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One of the factors that determines how hard it is to forge a MAC for a given message is how long the MAC is. If it's 1 bit long, you can definitely produce the correct MAC in two tries. $2^n$ is the number of possible bit-strings of length $n$; $1/2^n$ is the probability that any random bit-string happens to be the MAC (of length $n$) for a given message ...


Yes, that is sufficient. I realise this isn't the most helpful of answers, but I'm not quite sure how we're supposed to deal with questions that lend themselves so neatly to one word answers. Certainly it seems wrong that something that has been solved should continue to sit in the 'unanswered questions' column!

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