mfb said:
They have the same probability for every throw (assuming they are fair).
I might not have my terms correct. Craps is based on 7 happening more often (I think the term for that is probability). But if there were an infinite number of throws, won't all combinations happen an equal (that is infinite) number of times?
When you think about doing a process an infinite number of times, saying 'each pair of dice comes up an infinite number of times' is not good enough. It's nice to know but if you want to say that all possibilities really come up equally often, then what you are really interested in is the following:
Let X
n, Y
n be the results of the two dice on the nth throw, and let
S_N = \left( (X_1,Y_1),(X_2,Y_2),...,(X_N,Y_N) \right)
To say that, for example the pair (2,2) occurs as often as the pair (4,1), what you are interested in is (my notation here isn't great but should be comprehensive)
\# \left( (2,2) \in S_N \right) /N = \# \left( (4,1)\in S_N \right)
that is that you want the number of times (2,2) shows up in S
N to be equal to the number of times (4,1) shows up. Now this is generally not going to be true, but what is true (and is what we mean when we say that (2,2) and (4,1) have equal probabilities) is that
\lim_{N\to \infty} \frac{ \# \left( (2,2) \in S_N \right)}{N} = \lim_{N \to \infty} \frac{ \# \left( (4,1) \in S_N \right)}{N}
those limits are just going to be the probability of (1,2) showing up in any roll, or (4,1) showing up in any roll, which is 1/36. If you had a loaded die which came up on a value of 1 with a probability of 9/10 and a 2,3,4,5,6 each with probability 1/20, then we would get that every pair (a,b) still shows up infinitely often as N goes to infinity, but you would not have
\lim_{N\to \infty} \frac{ \# \left( (2,2) \in S_N \right)}{N} = \lim_{N \to \infty} \frac{ \# \left( (4,1) \in S_N \right)}{N}
because the left hand side would be 81/100 and the right hand side would be 1/400.
