Consider shift cipher So here, we have K = {0,..., 25) with Pr[k +K]= 1/26 for each key k. Now, we are given the following distribution over M: Pr[M=a] = 0.7 and Pr[M= z) = 0.3. What is the probability that the ciphertext is B?

To solve this problem, we will need to account for both the key distribution and the message distribution. This is essentially a case of calculating the probability that a specific ciphertext (B) is generated, given the key distribution and the distribution over the plaintext messages \( M \).

### Step-by-step process:

#### 1. Ciphertext generation formula:
In a shift cipher, the ciphertext \( C \) is generated from the plaintext \( M \) by shifting it by some key \( k \) modulo 26. That is:

\[
C = (M + k) \mod 26
\]

Given that we are asked for the probability that the ciphertext is \( B \), let's translate \( B \) into its corresponding numeric value:
- \( B = 1 \) (since \( A = 0, B = 1, C = 2 \), etc.)

Now, we need to account for both possible plaintexts and the corresponding shift keys that could lead to a ciphertext of \( B = 1 \).

#### 2. Case 1: \( M = a \) (where \( a = 0 \)):
- The probability that \( M = a \) is given as \( Pr[M = a] = 0.7 \).
- We want \( C = B = 1 \), which means:
  \[
  1 = (0 + k) \mod 26
  \]
  This simplifies to \( k = 1 \). So, the only key that satisfies this is \( k = 1 \).

- The probability of selecting this key \( k = 1 \) is \( Pr[k = 1] = \frac{1}{26} \).

Thus, the probability that \( M = a \) and \( k = 1 \) results in \( C = B \) is:
\[
P(C = B | M = a) = Pr[M = a] \times Pr[k = 1] = 0.7 \times \frac{1}{26}
\]

#### 3. Case 2: \( M = z \) (where \( z = 25 \)):
- The probability that \( M = z \) is given as \( Pr[M = z] = 0.3 \).
- Again, we want \( C = B = 1 \), so:
  \[
  1 = (25 + k) \mod 26
  \]
  This simplifies to \( k = 2 \). So, the only key that satisfies this is \( k = 2 \).

- The probability of selecting this key \( k = 2 \) is \( Pr[k = 2] = \frac{1}{26} \).

Thus, the probability that \( M = z \) and \( k = 2 \) results in \( C = B \) is:
\[
P(C = B | M = z) = Pr[M = z] \times Pr[k = 2] = 0.3 \times \frac{1}{26}
\]

#### 4. Total probability:
Finally, we sum the probabilities from both cases to find the total probability that the ciphertext is \( B \):

\[
P(C = B) = P(C = B | M = a) + P(C = B | M = z)
\]
Substituting the values:
\[
P(C = B) = 0.7 \times \frac{1}{26} + 0.3 \times \frac{1}{26}
\]
\[
P(C = B) = \frac{0.7 + 0.3}{26} = \frac{1}{26}
\]

Thus, the probability that the ciphertext is \( B \) is \( \frac{1}{26} \).

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Let me know if you'd like further clarification or have additional questions!

Here are 5 related questions:
1. How does the key distribution affect the overall probability in shift ciphers?
2. How would the probability change if the key space were larger?
3. What is the probability of obtaining any other letter as the ciphertext?
4. How would the solution change if the message distribution were uniform?
5. Can this approach be generalized to other ciphers, and how?

**Tip**: When working with modular arithmetic, always double-check edge cases where values might "wrap around" (e.g., going from 25 back to 0 in the alphabet).

Consider shift cipher So here, we have K = {0,..., 25) with Pr[k +K]= 1/26 for each key k. Now, we are given the following distribution over M: Pr[M=a] = 0.7 and Pr[M=z] = 0.3. What is the probability that the ciphertext is B?

Explore how to calculate the probability of getting ciphertext B in a shift cipher. Learn how key and message distributions affect cryptographic outcomes, using modular arithmetic and probability concepts. Suitable for students in Grades 11-12 and undergraduate level.

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