There are n voters in an upcoming election in a certain...

There are $n$ voters in an upcoming election in a certain country, where $n$ is a large, even number. There are two candidates: Candidate A (from the Unite Party) and Candidate $\mathrm{B}$ (from the Untie Party). Let $X$ be the number of people who vote for Candidate $\mathrm{A}$. Suppose that each voter chooses randomly whom to vote for, independently and with equal probabilities. (a) Find an exact expression for the probability of a tie in the election (so the candidates end up with the same number of votes). (b) Use Stirling's approximation, which approximates the factorial function as $$ n ! \approx \sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n} $$ to find a simple approximation to the probability of a tie. Your answer should be of the form $1 / \sqrt{c n}$, with $c$ a constant (which you should specify).

There are $n$ voters in an upcoming election in a certain country, where $n$ is a large, even number. There are two candidates: Candidate A (from the Unite Party) and Candidate $\mathrm{B}$ (from the Untie Party). Let $X$ be the number of people who vote for Candidate $\mathrm{A}$. Suppose that each voter chooses randomly whom to vote for, independently and with equal probabilities.
(a) Find an exact expression for the probability of a tie in the election (so the candidates end up with the same number of votes).
(b) Use Stirling's approximation, which approximates the factorial function as
$$
n ! \approx \sqrt{2 \pi n}\left(\frac{n}{e}\right)^{n}
$$
to find a simple approximation to the probability of a tie. Your answer should be of the form $1 / \sqrt{c n}$, with $c$ a constant (which you should specify).

Key Concepts

Stirling's Approximation

Stirling's approximation provides an asymptotic formula for estimating factorials when n is large. This approximation simplifies the calculations involving factorials in binomial coefficients, yielding more manageable expressions that reveal the underlying behavior of probability functions in the limit of large n.

Asymptotic Analysis

Asymptotic analysis deals with the behavior of functions as the variable grows large. In this context, it is used to approximate the probability of a tie in an election when the number of voters is very large, allowing the derivation of a simplified expression that captures the essence of the behavior without the complexity of exact factorial expressions.

Binomial Coefficient

The binomial coefficient, often represented as 'n choose k', counts the number of ways to choose k successes (or votes) out of n independent trials. It is a key combinatorial concept that appears in the expression for the probability of obtaining exactly k votes in a binomial distribution.

Binomial Distribution

The binomial distribution describes the number of successes in a fixed number of independent Bernoulli trials where each trial has the same probability of success. In the context of voting, each voter’s choice is treated as an independent trial with two outcomes, making the binomial distribution fundamental for computing the probability of any specific number of votes for a candidate.

Factorial Function

The factorial function (n!) is central to the calculation of binomial coefficients as it defines the number of possible arrangements of n items. Its properties are essential when working with combinatorial formulas in probability theory, especially when exact expressions involve large numbers.

Transcript

00:01 So we're to assume that the probability or the proportion of people that voted for the winning was 43%.

00:08 And we're going to take a sample of 611 people.

00:13 And we know that the outcome, x being the number that voted for x equal the number who voted for the winner.

00:22 And we're saying that that sample came out to be 308.

00:26 And we want to use our rule of thumb to find out what would.

00:30 Be significantly lower high.

00:33 So we know we need the mean and the mean is n times p, which is that 611 times 0 .43.

00:40 So 611 times 0 .43 that comes out to be 262 .73.

00:47 That would be the average number that would say yes if you voted for, if you took a random sample of 611, and they truly had a 43 % of the people who voted for the winner.

00:59 And then we have a 3 % of the people who voted for the winner.

01:00 Square root of n times p times 1 minus p and so i can multiply this value that i already have in my calculator times that point 57 and then square root it and that gives me a standard deviation of 12 .24 roughly and i'm going to store that value in my calculator and so now we want to find the cutoff for significantly low and the cutoff for significantly high and so if i take that mean that 262 .73 and subtract away two times the standard deviation, 262 .73 minus two times that standard deviation.

01:49 Anything that is lower than or equal to 238 point, i'll say 3, 2 ,6, that's considered to be significantly low.

01:59 And now likewise we'll do the same thing.

02:04 We'll put an addition sign between and anything that is greater than or equal to 287 .2 is considered significantly high.

02:18 So we can see that the 308 that we had, that 308 right here, is significantly high compared to this value.

02:28 Now, we want to find on part b, what is the probability of getting exactly 308 people saying they voted for the winner if the actual proportion is 43%.

02:44 Correct, 43%.

02:46 So we have a combination of, and we have the 611 as our value here, and we want to choose 308 of them to say, yes, i voted for the winner.

03:00 To the 308 and 0 .57 to the difference between those two, which looks like 303 to me.

03:09 Some of those two would add up to 611...

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