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Consider having the worst possible communications channel, where bits are flipped with probability 0.5.

Is there any probability that channel errors make the change in a frame go un-detected through CBC-MAC?

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  • $\begingroup$ What is a worst channel? $\endgroup$
    – Ella Rose
    Commented Aug 2, 2018 at 14:27
  • $\begingroup$ Sorry for worst channel, I know it sounds ridiculous. Consider a channel with bit error probability of 0.5. Or Is it possible if channel errors go un-detected through CBC-MAC? if yes what could be the probability. $\endgroup$ Commented Aug 2, 2018 at 14:52
  • $\begingroup$ Considering a channel flips the bits in a way, that MIC check results in Valid output. Is it possible? If yes then with which probability. Considering AES-128 and tag length of 4 bytes. $\endgroup$ Commented Aug 2, 2018 at 14:54

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Is there any probability that channel errors make the change in a frame go un-detected through CBC-MAC?

For any Message Authentication Code, there is always a probability [1] that the modified message will just happen to verify against the modified tag.

In your extreme "worse possible case" scenario, it's quite simple to compute; your communication channel effectively replaces the message/tag with a completely random message/tag. And, if we compute the MAC on a random message, the probability that it matches a random tag is $2^{-32}$ for a 32 bit tag.


[1]: Unless the errors are tightly constrained. For example, with CBC-MAC with a full length tag, we will always detect any error if the error is constrained to be always within one block, with all the other blocks being transmitted correctly. However, random errors (and active attackers) are generally not so accommodating with how they modify the message.

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  • $\begingroup$ ThankYou poncho, This is valid answer. I have one more small question, will the number of messages sent will change that probability. Consider an example of BLE, where CCM is used. BLEv4.0 has frame length of 36 bytes. If i am sending 1MB of data. With what probability the data will go un-authenticated? Thanks $\endgroup$ Commented Aug 3, 2018 at 14:16
  • $\begingroup$ @faheemawan: if you're asking whether any of the messages will be accepted, well, the probability that no messages will be accepted is $(1 - 2^{-32})^n$ (where $n$ is the number of messages). So, if you are sending 29127 36 byte messages (which is approximately 1MB), the probability that all messages will be rejected is about $1 - 6.78 \times 10^{-6}$ $\endgroup$
    – poncho
    Commented Aug 3, 2018 at 20:42

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