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 /images/graemlins/smile.gif

[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?

[/ QUOTE ]

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 (http://www.bradykiesling.com/election_fraud_analysis.htm)

[latefordinner]

just the reverse of how many trials do I have to do before I can say with x amount of certainty that this coin is rigged.

[Wyman]

BTW, the electoral college changes your answer to this question.

[pvn]

[ QUOTE ]
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?

[/ QUOTE ]

Yes. I posted exactly what you're looking for in politics a while back, in a discussion about how worthless your vote is.

[jstnrgrs]

Sort of unrelated, but I was reading in the paper the other day that in Natck, MA, the vote for town moderator was 1772-1772.

[hmkpoker]

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Sort of unrelated, but I was reading in the paper the other day that in Natck, MA, the vote for town moderator was 1772-1772.

[/ QUOTE ]

There's a big difference between local elections and federal elections. Do you think a national popular election would ever come down to 50,000,000-50,000,000?

[CallMeIshmael]

I think an interesting question is:

once we all concede that voting, as is now, is clearly irrational (which I certainly agree with), then when does it become rational?

I mean, if all I had to do was click one button on the internet, then I really think the decision becomes rational. IF this is true, then where is the line crossed?

[hmkpoker]

[ QUOTE ]
I think an interesting question is:

once we all concede that voting, as is now, is clearly irrational (which I certainly agree with), then when does it become rational?

I mean, if all I had to do was click one button on the internet, then I really think the decision becomes rational. IF this is true, then where is the line crossed?

[/ QUOTE ]

There really is no universal way to calculate this. Not only would we have to factor in a number of other annoying variables (for example, what is the probability that the candidate you elect will do what you want him to do?), but the pure, utilitarian EV of voting is going to be almost infinitely outweighted by the satisfaction of voting (or not voting) independently of any results. Most voters are completely unaffected by the actual chances of their vote mattering, because it is "the right thing to do." My brother's roommate (at Penn State) drove back to the southern border in his home state of Virginia in '04 to vote for Bush. That's a long drive, and a solid red state. Even if his odds of affecting the outcome were 1/infinity (zero), he'd still do it.

This should actually concern people. When voting populations become large and rational consequence-driven incentives diminish, the system must select against rational, consequence-oriented people. The winning candidates in these elections are selected by deontologically/morally-oriented people, which is not very representative of the population.

http://www.dadsdayoff.net/bush-church.jpg

[pvn]

[ QUOTE ]
This should actually concern people. When voting populations become large and rational consequence-driven incentives diminish, the system must select against rational, consequence-oriented people. The winning candidates in these elections are selected by deontologically/morally-oriented people, which is not very representative of the population.

http://www.dadsdayoff.net/bush-church.jpg

[/ QUOTE ]

gold star.

[jstnrgrs]

[ QUOTE ]
[ QUOTE ]
Sort of unrelated, but I was reading in the paper the other day that in Natck, MA, the vote for town moderator was 1772-1772.

[/ QUOTE ]

There's a big difference between local elections and federal elections. Do you think a national popular election would ever come down to 50,000,000-50,000,000?

[/ QUOTE ]

No, I understand the difference. But even with 3000 voters, I would have thought that the chance of a tie would be nearly 0.

[hmkpoker]

[ QUOTE ]
[ QUOTE ]
[ QUOTE ]
Sort of unrelated, but I was reading in the paper the other day that in Natck, MA, the vote for town moderator was 1772-1772.

[/ QUOTE ]

There's a big difference between local elections and federal elections. Do you think a national popular election would ever come down to 50,000,000-50,000,000?

[/ QUOTE ]

No, I understand the difference. But even with 3000 voters, I would have thought that the chance of a tie would be nearly 0.

[/ QUOTE ]

If 3,000 voters flip coins to select their candidate, the probability of a tie is a little better than one in a hundred. Now granted the true probability in these elections is probably not an exact coinflip, so the odds will be a bit worse, but even with a few powers of ten added on against the unlikelihood, you still have to account for the fact that there are TONS of these local elections taking place throughout the country. The USA has tons of small towns with tiny populations that are all having these elections. The odds that one of these four-digit populations will produce a tie now and then is actually very, very high.

For all practical purposes, you are guaranteed to see a tie in a local election (over a period of, say, several years), and guaranteed never to see a tie in a federally-sized election, even if it is purely popular, ever.

[Al6Jets18]

Understandable explanation of the Banzhaf Power Index found Here (http://www.cs.unc.edu/~livingst/Banzhaf/index.html)

The inherent problem of using voting power, or probability, to analyze voter behavior/turnout, is that it assumes a high percentage of totally rational voters, when in fact a majority of voters subscribe to "expressive voting," or voting to express support or dislike for a candidate, not necessarily to get that candidate elected. For instance, any 3rd party voters in federal elections are expressive, because someone like Nader is not going to win, yet these voters can have a great influence on the outcome (See 2000 Presidential Election, Florida). Others vote for the sake of voting in a democratic society.

PM me for more information, I have done research in this area.