Why does a single bit error when using the CBC block cipher mode of operation affect the decryption of a block (Oj) with a probabilty of 50%? And in the following block, why is the one bit error in the exact position as in cipher block? Differently worded: why is the 1 bit error propagation 1:1?

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    $\begingroup$ I'm not sure the first sentence has a correct assumption. It seems to to me that this is about a bit having a random value. The change of the bit being erroneous is then 50%, which means that in 50% of the time - on average - the decryption of the block is affected. If the bit is indeed in error the chance would be 100%, right? $\endgroup$
    – Maarten Bodewes
    Commented Oct 30, 2015 at 11:37

1 Answer 1


If you have an error in a cipher text block you can generally represent this as:


Now if you try to decrypt this block using the previous ciphertext block $IV$ as IV you get $P'=IV\oplus D_K(C\oplus\Delta)$ which is completely unrelated to $P=IV\oplus D_K(C)$ assuming the block cipher acts as a pseudo-random permutation. As the input changed the chance for each plain text bit flipping are 50% (independently), everything else would be considered a weakness of the cipher.

Now for the second part:
Assume you have the errorneous message $C_1'=C_1\oplus\Delta$ and a normal $C_2$. The decryption $P_2'$ is then $P_2'=C_1'\oplus D_K(C_2)=C_1\oplus\Delta\oplus D_K(C_2)$ and as $P_2=C_1\oplus D_K(C_2)$, you see that $P_2'=P_2\oplus\Delta$ meaning the error of the previous block propagated perfectly into the next plain text.


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