Breaking a three-way tie

[bonanova]

Al, Bert and Charlie are on the ballot for secretary of the local yacht club, and they finish in a three-way tie. To break it, they solicit the members' second choices, and again it's a three-way tie. Al notes that since the number of members is odd, a series of two-way votes would be decisive. As a reward for finding a way past the impasse, Al receives the bye and will take on the winner of a Bert vs Charlie vote to decide who gets the position as secretary. Assuming voter preferences do not change, what are the winning probabilities for the three gentlemen?

 

[Cygnet]

Surprisingly

Spoiler

Al always wins. Always. 

There are 6 "types" of voters possible, I'll notate them by their first and second choices: AB, AC, BA, BC, CA, CB

If this Yacht club is only 3 people, Al, Bert and Charlie themselves, voting for themselves first, then based on the second place vote, they're either AB, BC, CA or AC, BA, CB. We can run these through the bracket easy enough.

Option 1:

First B vs C: AB:B BC:B AC:C, B Wins. Followed by B vs A: AB:A BC:B CA:A, A Wins.

Option 2:

First B vs C: AC:C BA:B CB:C, C Wins. Followed by C vs A: AC:A BA:A CB:C, A Wins.

But, if this Yacht club is more than 3 people, It has to be a number of people divisible by 3. 9, 15, 21, etc are all valid, but work out the same way as 3 people ultimately. 

Since AB+AC and BA+BC and CA+CB are all equal, but the total number of people is known to be odd, we know that each of those three pairs of voter types has an odd number of people in it (because odd+odd+odd=odd and even+even+even=even), which means that out of each pair of voter types, one of the two is at least 1 person bigger than the other. The same is true for the pairs AB+CB, AC+BC and BA+CA, that each pair is equal to the others, but the groups inside the pair are not equal. This means that there will always be an advantage of at least one vote, in the same rotational manner as with 3 people, so the bracketed vote will turn ou tthe same way.

 

Edited March 10, 2018 by Cygnet

[CaptainEd]

Naively:

Al 1/2, Bert and Charlie 1/4 each

[Molly Mae]

interestingly:

The fact that Al pairs off with the winner is a red herring.  They each have a 1/3 chance of winning, since the distribution of votes is decidedly equal.

As an example, take 3 voters (the first odd number divisible by 3).  Their preference will be expressed as XY, where X is their first choice and Y is their second.  There are two possibilities:
1. AB or AC
2. BC or BA
3. CA or CB

Interestingly, I originally ran with 9 voters and figured the odds would carry over to a smaller (easier) number.  That isn't the case.  I'd like to try with 15 voters.  The odds definitely change based on the number of voters, though.  With 3 voters, Al always wins.  With 9 voters, I have more than one case where Al doesn't win.

[Tylerj21]

 

 

Spoiler

Nevermind...reread

 

Edited March 7, 2018 by Tylerj21