How often is data corrupted given how often a 3-way vote passes incorrect data?

Question

Assume we have a noisy system where data is available via sample() and in order to filter out the noise someone has implemented the following voting algorithm:

sample_a = sample();
sample_b = sample();
if (sample_a != sample_b) {
    sample_c = sample();
    if (sample_a == sample_c)
        return sample_a;
    else if (sample_b == sample_c)
        return sample_b;
}
return sample_a;

So it's almost a voting algorithm where we select the result appearing at least 2 out of 3 times. In the case where the first two match there is no third sample, but since the samples are not selected from a set but by adding a new "trial" I don't know if it matters that the third test is sometimes omitted.

My question is: If I have an estimate of how often the result of the above algorithm still yields an incorrect sample (based on outside knowledge of what makes a valid sample) can I reason back from that to the failure rate of the base sample() function?

My thinking: Take $P$ = probability of a bad sample. Consider the 8 combinations of right/wrong results in a 3-way vote: half of those cases have at least 2 failures. The 3 with two failures each have $P^{2}(1-P)$ chance of being wrong (two wrongs, one right) and the all-wrong case is $P^{3}$. The probability of the vote being wrong is then $3P^{2}(1-P)+P^{3}$ which simplifies to $-2P^{3}+3P^{2}$ which at least goes to the right limits (0 for $P\rightarrow 0$ and 1 for $P\rightarrow 1$). Leaning on Wolfram Alpha to solve that for me I get:

(I'm sure my tag selection is poor -- If I knew the right tags I would probably know the answer!)

Answer 1

Everything you say is reasonable. Your binomial method works as whuber has said. Alternatively, you could add the probability of the first two being wrong, the first being wrong the second right and the third wrong, and the first right and the second and third wrong to get the same result: $P^2 + P(1-P)P + (1-P)P^2 = 3P^2-2P^3$.

I personally would not try to solve a cubic equation that way using complex numbers. If you have a particular value of $X$, you could find $P$ through numerical methods, or using a look-table, or very approximately using a graph like the following.