Probability that wrong candidate was selected

[eddybob123]

Hi everyone. It won't be that common to see me actually post my own thread, but I am having a little trouble with this problem. My friend was accepted into the University of Berkeley, and this was one of the questions on his first exam. Apparently, no one in his class got this one right, and I told him I would try to get the answer. Here is the problem:

There are two candidates: A and B. Candidate A got m votes and B got n votes. If the probability that a vote was incorrectly counted is p, then what is the probability that the wrong candidate was selected (in terms of m, n, and p).

(Starting to copy agent here)

(Bandit)

 

[Nono713]

I don't have time to solve this right now but I would guess that it would involve the binomial distribution, as in "probability of at least $m - n$ incorrect votes" or something close to that (in order to push the wrong candidate ahead of the legitimate one). What do you think?

 

[I like Serena]

Hi everyone. It won't be that common to see me actually post my own thread, but I am having a little trouble with this problem. My friend was accepted into the University of Berkeley, and this was one of the questions on his first exam. Apparently, no one in his class got this one right, and I told him I would try to get the answer. Here is the problem:

There are two candidates: A and B. Candidate A got m votes and B got n votes. If the probability that a vote was incorrectly counted is p, then what is the probability that the wrong candidate was selected (in terms of m, n, and p).

(Starting to copy agent here)

Hi eddybob!

To be honest, this doesn't fit in the regular statistical testing theory that I'm used to.
But with a bit of creativity we can get an answer.
As I see it, your problem statement calls for a couple of approximations and assumptions.

I would assume the number of voters is big enough and the probability on a proper vote p is big enough that we can approximate the number of votes by a normal distribution.

Suppose A won the vote, that is, m > n.
And suppose this is representative for the real situation, what is then the probability of a vote in which A would have lost?
Due to symmetry, this should give approximately the same probability as when A should have lost, but scored m > n votes anyway.

We can approximate the corresponding binomial distribution with a normal distribution
with mean $m$ and variance $\sigma^2 = (m+n)p(1-p)$.
See wiki.

Now we're looking for the probability that the number of votes is less than half.
The corresponding z value is:
$$z = \frac {\bar x-\mu}{\sigma} = \frac {(m+n)/2 - m}{\sqrt{(m+n)p(1-p)}}$$
The desired probability is P(Z < z).
That is, look up the z-value in a z-table and find the corresponding cumulative probability, which will be approximately the probability that A won while he should have lost.

 

[eddybob123]

Thanks a lot! However I think I have some problems with your response (or it just might be me being really naive again (Tongueout).

1. This only happens when m and n are large enough
2. This is an approximation

Can you please justify these?

(Bandit)

 

[HallsofIvy]

A binomial distribution with mean \mu and standard distribution \sigma can be approximated with a normal distribution with the same mean and normal distribution. The larger "n" is in the binomial distribution, the better the approximation is. And, of course, it is precisely in the case of large "n" that it is much easier to evaluate the normal distribution than the binomial distribution.