Subject: Re: Probabilities with two die
Date: Wed, 3 Dec 1997 18:07:33 -0500
From: Alex Bogomolny
A discussion may have no end unless a question at hand has been unambiguously formulated. I take it, in case of two die, the question is:
What is the probability of having at least one 1?
The answer to this question is 11/36. This follows from the Inclusion-Exclusion principle (6+6-1) or by direct enumeration. In all, there are just 36 possibilities. Arrange them in a 6x6 square like matrix indices. You will have 1s in the first row and in the first column. But the first diagonal square will be counted twice which results in -1 above.
A different question would have a different answer. For example, What is the probability of having exactly one 1? The answer is 10/36.
The only question that leads to 1/6 I can think of is What is the probability that the first die will show 1?