# A single transferrable vote question

I'm preparing for the Oxford TSA, and am using past papers as a way to practice. One of the questions, I thought I had right, but turned out incorrect. Would love it if you were to analyze my reasoning here -->

Question:

In the elections for the mayor of Bitton, the single transferable vote system is used. In this, each voter places a '1' beside the candidate they most want and a '2' beside their second choice. In the first round of votes, all the '1' vote of candidates are counted. If no candidate has over 50% of the votes, the bottom candidate in the poll drops out, and the '2' votes for the candidate are added to the appropriate other candidates. This is repeated until one person has over 50% of the votes.

The votes for the first round of counting were as follows:

MG 87

SJ 63

PG 45

IM 36

WD 18

RM 17

How many candidates can still win?

a) 1 b) 2 c) 3 d) 4 e) 5

My reasoning may have been off. I tried subtracting the voters at the bottom, until at least the highest voter could win. This got me to subtract the bottom 3, leaving 3 more people. Hence, I said 3 more people could win. I didn't take into consideration though if other people would be able to win the the added votes.

However, even if I were, my answer would be wrong, because, the correct answer is 4 people.

How can 4 people have a chance of winning here?

Suppose that IM is the first or second choice of every voter. When RM is dropped, IM picks up $17$ votes, and the table is now:

$$\begin{array}{cc} \text{MG}&\text{SJ}&\text{IM}&\text{PG}&\text{WD}\\ 87&63&53&45&18 \end{array}$$

Then WD is dropped, and IM picks up another $18$ votes:

$$\begin{array}{cc} \text{MG}&\text{IM}&\text{SJ}&\text{PG}\\ 87&71&63&45 \end{array}$$

Then PG is dropped, giving IM another $45$ votes:

$$\begin{array}{cc} \text{IM}&\text{MG}&\text{SJ}\\ 116&87&63 \end{array}$$

And now when SJ is dropped, IM gets $63$ more votes and wins.

Clearly if IM can win, so can anyone above IM in the original table. It’s also clear that RM cannot win. Neither can WD, who would still be at the bottom of the table in round two even after receiving all of RM’s second choice votes.

(By the way, the actual total is $266$, not $254$.)

• Gotcha Brian, but why would the answer be 4 people? I also came up with the answer of 3 people – Stepan Parunashvili Oct 27 '13 at 6:59
• @Stepan: See the bit that I just added; does it answer the question for you? – Brian M. Scott Oct 27 '13 at 7:02
• Brian. Mind. blown! Okay, I get it! I wasn't looking at multiple ways that the votes can run. Thank you – Stepan Parunashvili Oct 27 '13 at 7:07
• @Stepan: You’re welcome. Just to make it explicit, the general principle is that you should look at the situation that is most favorable to candidate X and consistent with the data in order to see whether X has a chance to win. – Brian M. Scott Oct 27 '13 at 7:09