Miller Rabin - Error probability of .5 a possibility?

I'm testing the property of Miller Rabin that the error probability is at most 1/4 when only a single base a is chosen and we iterate only one time. We are testing odd integers 90,000 to 100,000.

I've written up the implementation in Java and as the test is running, I'm seeing a lot of probabilities of .5. This leads me to believe that there is an issue with my implementation.

Some of the odd integers in which I'm seeing a .5 error probability are:900079157194343

There are plenty more (the test is still running).

Update: Here is the Algorithm I've implemented

 Miller–Rabin Primality Test Input: prime candidate ˜ p with ˜ p−1 = 2ur and security parameter s Output: statement “˜ p is composite” or “˜ p is likely prime” Algorithm: FORi = 1 TO s    choose random a ∈ {2,3, . . . , ˜ p−2}    z ≡ ar mod ˜ p       IF z ≡ 1 and z ≡˜ p−1          FOR j = 1 TO u−1             z ≡ z2 mod ˜ p             IF z = 1                RETURN (“˜ p is composite”)          IF z = ˜ p−1             RETURN (“˜ p is composite”) RETURN (“˜ p is likely prime”)

Here is the implementation, if anyone could take a look and determine what the problem is I would really appreciate it.

Thanks

public BigInteger mr(int x, int y){    int u = 0;    BigInteger p = BigInteger.valueOf(x);    BigInteger r = p.subtract(ONE);    BigInteger a = BigInteger.valueOf(y);    while (r.mod(TWO).equals(ZERO)){        u++;        r = r.divide(TWO);    }    BigInteger z = a.modPow(r, p);    if ((!z.equals(ONE) && !z.equals(p.subtract(ONE)))){        int j = 1;        for (; j < u; j++){            z = z.modPow(TWO, p);        }    }    return z;}public boolean isPrime(int n){    if ( n % 2 == 0)        return false;    for (int i = 3; i <= Math.sqrt(n) + 1; i+=2){        if (n % i == 0)            return false;    }    return true;}public static void main(String[] args) {    double ea;    MillerRabin mr = new MillerRabin();    int count = 0;    BigInteger ans;    for (int n = 90001; n< 100000; n+=2){        count = 0;        for (int a = 1; a < n; a++){            ans = mr.mr(n, a);            if (mr.isPrime(ans.intValue())){                count++;                }        }        ea = ((double)count) / (n-1);        System.out.println(ea);    }}