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Canberra Today 5°/9° | Saturday, April 20, 2024 | Digital Edition | Crossword & Sudoku

Counting the Speaker’s vote

Canberra legal identity Ric Lucas writes:

Rob Oakeshott must be an independent because he cannot count.

That’s the only conclusion I can draw about the fuss he is making about pairing arrangements for the Speaker.

In fact, pairing the Speaker can never make any difference to the outcome of a vote.

This is because s40 of the Constitution provides that “questions arising in the House of Representatives shall be determined by a majority of votes other than that of the Speaker.  The Speaker shall not vote unless the numbers are equal, and then he shall have a casting vote.”

But of course the only time the government needs the Speaker’s vote will be when the numbers are equal.

Pairing cannot apply in this situation, because a pair is granted on the assumption that the Speaker and the pair balance each other out.  In other words, the Speaker’s deliberative vote is already cast.  He cannot in those circumstances also have a casting vote!  That would be like letting him vote twice.

Another way of analysing the problem is to consider what happens if a government supporter (Mr Rudd perhaps?) is stuck overseas when an important vote is taken.

The numbers are then 75 to 74.  Without a pair for the Speaker, the vote is tied at 74 each.  The Speaker then must exercise his casting vote and the measure passes.

With a pair, the only difference is that the Speaker is not called on to exercise his casting vote.

One of the interesting points in the Solicitor-General’s advice is his view that if the the numbers are equal, the Speaker must exercise his casting vote.  As I see it, that would mean that a pairing arrangement would be directly inconsistent with the Speaker’s duty, if the pair resulted in a tied vote.

That could occur if two government supporters were absent from a vote.

It seems we may soon live in interesting times!

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Ian Meikle, editor

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