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# 2/3 vote requirement ## Recommended Posts Our 9 member board will be voting on a new code of regulation soon that requires a 2/3 vote to pass. Is the correct number required to pass 6 votes or7?  2/3 of 9 is six but .667% of nine comes out to more then 6.  Which calculation is correct?

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Our 9 member board will be voting on a new code of regulation soon that requires a 2/3 vote to pass. Is the correct number required to pass 6 votes or7?  2/3 of 9 is six but .667% of nine comes out to more then 6.  Which calculation is correct?

See FAQ #5

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Is the correct number required to pass 6 votes or 7?

It might be neither (and it's certainly not 7). Some members might be absent and some members might abstain (i.e. choose not to vote).

A vote of 1-0 would meet the requirement for a two-thirds vote. As would a vote of 6-3.

All that's needed is at least twice as many "yes" votes as "no" votes".

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Our 9 member board will be voting on a new code of regulation soon that requires a 2/3 vote to pass. Is the correct number required to pass 6 votes or7?  2/3 of 9 is six but .667% of nine comes out to more then 6.  Which calculation is correct?

Neither. It does not depend on the size of the board, but on how many are present and vote.

For a 2/3 vote to pass, there must be at least twice as many Yes votes as No votes.  So 6-3 is sufficient.  But so is 2-1 or 1-0

Abstentions and absentees make no difference as long as a quorum is present.

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The easiest way to know if you have a two thirds vote, though, is what Edgar Guest said:  "Twice as many yes votes as no votes".  No need for all the mental gymnastics unless the vote requirement is two thirds of the members present or two thirds of the entire membership.   Then you have to do the math.  The link Mr. Honemann provided will help you with that.

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As a math major I just would like to point out to the OP that 2/3 does NOT equal 0.667 which is why you got two different results.

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As a math major I just would like to point out to the OP that 2/3 does NOT equal 0.667 which is why you got two different results.

But in a group of less than 10 000 people, the number 0.666667 should give you acceptable results--but still not as good as multiplying by 2 and dividing by 3... or just using the "at least twice as many Yes votes  as No votes" rule, which yields a value of True or False, rather than a fraction.

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But in a group of less than 10 000 people, the number 0.666667 should give you acceptable results--but still not as good as multiplying by 2 and dividing by 3... or just using the "at least twice as many Yes votes  as No votes" rule, which yields a value of True or False, rather than a fraction.

So if 9 votes are cast, a two-thirds vote is 6.000003?

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So if 9 votes are cast, a two-thirds vote is 6.000003?

I certainly hope not!!! ##### Share on other sites

Our 9 member board will be voting on a new code of regulation soon that requires a 2/3 vote to pass. Is the correct number required to pass 6 votes or7?  2/3 of 9 is six but .667% of nine comes out to more then 6.  Which calculation is correct?

As a math major I just would like to point out to the OP that 2/3 does NOT equal 0.667 which is why you got two different results.

Not to mention that it doesn't equal 0.00667 (or .667%). ##### Share on other sites

But Mr. Novosielski's suggestion to multiply by 2 and then divide by 3 does work.... as does multiplying 9 by .6666666.

But the most accurate is obtained by multiplying by 2 and dividing by 3.   Or just making sure you have twice as many yes votes as no votes.

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But the most accurate is obtained by multiplying by 2 and dividing by 3.

Oh, yeah? I say you gotta' first divide by 3 and then multiply by 2. ##### Share on other sites

Oh, yeah? I say you gotta' first divide by 3 and then multiply by 2. Well, with a committee of 9 they both seem to work.  I'll leave it to the math majors to sort this out!

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But the most accurate is obtained by multiplying by 2 and dividing by 3.

So, to find two-thirds of nine you first multiply nine by two to get eighteen? What significance does eighteen have?

What you want to do is first find the value of the denominator (i.e what size the pieces are) and then multiply it by the numerator (i.e. how many pieces you have).

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So, to find two-thirds of nine you first multiply nine by two to get eighteen? What significance does eighteen have?

What you want to do is first find the value of the denominator (i.e what size the pieces are) and then multiply it by the numerator (i.e. how many pieces you have).

Well, I'm just quoting Gary Novosielski, but it works....at least with a board of nine.  Try it.  2 x 9 = 18.  Divide 18 by 3:  You get 6.  Six is two thirds of 9.

But, then, Dan Honemann's formula works, too.,... at least for a board of 9.

Like I said in the other post, I'm leaving it to the math geniuses (who told us that .6666667 works when it doesn't) to figure it out. ##### Share on other sites

But Mr. Novosielski's suggestion to multiply by 2 and then divide by 3 does work.... as does multiplying 9 by .6666666.

So if 9 votes are cast, a two-thirds vote is 5.9999994?

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So if 9 votes are cast, a two-thirds vote is 5.9999994?

No.  According to FAQ # 5, which you co-authored, it would be 6, right? http://www.robertsrules.com/faq.html#5

But, isn't that better than using Gary's formula of .6666667, which gives you 6.000000.3... or 7, as two thirds of nine?

I'm gonna leave it to the mathematicians, including the one who told us to use .6666667, to sort this out.

And I'll stick to hoping I have twice as many yes votes as no votes and hope that if I run across a case where the threshold is two thirds of those members present or two thirds of the entire membership, using both .6666666 and .6666667 both yield a clear winner.  I have a hunch that Gary's other formula of multiplying by 2 and dividing by 3 might work, too.  Or Dan's formula of dividing by 3 and multiplying by 2.   And I'll hope that the yes votes are at least twice the combined number of people present who either didn't vote or voted no. C'mon, you mathematicians:  Settle this!!!

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No.  According to FAQ # 5, which you co-authored, it would be 6, right? http://www.robertsrules.com/faq.html#5

But, isn't that better than using Gary's formula of .6666667, which gives you 6.000000.3... or 7, as two thirds of nine?

I'm gonna leave it to the mathematicians, including the one who told us to use .6666667, to sort this out.

And I'll stick to hoping I have twice as many yes votes as no votes and hope that if I run across a case where the threshold is two thirds of those members present or two thirds of the entire membership, using both .6666666 and .6666667 both yield a clear winner.  I have a hunch that Gary's other formula of multiplying by 2 and dividing by 3 might work, too.  Or Dan's formula of dividing by 3 and multiplying by 2.   And I'll hope that the yes votes are at least twice the combined number of people present who either didn't vote or voted no. C'mon, you mathematicians:  Settle this!!!

I thought it was fairly well settled already. For one side of a question to obtain two-thirds of the vote, it must have double the number of votes of the remaining side(s), since two-thirds to one-third is a ratio of two to one (2/3 : 1/3)  =  (2 : 1).

Also (and I am sorry to say that this requires no math beyond the elementary-school level),

two-thirds of a number N is equal to (2÷3)×N, which is equal to (N×2)÷3, which is equal to (N÷3)×2.

So you can multiply by 2 and then divide by 3, or divide by 3 and then multiply by 2, or -- and here's a concept -- actually just multiply 2/3 by the number.

It is not possible to exactly represent a fraction of two-thirds by any finite number of decimal digits, but if all you're looking to do is determine what the whole number of votes needed to meet a two-thirds threshold is, this really shouldn't present any difficulty.

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So if 9 votes are cast, a two-thirds vote is 5.9999994?

Yes.

Count the votes. Any vote that exceeds that number is sufficient to approve a motion that requires a 2/3 vote..

Now you might quibble that a vote equal to that technically would not, since it is less than two thirds.  But my guess is that you will either get more or fewer votes than 5.9999994..

If you happen to get exactly that number, we'll burn that bridge when we come to it.

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Yes.

Count the votes. Any vote that exceeds that number is sufficient to approve a motion that requires a 2/3 vote..

Now you might quibble that a vote equal to that technically would not, since it is less than two thirds.  But my guess is that you will either get more or fewer votes than 5.9999994..

If you happen to get exactly that number, we'll burn that bridge when we come to it.

Continuing with this frivolity, I hereby declare that 5.9999994 is not two-thirds of 9. (It's a little too windy to go fishing.)

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Now just suppose our bylaws allow us to split our vote and I vote 0.9999995 yes and 0.0000005 no.  Then it may get more than 5.9999994 votes but not 2/3.

Moral of the story: don't try to round 2/3.  Take the number of voters and multiply by 2 and divide by 3.

On a side note:  Gary was a math major too?

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Oh, yeah? I say you gotta' first divide by 3 and then multiply by 2. Not so fast.  Mathematically they're equivalent, but see below.

And by the way, it's worse than simply not being able to represent 1/3 in decimal. If you use a computer or calculator, the numbers are internally represented in binary.

Many (most) decimal fractions cannot be exactly represented in binary.  The number 2/3 can't be represented in either, but neither can common decimals like, for example, 0.1 will yield the repeating "bicimal"  0.0001100110011001100110011001100110011....

Division creates a loss of precision in the least significant digit(s) and multiplication can magnify the error, so it's better to multiply first and divide last.  That's an area where pure mathematics and computer science part company.

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Not so fast. Mathematically they're equivalent, but see below.

And by the way, it's worse than simply not being able to represent 1/3 in decimal. If you use a computer or calculator, the numbers are internally represented in binary.

Many (most) decimal fractions cannot be exactly represented in binary. The number 2/3 can't be represented in either, but neither can common decimals like, for example, 0.1 will yield the repeating "bicimal" 0.0001100110011001100110011001100110011....

Division creates a loss of precision in the least significant digit(s) and multiplication can magnify the error, so it's better to multiply first and divide last. That's an area where pure mathematics and computer science part company.

Yes, but if someone is using a decent calculator or computer, then they can simply multiply the number by 2/3, so there is no need to worry about any of this. Doing it in steps is more helpful when someone is calculating it by hand.

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Yes, but if someone is using a decent calculator or computer, then they can simply multiply the number by 2/3, so there is no need to worry about any of this. Doing it in steps is more helpful when someone is calculating it by hand.

There never was any need to worry about it, and no need to start now. But unless you have a 2/3 key on your calculator, you'll need to multiply it by 3, which will actually multiply by 2 and then divide by 3.

Or you can multiply by which will first divide 2 by 3 (giving an imprecise result) and multiplying by that.

Most calculators actually carry out their math two or three digits beyond what they show you, and round off, so nearly all of the time none of this will  be detectable.

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Yes, but if someone is using a decent calculator or computer, then they can simply multiply the number by 2/3, so there is no need to worry about any of this. Doing it in steps is more helpful when someone is calculating it by hand.

Well, I guess I'm using indecent calculators because none of mine have a 2/3 key.  They've never offered me the option of watching anything indecent, either, so I guess they are really cheap, indecent calculators!

Maybe I need to try pushing some more buttons. . . .