Using amplification lemma to decrease error probability gives 0 as divisor
If the Turing machine has an error probability of $1/2$, the Turing machine is useless and you cannot amplify to reduce the error bound.

Consider: if you just ignore the input and flip a coin and use that to decide whether to accept or reject, you get an error probability of $1/2$ -- yet that is an obviously useless procedure.  A Turing machine with an error probability of $1/2$ is no better than that.

Here's another way to see why it's hopeless.  One standard method of amplification is to run the Turing machine multiple times and take a majority vote.  Suppose we run the Turing machine 3 times, and take a majority vote.  What is the probability that the outcome of the majority vote is correct?  It's not hard to see that this probability is $1/2$ -- no better than running the machine once.  (Why?  Each time you run the Turing machine, it is either correct or incorrect.  Since you run it 8 times, there are 8 cases: CCC, CCI, CIC, CII, ICC, ICI, IIC, III, where C = correct and I = incorrect.  If the Turing machine's error probability is $1/2$, all 8 of those cases are equally likely.  The majority vote after 3 runs will be correct in the cases CCC, CCI, CIC, and ICC; thus, the majority vote is correct with probability $4/8 = 1/2$.)  The same thing happens no matter how many times you run the Turing machine.

To get amplification, you need the error probability to be strictly less than $1/2$.