Re: Confirmation of Shannon's Mistake about Perfect Secrecy of One-time-pad

Shannon misused Bayes' formula, similarly the above proof misused
Bayes' formula. From P(M = x)・P (K = (x?y)) = P(M = x) ・2-n, we can
see the condition that the ciphertext y is a fixed value is never
considered when computing P(M = xΛC = y). We can get that result by
reductio ad absurdum. Suppose for fixed y, if P (K = (x?y))=2-n (that
is used in the proof, but indeed it is wrong. It is used just to get
wrong conclusion), we can get P(M = x｜C = y)= 2-n because there is a
one-to-one correspondence between all the plaintexts and keys for the
fixed ciphertext in OTP. But it is obviously wrong, for the prior
probabilities of all plaintexts are seldom equally likely. So P(M =
x)・P (K = (x?y)) stand for the joint probability of x and y when y is
not fixed. But Shannon thought of the posterior probability as the
probability of plaintext when ciphertext had been intercepted, we can
see that there is a presupposition in P(M = x｜C = y) that y is fixed,
but in P(M = x), P (K = (x?y)) and P(C=y), y is not fixed, otherwise
we can get obviously wrong results. In such way, the Bayes's formula
was misused for the probability was not on the same presupposition and
the equation does not come into existence.
In OTP there are complex and crytic conditions that influence the
probability of plaintext, key and ciphertext, so it is essential to
cognize all the conditions and carefully use probability theory. The
proof did not realize the crytic condition that ciphertext was a fixed
value (even though unknown) rather than a random variable.

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