In the Australian 2016 federal election, the seat of Herbert was won by 37 votes out of 88,337 cast....

Question

In the Australian 2016 federal election, the seat of Herbert was won by 37 votes out of 88,337 cast. This problem explores the likelihood that a close election could be decided by purely random factors.

(a) Suppose that an election is to be held between two candidates, Alice and Bob. Assume that 100,000 people are eligible to vote, and that they are exactly evenly divided between the two can- didates. Assume that on voting day, a random 1% of the voters, i.e. 1,000 voters, do not vote due to illness, travel or other factors. We assume that each group of 1,000 voters is equally likely to be absent (that is, it is equivalent to choosing 1,000 balls out of an urn containing 100,000 balls without replacement). By making reasonable approximations and using the central limit theorem, estimate

(1)P

Xi=a=e-nb.

f(x)?(x)dx

3

the probabilities that Alice wins the election by at least 20, 40 or 100 votes. Clearly identify the approximation(s) that you are making.

(b) Do the same calculation using the numbers for the seat of Herbert in the Australian 2016 federal election (88,337 votes, won by 37) and the 2000 Florida US Presidential election (5,963,110 votes, won by 537). In both cases, assume that there were only two candidates to choose from, that the votes were exactly evenly split between the two candidates, and that on voting day, a random 1% of the voters (rounded to the nearest integer) did not vote.

David · Accepted Answer

Answer (a) 
There are 100,000-1,000= 99,000 voters who cast their votes. Since each group of 1,000 voters is equally 
likely to be absent, these 99,000 voters are equally likely to vote either of the candidates. 
Let X be the number of votes Alice gets and Y be the number of votes Bob gets. 
Hence X+Y = 99,000 
P(Alice wins by at least 20 votes) = P( X-Y>=20) = P[X-(99,000-X) >=20] = P(2X>=99,020) 
 P(X>= 49,510)   
Now X ~ Binomial (99,000, ½) Hence n=99,000 and p(probability of winning) =1/2  
Mean=n*p= 99,000*1/2 = 49,500 and variance = n*p*(1-p) =99,000*1/2*1/2= 24,750 
Since n is large, we can use Central limit theorem to approximate the Binomial distribution with the 
normal distribution, with continuity correction. 
Continuity correction means P(X>=a) = P(X>=a-0.5)  
  P(X>= 49,510) = P(X> 49,510-0.5) = P(X> 49,509.5) = P[z> (49,509.5-49,500)/ 24,750^0.5] 
 P(z>0.060386) = 1-0.5239 (Using z table) 
 P(Alice wins by at least 20 votes) = 0.4761 
P(Alice wins by at least 40 votes) = P( X-Y>=40) = P[X-(99,000-X) >=40] = P(2X>=99,040) 
 P(X>= 49,520) = P( X> 49,520-0.5) = P(X>49,519.5)= P[z> (49,519.5-49,500)/ 24,750^0.5] 
 P(z>0.12395) = 1-0.5478(Using z table) 
 P(Alice wins by at least 40 votes) = 0.4522 
P(Alice wins by at least 100 votes) = P( X-Y>=100) = P[X-(99,000-X) >=100] = P(2X>=99,100) 
 P(X>= 49,550) = P( X> 49,550-0.

In the Australian 2016 federal election, the seat of Herbert was won by 37 votes out of 88,337 cast. This problem explores the likelihood that a close election could be decided by purely random...

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