Probability that two random letters from a language will be the same? I'm trying to find the probability that two randomly-selected letters from "average" text in a language will be the same.  
For example, if my hypothetical language contains four letters which each occur on average with the following frequency:
A = 60%
B = 25%
C = 10%
D = 5%

What is the probability that selecting any two letters from a representational text will be the same?
My intuition for solving this is first to find the chance that they're different, so the sum over the probabilities that a letter is chosen and then some other letter is chosen next, over each letter in the alphabet:
(0.6  * (1 - 0.6)  + 
 0.25 * (1 - 0.25) + 
 0.1  * (1 - 0.1)  + 
 0.05 * (1 - 0.05)) = 0.565

Then the chance that they are the same: 
1 - 0.565 = 0.435

Is this reasoning sound? It seems like a very basic probability problem, but I always seem to be thinking about these things in the wrong way and would appreciate a sanity check (and any pointers to materials which would help me be more confident about this kind of thing in the future!)
 A: I think you can compute it by $A^2+B^2+C^2+D^2=0.6^2+0.25^2+0.1^2+0.05^2=0.435$
And $A^2$ means you will get $2A$s, $B^2$ means you will get $2B$s and so on.
A: You got the right answer but you kind of did it the hard way.
If $L_1$ is the first letter and $L_2$ is the second letter, then 
the probability is $\sum_x P(L_1=x) \times P(L_2=x|L_1=x)$. 
If you assume independence (which would be the case if you were selecting from the whole population at random for each letter, for example), that's
$\sum_x P(L_1=x) \times P(L_2=x)$.
But of course then the two probability distributions will be the same. Let's call the probability that a random letter has the value $x$ "$p_x$".
That is, the overall probability =  $\sum_x p_x^2$
In your case, that's 0.6*0.6 + 0.25*0.25 + 0.1*0.1 + 0.05*0.05 which comes to 0.435. 
-- 
generalization to $k$ letters:
To find the probability that $k$ letters selected from the population are all identical:
If $L_1$ is the first letter, $L_2$ is the second letter, ..., and $L_k$ is the $k^\mathrm{th}$ then 
the probability is $\sum_x P(L_1=x) \times P(L_2=x|L_1=x)\times P(L_3=x|L_1=x,L2=x)\times ... \times P(L_k=x|L_1=x,L2=x, ...,L_k=x)$. 
If you assume independence (which would be the case if you were selecting from the whole population at random for each letter, for example), that's
$\sum_x P(L_1=x) \times P(L_2=x)\times ... \times P(L_k=x)$.
But of course then all the probability distributions will be the same. Let's call the probability that a random letter has the value $x$ "$p_x$".
That is, the overall probability =  $\sum_x p_x^k$
