those two.
My point is that the question needs to be bounded with a confidence interval, i.e.
How many outcomes do you need to observe to determine which is which at a confidence at a level at/over 50%/75%/90%/95%?
Greenbank-
The way I calculated it was to take a sequence of rolls and calculate the likelihood of that series of rolls given that 11 and 12 are equally likely. If it was less than 0.05 then I would end the sequence there and count the number of rolls. Do that a bunch of times and then take the average. That gave me an answer of 28 rolls with value 11/12 which means about 336 rolls total.
But I guess you can take any confidence value you want; the calculation will presumably be the same.
@GreatSince78 this was before probability so Leibniz couldn't do that calculation. He thought that they were equally likely based on intuition. The question is, under the assumption that they're equally likely, how long would it take before you've proven empirically that they're not?
frankenbike
I may have misunderstood the question and be massively oversimplifying, but you wouldn’t actually need to roll any dice to calculate the expected probabilities of the two outcomes would you?
With two die there’s a total of possible 36 outcomes, only one of which produces 12, whereas there are two outcomes resulting in 11, so given infinite rolls you’d expect to see twice as many 11’s rolled as 12’s right?