How would you determine the probability that there strict majority of voters in a 3 party system? In particular, when there are 3, 4 and 5 voters.
In other words, if you have a randomized 3 people, who can have the same voting preferences, what is the probability there will be a majority?
Here is what I have tried so far:
1) Binomial Theorem
2) Combinatorics
So for 3 parties there are 3! = 6 different arrangements of voting preferences:
- A>B>C
- A>C>B
- B>A>C
- B>C>A
- C>A>B
- C>B>A
From this, I know that there a total of $6^3$ = $\binom{6} {1} $ $\binom{6} {1} $$\binom{6} {1} $ different possibilities with repetition.
Binomial Probability
The probability of any person having a given first choice is 1/3
Let p= prob of success = 1/3
Let q= 1-p= 2/3 i.e probability of not having the same vote
A majority of 3 is either 3/3 people or 2/3 people:
If 2 people of 3 have same 1st choice:
P (X=2) = $\binom{3} {2} $ $(1/3)^2$ $(2/3)^1$ = 6/27
If 3 people of 3 have same 1st choice:
P (X=3) = $\binom{3} {3} $ $(1/3)^3$ = 1/27
The sum of these = 7/27 = about 25.926 %
Therefore the probability of there being a majority with 3 voters is 7/27 = 25.93 %
FOR 5 PEOPLE:
I can extend this to 5 people where there is still p=1/3 and q=2/3, and a majority is obtained by 5 out 5 voters, 4 out of 5 or 3 out of 5.
Thus the probability for a majority is
P(X>2) = P (X=3) + P(X=4) + P(X=5)
= $\binom{5} {3} $ $(1/3)^3$ $(2/3)^2$ +$\binom{5} {4} $ $(1/3)^4$ $(2/3)^1$ +$\binom{5} {5} $ $(1/3)^5$
= 51/243
Combinatorics
If I want there to not be a majority with 3 people, then I need to pick, say, 1 person with A as a 1st choice, the next person with a B as their first choice, and the last person with C as their first choice So, any of the 6 preferences for the 1st person, any of the 4 remaining preferences that exclude A as their first choice, and then any of the 2 remaining preferences for the 3rd voter :
$\binom{6} {1} $ $\binom{4} {1} $ $\binom{2} {1} $ = 6*4*2 = 48
Thus the probability of there NOT being a majority is 48/216 which is about 0.2222.
Then the probability of there being a majority is 77.78%, which is contradicts with the binomial theorem results.
Any help is appreciated ! Thanks a ton!!
EDIT: I just realized that if I multiply my binomial result by 3, the number of voters, I get the same result as by combinatorics.
(Since there are 3 different initial voter choices i.e the 1st voter can pick A, and then I find the probability of at least one the other 2 picking A, then it's the same probability the first voter votes for B or C, similar to arranging people around a table).
How can I extend this result to 4 and 5 people though?