Strange behavior in elections and pie charts
So, a friend asked me the probability for a candidate to get at least 50% of the total votes in an election consisting of 5 candidates (let's pretend everyone picks at random 🙂). I thought for a bit and then presented an answer:
Let's first generalize so that the election has $n$ candidates.
We can think of the votes cast between candidates as a pie chart with $n$ slices and $n$ borders. We draw one fixed border from the center to the top, and then draw $n-1$ more with randomized positions in the chart. Furthermore, let's say the chart shows the votes acquired by candidate 1, candidate 2, etc from the top going clockwise.
Now, the probability of the first candidate getting $\geq50\%$ of all votes (shaded area) is equivalent to the probability of all $n-1$ lines being on the unshaded side of the disk, which is $\left(\frac12\right)^{n-1}$.
Since this probability should be the same for every candidate, the overall probability of someone getting at least half of all votes cast is $n\cdot\left(\frac12\right)^{n-1}=\frac n{2^{n-1}}$.
I found with some coding that for my pie chart analogy of the problem, this is indeed correct (yeay!).
However, for the original problem... I was absolutely wrong! With some coding (again), I found out that the actual probability is very close to, if not $\frac1{(n-1)!}=\frac n{n!}$.
Why's that? I'm certain it's because the two behave differently, but if that's so I don't get why they do.
What causes this behavior? Has this problem been studied before, and is this perhaps a branch of probability theory?
Thanks in advance!
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| User | Comment | Date |
|---|---|---|
| TheCodidacter, or rather ACodidacter†| (no comment) | Sep 28, 2023 at 08:33 |
with randomized positions in the chart
Implicit in this description of your model is the notion that every location for a separator line is equally likely. This doesn't describe the actual distribution of vote totals; with random voters, lines are much less likely to appear near other lines than farther away. This is because to have two lines very close to each other, some candidate would have had to have received only a handful of votes; if each voter votes randomly, this is much less likely than a distribution where all candidates have roughly equal votes.
The actual probability that one candidate obtains a majority of votes, for $n > 2$ candidates, depends on the size of your voting population and approaches zero as the population grows, so your other formula isn't correct either. (The number of votes a single candidate receives is a random variable with a binomial distribution, and the standard deviation of a binomial distribution is proportional to the square root of the number of voters, which grows more slowly than the number of votes needed for a majority.)
My approach is, I suppose, simpler.
There are 4 voters: A, B, C, D. They vote randomly
There are 4 candidates vying for A, B, C, D's attention. Let's call them W, X, Y, Z
How many different possible electorate results are there?
1 candidate gets all 4 votes: $^4C_1 = 4$
1 candidate gets 3 votes: $^4C_1 \times ^3C_1 = 12$
1 candidate gets 2 votes (half the votes): $^4C_1(^3C_1 + ^3C_2) = 24$
All candidates get 1 vote each: $^4C_4 = 1$
Total = $4 + 12 + 24 + 1 = 41$
Probability that one candidate gets at least 50% of the votes = $P(50\%)$
$P(50\%) = \frac{40}{41}$
I don't know how to generalize this "finding" though. What I can perhaps say that I know is the electorate is going to be myriad and the candidates, comparatively few.
How does that affect the probability? Feels very tower of Hanoish and I don't have what it takes to solve that problem.
Side quest: Despite the fact that this is a discrete distribution (1 or 2 votes, not 1.5 votes), the OP has chosen a continuous model (area). Does the fact that the answer is in combinatorics form, $\frac{n!}{(n - 1)!}$ mean anything?
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