The faces are equally likely, they just add up differently.
Really? You think that is the answer?
Let X1 be r.v. for outcome of die 1
Let X2 be r.v. for outcome of die 2
P( X = 2 ) = P( X1 = 1 ) * P( X2 = 1 ) = 1/36
P( X = 3 ) = P( X1 = 1 ) * P( X2 = 2 ) + P( X1 = 2 ) * P( X2 = 1 ) = 2/36
and so forth.
E[X] = 7
Edit: I clearly misunderstood the OP
This was solved this past summer over at 2+2 forum by the handsome math guru BruceZ.
I used the BruceZ method and agree with his answer.
For now, I don't think I can explain his answer any better than he did, but will try for my next Ask the Wizard column. Meanwhile, the following table shows the expected number of rolls to get a 2, 2-3, 2-4, 2-5, ...
|Highest Number||Prob||Exp wait||Prob rolled before last||Prob rolled last||Exp rolls|
You want to know the average wait time for all 11 numbers to roll from 2 dice. Excellent math question.
This is not even a normal distribution.
The mean of 61.2 is not even close to a mode of around 35-38 and they are not close to the median of 52.
I would like to see someone here at WoV create a formula that would solve for the distribution.
It's not normal, it's Poisson. See this post for the formula (from a textbook by Sheldon Ross).