Jump to content
The Official RONR Q & A Forums

computing 2/3 majority


Guest iwilliams

Recommended Posts

before a vote it was established that 33 people were eligible to vote. Thus, 2/3 is 22. The vote was 21 for and 7 against. Some people (5) didn't vote. Is the 2/3 determined on the 33 before the vote or do you recompute the 2/3 based on the 28 people who actually voted.

Link to comment
Share on other sites

Is the 2/3 determined on the 33 before the vote or do you recompute the 2/3 based on the 28 people who actually voted.

It's based on the number of members present and voting. So a vote of 1-0, with all other members abstaining (i.e. not voting) would constitute a two-thirds vote.

And there's no "re-computation" since it doesn't matter how many members are present.

The simplest way to determine a two-thirds vote is if there are at least twice as many "yes" votes as "no" votes.

Link to comment
Share on other sites

before a vote it was established that 33 people were eligible to vote. Thus, 2/3 is 22. The vote was 21 for and 7 against. Some people (5) didn't vote. Is the 2/3 determined on the 33 before the vote or do you recompute the 2/3 based on the 28 people who actually voted.

A majority vote is one thing. A 2/3 vote is another. There is no "2/3 majority" vote. It's like colorless pink unicorns.

For a "2/3 vote", this refers to 2/3 of the members present and voting. People who don't vote don't count (except toward a quorum).

As long as there are at least twice as many Yes votes as No votes, then 2/3 approval has been achieved. So 21-7 is more than enough. In fact that even qualifies as a 3/4 vote. You don't "recompute" anything, because computing the number in the first place was meaningless. You just vote, and count the votes.

It would have to be qualified, such as "2/3 of those members present" in order to matter how many didn't vote. And most well-written rules don't do things like that.

Link to comment
Share on other sites

Everyone knows what a two-thirds majority is. It's one of an infinite number of possible majorities.

Exactly. <chortle>

But as a rational number, it's merely countably infinite. Furthermore, if we restrict ourselves to fractions like 2/3, 3/4, 4/5, that is, fractions of the form n/(n+1), then it's--well, never mind, it's still countably infinite. But who's counting?

Link to comment
Share on other sites

Archived

This topic is now archived and is closed to further replies.

×
×
  • Create New...