I am very lost. I would like to know how I calculate the probability of finding errors in repetition code.
For example say the codewords are 111 or 000 and probability of an error is 0.1. How would I find the probability of having one error in the transmitted codeword? i.e the chance that 000 becomes 001, 010, 100 or 111 becomes 110, 101, 011?
Am I correct to believe that this formula is the answer?
Where P is the probability 0.1
$$3p^{2}(1 − p) + p^{3} = 0.028$$
If so, how would I compute the chance of receiving a codeword with one error with a repetition code of (4,1) or (5,1), (6,1) etc. I cannot find any examples that make this clear or if I'm even on the right tracks.
Best Answer
What you need is not to know "the formula", but where it comes from.
In $(3,1)$ repetition code, we code the
0as000. The decoding (assuming a BSC channel with crossover (bit error) probability $p<1/2$) is by majority: if, say, we receive001we decide that it's more probable that the trasmitted codeword was000than111. Why? Because it's more probable to have a single bit error (000$\to$001: last bit was changed) than having two (111$\to$001: two first bit changed).When does this decoding strategy fail? It should be clear that the decoding is wrong where there are indeed two bit-errors ... or three. If we trasmit
000and the first and last bit are flipped, so we receive101, the decode will guess :111and we'll have decoding error. Conclusion: the event "decoding error" is equivalent to the event "having more than half bit errors". The formula you copied computes this, the first term correspond to the 2-bits errors (there are 3 possible cases) and the last term for the 3-bits error.If you understood the above, then you should be able to compute the probability of bad decoding for a repetition $(5,1)$ code (I've never seen a $(4,1)$ repetition code ; even lengths are avoided to get rid of ties).