Calculating Majority Chances in a Three-Way Election

[Loren Booda]

Consider an election with only one Democrat, one Independent and one Republican candidate running. Assuming a random vote, what is the chance that all pairs of parties attain a majority?

 

[g_edgar]

Depends on the number of voters. For example, with 1 voter, the probability is 0. In general, look up "binomial distribution" I guess: you are asking for the probability that some party attains at least half the votes.

 

[Loren Booda]

Depends on the number of voters. For example, with 1 voter, the probability is 0. In general, look up "binomial distribution" I guess: you are asking for the probability that some party attains at least half the votes.

Thank you for the modification. Try answering what the probability is as the number of votes approaches infinity.

But not just "some" party of the three. I am asking what the chance is that all pairs of parties attain a majority, i.e., when the number of votes for all parties taken two at a time are greater than 50%.

 

[Nemorad]

Hi,

The probability, that all pairs attain a majority approaches 1 as the number of votes tends to infinity for random votes. A random vote means, that the probability to choose a certain candidate is 1/3. So as the number of votes tends to infinity, 1/3 of the votes goes to each candidate. That implies that no matter which two parties are chosen they always get 2/3 of the votes, i.e. gain a majority.
The formula for sample size n (>= 5) is in the appendix. Hope I got you right!

 

[jambaugh]

Thank you for the modification. Try answering what the probability is as the number of votes approaches infinity.

But not just "some" party of the three. I am asking what the chance is that all pairs of parties attain a majority, i.e., when the number of votes for all parties taken two at a time are greater than 50%.

In short the probability that no one candidate gets 50% or more of the vote.

The law of large numbers says that the average vote will approach the expectation value in probability, i.e. the variance for the percentage of votes approaches zero as N approaches infinity. So in the limit, assuming equally likely voting between candidates, the probability of any a priori deviation from a 1/3 1/3 1/3 distribution approaches 0.

For a finite number of voters with some aribitrary probabilities you have a multinomial distribution. However anyone candidate's % vote can be treated as a straight binomial distribution.

Since the cases of each candidate getting >= 50% are mutually exclusive you can add each probability which can be calculated from a binomial distribution.

The answer to your question will then be 1 minus these three probabilities.

In particular for probabilities 1/3 by 1/3 by 1/3 :
Answer = 1 - 3* P(X> or = N/2) for X ~ Binom(N,1/3).
The mean being N/3 and the stdev = sqrt(2N)/3,
where Answer = 1 - 3P(X > or = x) we have
x= N/2 = mu + z sigma
where z = sqrt(N/8).

we can approximate this probability for large N with a normal distribution:
1 - 3*P(Z > sqrt(N/8) ).
In the limit this approaches certainty as P(Z > infinity) = 0.

The z value giving 50% probability i.e. where P(Z>sqrt(N/8) = 1/6 is z=0.967422,
so N about 7.5. (too small for the normal approximation to be very good).

A little spreadsheet work...
Prob N (normal approximation)
0.5 7.487235892
0.9 26.90594313
0.99 58.8852044
0.999 92.63961506
0.9999 127.2254273
0.99999 162.2926135
0.999999 197.6732578
0.9999999 233.2740074
0.99999999 269.037677
0.999999999 304.9267328

N Prob (Binomial calculation)
4 0.666666667
8 0.736168267
16 0.850112556
32 0.950815451
64 0.99408042
128 0.999896477
256 0.999999959

 

[Loren Booda]

Thank you all for addressing this problem, which I thought was paradoxical (in the sense that the answer would seem independent of the initial conditions), but showing it to involve straightforward mathematics.

 

[bpet]

I ran a few Monte Carlo simulations and the probability seems to converge to between 0.085 and 0.09 as the number of voters increases, which disagrees with every answer here so far.

Consider the voter preferences, one of 6 equally-likely possibilities (ABC, ACB, BAC, BCA, CAB, CBA) and let N1,...,N6 be the number of voters with each, so we have a multinomial distribution. A has a majority preference over B if N1+N2+N5>N3+N4+N6, B beats C if N1+N3+N4>N2+N5+N6 and C beats A if N4+N5+N6>N1+N2+N3. We need to work out the probability of all three inequalities or all three reverse inequalities (for the case A<B<C<A) being satisfied simultaneously.

BTW this is an example of the Condorcet paradox and you're effectively looking for the probability that an election has no Condorcet winner.

 

[Loren Booda]

Much appreciation for your heads up, bpet!

 

[jambaugh]

I ran a few Monte Carlo simulations ... which disagrees with every answer here so far.

I don't see how the answer can't be 100% when the number of voters reaches infinity.

The law of large numbers says as the sample size grows large, the averages % of vote each candidate gets will "converge in probability" to their expectation values. That means each candidate gets 33.3333% +/- epsilon with probability 100% - delta for arbitrarily small epsilon and delta once n passes some finite value.

In your simulation are you parsing the condition correctly?

"Every pair is a majority = no pair is a minority = no third candidate is a majority = all candidates are a minority"

That means say for 100 voters you can have a voting distribution of 49,49,2.

I did a quick spreadsheet simulation using excel's random number generator and with about 1500 samples for 100 voters I got 0.0988% of the time one candidate got a majority or .999 frequency of no 1 candidate majority. That is quite consistent with my calculations.
(to be precise one candidate got majority 15 out of 15183 times.)

[EDIT: Here is the excel formula I used for voters: =IF(RAND()<0.3333333333333,"A",IF(RAND()<0.5,"B","C"))

 

[bpet]

I ran a few Monte Carlo simulations and the probability seems to converge to between 0.085 and 0.09 as the number of voters increases, which disagrees with every answer here so far.

Consider the voter preferences, one of 6 equally-likely possibilities (ABC, ACB, BAC, BCA, CAB, CBA) and let N1,...,N6 be the number of voters with each, so we have a multinomial distribution. A has a majority preference over B if N1+N2+N5>N3+N4+N6, B beats C if N1+N3+N4>N2+N5+N6 and C beats A if N4+N5+N6>N1+N2+N3. We need to work out the probability of all three inequalities or all three reverse inequalities (for the case A<B<C<A) being satisfied simultaneously.

BTW this is an example of the Condorcet paradox and you're effectively looking for the probability that an election has no Condorcet winner.

Oh I misinterpreted the question and made it more complicated than it was. I agree with Nemorad's formula and that the probability should reach 100% as N becomes large.

 

[Loren Booda]

Thank you all for your time and expertise.