Any maths buffs?

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frankenbike

Problem that I came across. Leibniz thought that rolling two die and getting 11 was as likely as getting 12. If he took two die and started rolling them,

"how many 2d6 rolls is Liebniz expected to need before he can say they're not [equally likely]?"

I bodged an answer with Python but maybe a proper stats/Bayesian person can give a real answer?

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GreatSince78 in reply to @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?

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Brun in reply to @GreatSince78

Dices.

Grammar buffs thread >>>>>>>>>>>>>>
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Greenbank in reply to @GreatSince78

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%?
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Drakien in reply to @GreatSince78

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...

Any maths buffs?

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