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You are correct, but Leibniz, despite being a mathematical genius, got it wrong and assumed 6,5 was the same outcome as 5,6 and therefore equally likely to get as 6,6.
In his defence probability was a new field at the time, and we only understand it better with the benefit of centuries of history on our side.
The question asks
"how many 2d6 rolls is Liebniz expected to need before he can say they're not [equally likely]?"
But it doesn't say what degree of certainty you'd need to have in order to say that.
Consider the question
how many throws of a die would you need before you could say that it was a loaded die?
This is a similar kind of question where you compare your hypothesis with the observations.
So you'd need some way of comparing your observed results against the expected results for that many throws.
If, after n throws, our outcome is x instances of 11 and y instances of 12, does that fall within the realms of chance, or do we need to revise our model?
edit: This is a slow response to a comment far upthread...
Drakien-
Yes, an extremely clear statement of the problem.
It occurs to me now that the question is a bit nonsensical; ironically Leibniz wouldn't have been aware of confidence intervals, certainty, etc., so wouldn't actually have been able prove anything empirically anyway! But it's still an interesting question.
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?