Understand LWE(Learning With Error) negligible error probability According to Regev's paper, p15 


  Correctness. Note that if not for the error in the LWE samples, $b-⟨a, s⟩$ would be either 0 or ⌊ q ⌋ depending on the encrypted bit, and decryption would always be correct. Hence we see that a decryption error occurs only if the sum of the error terms over all S is greater than q/4. Since
  we are summing at most m normal error terms, each with standard deviation αq, the standard deviation of the sum is at most $\sqrt{m}\alpha q &lt; q \log{n}$; a standard calculation shows that the probability that such a normal variable is greater than q/4 is negligible.


I did a little experiment with mathematica and the resulting probability of decryption error is around 0.28. Can you point out where I was wrong?

Firstly I build the the following variables as in the paper:

In[1]:= n=10
        q=RandomPrime[{n^2,2n^2}]
        m=1.1*n*Log[q]
        α=1/(Sqrt[n]*Log[n]^2)
        σ=α * q
        Floor[q/2]
        q/4
Out[1]= 10
Out[2]= 131
Out[3]= 53.6272
Out[4]= 1/(Sqrt[10]*Log[10]^2)
Out[5]= 131/(Sqrt[10]*Log[10]^2)
Out[6]= 65
Out[7]= 131/4


Then I calculate the maximum probability that the error terms sum is greater than q/4 by calculating the value of the Cumulative Probability function when x is -q/4:

In[323]:= (*Probability that sum is greater than q/4*)
          CDF[NormalDistribution[0,Sqrt[m]*α*q],-q/4]
Out[323]= 0.279785


As you can see, it's too large for "negligible" so I think there's something wrong with my calculation.

Here are the PDF for a single error $e$ and the sum of errors:


Single error


Sum of errors
The probability of error is negligible "as a function of $n$", meaning that the probability of error will decrease (quickly) as $n$ grows. Increasing $n$ should solve your issue.