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The probability of the undetected errors for an n bit CRC is said to be $1/2^n$. But in practice, CRC also has bit errors in itself. Could I say that a 16 bit CRC reduces to a 13 bit CRC if the BER is 20%?

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For a random draw of bits, the probability of undetectable errors is $1/2^n$, but if you have non-random bits - for example, if you have a valid codeword that is divisible by the CRC generator polynomial and have a BER that is not 50% - then the probability changes and is dependent on the codeword size and BER.

For example, suppose the number of user bits is 1 and the CRC remainder is length p and these are detected with BER=20%. The codeword is then length $p+1$, and the only way you get an undetectable error is if all of the bits are in error. Aside from some bug that happened to flip all of the bits, this is less likely than an undetectable error from random draws, since the BER is less than 50%.

Example: p=3 with BER=0.2 $$ P_{undetectable} = 0.0016 = (1/5)^{p+1} < (1/2)^{p+1} = 0.0625 $$

Of course, this is a trivial case, but the notion is that it is harder to come up with an error polynomial divisible by the CRC generator polynomial than to do a random draw of bits (same code block size) that is divisible by the CRC generator polynomial.

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  • $\begingroup$ Thank you. In communication systems valid codeword is always sent. Then the probability of undetected errors depends on BER. Do I understand it correctly? $\endgroup$
    – c1119
    Commented Apr 21 at 2:37
  • $\begingroup$ Yes, depends on BER and on codeword size. Even though only valid codewords are sent, there can be cases in practice where the receiver expects a transmission but the transmitter missed the allocation and didn't send anything, or there is a false alarm in detecting a preamble/sync/training-sequence, and the follow-on reception is only noise. $\endgroup$
    – vml
    Commented Apr 21 at 16:59

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