Single Transferable Vote
As our election system in Rotary is largely based on what is known as the “single transferable vote” system, it is most important that you understand how this system works. It may also be used by districts if there are three or more nominees for District Office. Let us assume that you have to select one candidate from among five nominees for a Rotary office. The five nominees, let us say, are Rotarians Black, Brown, Green, Grey and White. Their names will be printed in that order on a voting slip and you will be required to vote by putting a “1” against the candidate of your choice, a “2” against your second choice, a “3” against your third choice, and so on. You must not vote with a cross. You will be wise to number every candidate but need not do so if there are some about whom you have no opinion.
Let us assume that 100 valid votes are cast. The scrutineers will then sort out these 100 into five piles, each containing all the votes cast as first choice for a particular candidate. After counting, let us say the result is as follows:
Green 33 (i.e., 33 papers with a “1″ against Green)
White – 27 (i.e., 27 papers with a “1” against White)
Grey 15 and so on
Brown 15
Black 10
None has secured a majority over all the others combined, so some might urge that the leader, Green, should be elected and others that he should not on the grounds that, for instance, if Black had not stood all Black’s supporters would have voted for White, putting him ahead. To find out whether this is so or not, we ask Black’s supporters what they would have done, Black is declared defeated and each of his papers is transferred to the candidate that voter has marked “2”. The distributed votes are then counted and added to the first figures. We might then get this:
Green 33 + 2 = 35 (Two delegates voted Black 1, Green 2)
White 27 + 3 = 30 (Three delegates voted Black 1, White 2)
Brown 15 + 3 = 18 (Three delegates voted Black 1, Brown 2)
Grey – 15 + 2 = 17 (Two delegates voted Black 1, Grey 2)
Grey is now eliminated and each of his votes is transferred to whichever of the three remaining candidates is marked by that voter as his next choice. For the original 15 papers this will usually be the second choice, but if there is any paper marked 1 for Grey and 2 for Black it will go to the third choice; the additional votes picked up from Black will also go to the third choice. After counting again, we would reach this result:
White – 30 + 8 = 38 (Eight delegates voted Grey 1, White 2)
Green – 35 + 2 = 37 (Two delegates voted Grey 1, Green 2)
Brown 18 + 6 = 24 (Four delegates voted Grey 1, Brown 2 and two voted Black 1, Grey 2, Brown 3)
Non-transferable – 1 (One delegate voted Grey 1 and no more)
Nobody has yet secured an absolute majority over all the others combined. Had the result after the second elimination been:
Green 35 + 16 = 51;
White 30 + 0 = 30;
Brown 18 + 1 = 19
there would have been no need to eliminate Brown and redistribute his votes. Green would have had an absolute majority and if all Brown’s 19 voters had White as their next choice, White could not overtake Green’s 51 votes.
To complete our example Brown’s votes are reallocated each being transferred to Green or White according to which is marked by that voter as his earlier preference. After the last count the position might be:
White 38 + 13 = 51 (Thirteen delegates put White before Green)
Green 37 + 11 = 48 (Eleven delegates put Green before White)
Non-transferable 1
The one non-transferable paper could not have altered the result but had there been more they might have done; it is unwise to stop numbering so long as there are any candidates left about whom you have any opinion.
Suppose the final result had been a tie between Green and White, then whichever of those two had received the more votes on the first count would have been declared elected. If they were equal on the first count, the one leading on the second count would have been declared elected, and so on. Similarly, if there is a tie for bottom place, that one of the equal candidates is eliminated who was lowest on the first count (or on the earliest count in which they were unequal). In the unlikely event of their tying all the way through, the returning officer must draw lots.