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By that statement, does that mean adding a random character to a random position in a Diceware word adds 10 bits to each word? No. The ten bit estimate is for adding a random symbol from the 36-item table to a random position in the passphrase. The entropy in the character choice is about five bits and the entropy in the choice of position is another ...


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 ...

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