Q:

What is the entropy per Diceware word if a random symbol is inserted into a random position in the word?

On the Diceware page is this little gem:

For extra security without adding another word, insert one special character or digit chosen at random into your passphrase...  Inserting a letter at random adds about 10 bits of entropy.

By that statement, does that mean adding a random character to a random position in a Diceware word adds 10 bits to each word?
What would be the entropy per Diceware word if random printable ASCII character is used instead?
What would be the entropy per Diceware word if a random Diceware word is used as the random symbol in a random position per Diceware word?
How exactly can this be calculated?

A:

If you add a truly random character into a truly random position of a word (uniformly chosen), you get "entropy of position" + "entropy of character" as addition to the entropy of the word. (Not exactly, it's a bit less). The entropy of character is the size of the possible characters. 64 possible characters would be $log2(64) = 6$ bits of entropy. Entropy of position are the numbers of possible positions where the character can be placed. With a word with a length of 4 characters you can get 5 positions. (length of the word plus 1. "foo" has 4 pssibilities: "?foo", "f?oo", "fo?o" and "foo?".)
Example: You got the word "hello". Now you want to add one character of a possiblity of $26+26+10 = 62$ characters (small and big letters and numbers) into the word. Possible outcome: "h9ello", "helalo", "heXllo", ...
Entropy of "hello": Let say 0, because it is a constant value in this example. Entropy of character: $log2(62) = 5.95$ bits. Entropy of position: $log2(4+1) = 2.32$ bits. Whole entropy: 8.27 bits. 
Problem: That's not entirely true, because you lose a bit entropy to a specific effect. Look at the example "hello": You could add an "e" after, but also before the e in the word. It's not possible to distingush this two possibilities ("heello" and "heello"), they are the same. That's one possibility less than we have calculated. You can do this for every letter of the word. Words like "hhhhh" are even worse, because every time the random character is a "h" you can't tell where in the word it was added.

