Analyzing a Voting System (Math Geeks Please Help!)

[hmkpoker]

I'm testing a theory about the US voting system and I'd like a little help devising a formula to tell me the likelihood that one vote will make or break the election.

The assumptions are that there are n number of participating voters, two candidates A and B, and all voters will vote with a 50-50 likelihood of either candidate. What is the formula to determine the likelihood that the same number will show up for both sides?

There is probably going to be a resolution problem, because for even numbers a vote count of n/2 on either side yields, and ((n/2) +/- 1) depicts voting situations where one vote did make a difference...but for odd numbers, the countable outcomes would be ((n/2) +/- .5) with no stalemate.

Can anyone figure out a simple formula for this?

[Siegmund]

In a pure popular election, with every voter independent and equally likely, the chance of a tie is, of course, zero if there are an odd number of votes, and (n choose n/2) * 2^-n if there is an even number.

For large n, 1 in sqrt(n*pi/2) is an excellent approximation. (About 1 in 12.5 for n=100, 1 in 17.7 for n=200, etc.)

For the presidential election, it'd be a much harder calculation, since you'd have to calculate the chance of each state having its vote swung by 1 vote, and compare it to the chance that that state's electoral votes would swing the election. (Also a much less meaningful calculation, since there are several states in which the vote split will never be remotely close to 50-50.)

[hmkpoker]

I'm not looking for the odds of a tie, I'm looking for the odds of one vote mattering, and only for popular elections (not factoring in the electoral college just yet)

Interesting formula, btw. Where did you come up with it?

[arahant]

As one enters the real world, one starts using excel for these things and forgetting the analytic formulas behind it, which at any rate are pretty messy.

e.g., I'm thinking "1 vote matters" = twice the prob. density at, say, 100 million, of a binomial distribution with 200 million trials and p=.5...

Of course, this gives me a surprisingly high 1 in 10,000...either my intuition is off or my calculations.

[hmkpoker]

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Of course, this gives me a surprisingly high 1 in 10,000...either my intuition is off or my calculations.

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This is what I'm thinking also. 1 in 10,000 seems WAY off, but this isn't the kind of thing that I trust my intuition to.

[Siegmund]

Well, "What is the formula to determine the likelihood that the same number will show up for both sides?" sure sounded like "what is the probability of a tie?" to me!

If the number of votes in the election is even, a change of one person's vote, or one additional voter casting a vote, breaks the tie; if the number of votes is odd, you want it to be as close to even as possible.

I guess you can argue as to whether the chance that "one vote matters" is equal to the probability of a tied vote, or twice the probability of a tied vote (but, given an off-by-one vote, only half the people can change the outcome by changing their vote.)

The approximation comes from the formula for the pdf of a normal distribution: the number of votes for A is approximately normal with mean n/2 and variance n/4. The "1/sqrt(2pi)/sigma" bit determines the maximum of the fuction, with the e to the minus yadayada part after that ranging from 1 at the mean to 0 at the extremes.

Yes, 1 in 10,000 to 1 in 20,000 is a good ballpark figure.

[hmkpoker]

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Well, "What is the formula to determine the likelihood that the same number will show up for both sides?" sure sounded like "what is the probability of a tie?" to me!

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True, never mind, I shouldn't nitpick. I'm just looking to be within a factor of ten anyway [img]/images/graemlins/smile.gif[/img]

[jay_shark]

Just to add on Siegmunds solution ,the pdf is usually denoted as f(x) where

f(x) = 1/[sqrt(2*pi)sigma]*e^[-(x-mu)^2/(2*sigma^2)]

where sigma =standard deviation
mu is the mean and x takes on the values from negative infinity to positive infinity . Think of f(x) as the y-value and naturally x is on the x-axis . We can evaluate the height of this function by plugging in the appropriate numbers .

In this case x takes on the mean mu , then all you have to do is compute the standard deviation which is simple to do .

[hmkpoker]

I just thought of a new variable that will probably explain why the 1/10,000 tie seems so unlikely. What if the probabilities of each voter voting for a different candidate are changes? What if each voter has a 45% chance of voting for candidate A and a 55% chance of voting for candidate B? What would the odds then be for a tie?

[arahant]

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I just thought of a new variable that will probably explain why the 1/10,000 tie seems so unlikely. What if the probabilities of each voter voting for a different candidate are changes? What if each voter has a 45% chance of voting for candidate A and a 55% chance of voting for candidate B? What would the odds then be for a tie?

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Yeah, good point...
The real probability of your vote counting is a compound function that has to take into account the underlying distribution. One would need to integrate the binomial distribution (which gives the 1/10,000) over the underlying.

odds of a tie actually fall to 1 in a million if you change p to .499....by .490, it's zero for any practical purpose. (on 200 million votes).

The underlying distribution pretty much has to be normal, and I would guess it is so wide as to make the actual odds of affecting the outcome neglible.

Edit: This is a really cool question to think about...nice post. I think I found a new career!

Here's an interesting link on statistical analysis pertaining to voter fraud Link