# Simple One time Pad Probability

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?

• 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. – Marc Ilunga Mar 7 '19 at 12:30
• Hello! Yes exactly that was the case. Key is 0 or 1 for the whole block of 8-bit text to cipher! – baskon1 Mar 7 '19 at 16:31

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

• Thanks Tjaden Hess! What I cant understand is how P(B) was calculated. Why 1/2 in the front? – baskon1 Mar 8 '19 at 7:30
• 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. – Tjaden Hess Mar 8 '19 at 13:43