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We have a one time pad where:

  1. We have 8 bits. probability that a bit=0 is p.

  2. The key is common for all 8 bits. It can have a value of 0 or 1 with equal probability.

  3. After applying the XOR operation we take the encrypted result that has n zeros and 8-n ones.

What I am asked to do is calculate the probability that the key used is 0.

Until now I am thinking like this:

There are 2 events:

EventA. We have k zeros in the original 8 bits

EventB. We have k zeros in the encrypted data.

P(A) can be calculated from the binomial distribution P(B) = 2P(A) I am searching the probability P(key=0|P(B)).

Is my thinking correct until now? How can i procede from there?

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  • $\begingroup$ Welcome to CryptoSE! :), Just to clarify, your message is 8 bits long but your key is only 1 bit? and you use that one bit to xor with the 8 bits of the message and then get 8 bits ciphertext? If that's the case then consider that the key is sampled uniformly at random and independently of the message. $\endgroup$ – Marc Ilunga Mar 7 at 12:30
  • $\begingroup$ Hello! Yes exactly that was the case. Key is 0 or 1 for the whole block of 8-bit text to cipher! $\endgroup$ – baskon1 Mar 7 at 16:31
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Assume the message is an $n$-bit string with each bit drawn independently with $P(0) = p$.

Let $A$ be the event that the key bit is $0$.

Let $B$ be the event that there are $k$ zeroes in the ciphertext.

Then $P(A) = \frac{1}{2}$ and $P(B) = \frac{1}{2}{{n}\choose{k}}\left(p^k(1-p)^{n-k} + p^{n-k}(1-p)^{k}\right)$, since with equal probability we had $k$ zeroes or $n-k$ zeros in the message.

$P(B | A) = {{n}\choose{k}}p^k(1-p)^{n-k}$ since this occurs exactly when the message originally had $k$ zeros

Then $P(A | B) = \frac{P(A)P(B | A)}{P(B)} = \frac{p^k(1-p)^{n-k}}{p^k(1-p)^{n-k} + p^{n-k}(1-p)^k}$

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  • $\begingroup$ Thanks Tjaden Hess! What I cant understand is how P(B) was calculated. Why 1/2 in the front? $\endgroup$ – baskon1 Mar 8 at 7:30
  • $\begingroup$ It may make more sense if you distribute it out. You can also think of it as averaging over the two possibilities of the key bit. $\endgroup$ – Tjaden Hess Mar 8 at 13:43

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