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I'm trying to solve a question about one time pads but I'm not sure if my assumptions are correct.

I am to assume that I have a one time pad with perfect secrecy. This one time pad is used to encrypt a message of 7 bits in length (message space size is 2^7). This encryption yields a ciphertext of "0110011". What is the probability that the encrypted message was "1001001".

My understanding of one time pads is that if we assume the pad has perfect secrecy, then we can no nothing about the message given the ciphertext and thus we cannot determine the probability. Am I right in my assumption?

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  • $\begingroup$ Well, no, not exactly. We can determine the probability: It's the same for every ciphertext. $\endgroup$ – Nova Jan 31 '15 at 2:59
  • $\begingroup$ Hmm. Then my only other guess would be that the probability would have to be 1/2^7. $\endgroup$ – user2276280 Jan 31 '15 at 3:20
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    $\begingroup$ Yes, correct. That's the basic principle of the one time pad. You can't say which one of the encrypted messages is right, because the key is fully random and every message is equally likely. $\endgroup$ – Nova Jan 31 '15 at 3:37
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Actually, you are quite correct; you have not been given enough information to determine the probability.

The piece of information that you have not been given is the probability distribution that the plaintext was originally chosen from. Perhaps all 128 possible plaintexts were chosen with equal probability (that is, with probability $1/128$). Perhaps the plaintext 1001001 was chosen with probability 0.25, and the plaintext 0110110 was chosen with probability 0.75, and all other plaintexts had zero probability. The point is that you do not know.

The entire point of a one time pad is that knowledge of the ciphertext gives the attacker no new information about the probability distribution of the plaintext; that is, the ciphertext does not make any specific plaintext any more or less probable. The question asks you about the probability distribution given a specific ciphertext; we know that it is the same as it was without the ciphertext - however, we don't know what the original probability was.

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