majority puzzle

[puzzling]

suppose a 7 members meeting has to elect 2 directors.

Each member may vote for two different candidates.

* 2 members vote for A only

* 3 members vote for B only (one of them votes for B twice but this second vote for B is discarded, no culumative voting)

* 2 members vote for A and C

 

It is clear that A is elected (4 votes is a majority of 7 members )

But is B also elected?

Yes,  After the members voting for candidate A only  are removed B has 3 of the 5 remaining votes and 3 is a majority of 5.

or :

No, B has less than the 4 votes needed to get a majority of the 7 members voting, a new round is needed.

 

 

[Rob Elsman]

it is not clear to me whether the 3 votes for B only includes the illegal vote or not.  Why make this so confusing?  Did B garner 3 votes or two?  Skip the illegal vote.  How many legal votes were cast?

[Richard Brown]

No.  There were seven votes cast according to the rules in RONR.  So, B has less than the four votes needed of the seven members who cast votes.  

Edited to add:  See 45:32 and the table on tinted page 52

Edited August 26, 2022 at 01:05 AM by Richard Brown
Added last paragraph

[Gary Novosielski]

On 8/25/2022 at 8:40 PM, puzzling said:

suppose a 7 members meeting has to elect 2 directors.

Each member may vote for two different candidates.

* 2 members vote for A only

* 3 members vote for B only (one of them votes for B twice but this second vote for B is discarded, no culumative voting)

* 2 members vote for A and C

 

It is clear that A is elected (4 votes is a majority of 7 members )

But is B also elected?

Yes,  After the members voting for candidate A only  are removed B has 3 of the 5 remaining votes and 3 is a majority of 5.

or :

No, B has less than the 4 votes needed to get a majority of the 7 members voting, a new round is needed.

 

 

Number of ballots cast……7
Necessary for election (majority)……4
A received……4
B received……3
C received……2

I would say A is elected.  B and C are not, and another round will be required.

[Rob Elsman]

Well, if the illegal vote is one of the three cast for member B, then there would seem to have been six legal voters, not seven.  Either way, it seems to me that the only person to have received the necessary four votes was A.

When tabulating the results you should first count the number of voters expressing any judgment.  From this number, it is possible to determine how many votes a person must garner to be elected.  In this case, the two voters who voted for A counts as two legal voters; depending on whether the three votes for B includes an illegal vote or not, the total so far would either be four or five legal voters; and, the two members who voted for A and C makes a total of either six or seven legal voters

After the number of total legal voters is determined, the tabulation of the votes garnered by each person can be determined.  In this case A gets four votes; B gets either two or three votes, depending on whether the three votes includes the illegal vote; and, C gets two votes.

Since four votes are necessary to be elected, only A has been elected.  B's two or three votes are insufficient.  C's two votes are also insufficient.

 

[Atul Kapur]

I agree with Messrs. Brown and Novosielski.

I also note that this option makes no sense to me. 

On 8/25/2022 at 8:40 PM, puzzling said:

After the members voting for candidate A only  are removed B has 3 of the 5 remaining votes and 3 is a majority of 5.

Why would you remove the ballots that voted for A?

[Gary Novosielski]

 

Edited August 26, 2022 at 01:35 AM by Gary Novosielski

[Rob Elsman]

This is why the original poster should not have been so confusing.  I still do not understand whether the three votes cast for member B were cast by two or three legal voters.  The way the origianl poster phrased it, it could be either.  In other words, were four votes (one of which was illegal) cast by three legal voters, or were three votes (one of which was illegal) cast by two legal voters?  I cannot tell from what was written.

[Gary Novosielski]

 

Sorry for all the duplicates.  The window was not showing me a correct representation of what it was about to save.

Edited August 26, 2022 at 01:42 AM by Gary Novosielski

[Gary Novosielski]

On 8/25/2022 at 9:23 PM, Rob Elsman said:

This is why the original poster should not have been so confusing.  I still do not understand whether the three votes cast for member B were cast by two or three legal voters.  The way the origianl poster phrased it, it could be either.  In other words, were four votes (one of which was illegal) cast by three legal voters, or were three votes (one of which was illegal) cast by two legal voters?  I cannot tell from what was written.

We are told that 3 different members cast bullet ballots for B.  They were counted as 3 for B.  So the one person who tried to vote twice (presumably placing a double X next to B's name) would be counted as 1 vote for B.  If you like, a line could be added to the report under Illegal Votes:  Duplicate vote for B, rejected.......1, but nothing else would change.  The total cast would still be 7, and the number required to elect would still be 4.

[Tapestry]

If the one person who tried to vote twice, marking two votes for B, the entire ballot is invalid, correct?

Edited to add:

RONR (12th ed.) 45:32 "... Similarly, a ballot that contains votes for too many candidates for a given office is counted as one illegal vote cast for that office, because it is not possible for the tellers to determine which candidate(s) the voter prefers."

There shouldn't be any confusion for the teller to determine the candidate here, but does that then mean, the teller can simply subtract the one vote and leave one vote or is the entire ballot still considered illegal?

Edited August 26, 2022 at 03:12 AM by Tapestry
Edited to add 45:32

[Richard Brown]

On 8/25/2022 at 10:06 PM, Tapestry said:

If the one person who tried to vote twice, marking two votes for B, the entire ballot is invalid, correct?

The ballot does not count as a vote cast for any candidate, but it still counts as one vote cast. In other words, it is a vote cast, but it is not credited to any candidate. See the chart on tinted  page 52 and also see section 45:32

Correction:  I think that one vote would be credited toward candidate B and the ballot would also count as one vote cast.

Edited August 26, 2022 at 03:22 AM by Richard Brown
Added last paragraph

[Tapestry]

@Richard Brown  Thank you! 

[Richard Brown]

On 8/25/2022 at 10:16 PM, Tapestry said:

@Richard Brown  Thank you! 

Not so fast, tapestry, I think that one vote would be credited for candidate B. See my corrected answer immediately above.
 

Edited August 26, 2022 at 03:27 AM by Richard Brown
Added last sentence

[Tapestry]

@Richard Brown I understand the ballot remains in the count of ballots cast.  I'm not certain one vote should be credited to B and I am of the mind (right now) it is an illegal vote.  

[Richard Brown]

On 8/25/2022 at 10:25 PM, Tapestry said:

@Richard Brown I understand the ballot remains in the count of ballots cast.  I'm not certain one vote should be credited to B and I am of the mind (right now) it is an illegal vote.  

I think it would count as one vote cast for candidate B because it is clear that he intended to vote for candidate B.

[Atul Kapur]

On 8/25/2022 at 11:06 PM, Tapestry said:

If the one person who tried to vote twice, marking two votes for B, the entire ballot is invalid, correct?

No. I agree with Mr. Brown's corrected response.

You quote 45:32, but it does not apply in this situation for two reasons

On 8/25/2022 at 11:06 PM, Tapestry said:

"... Similarly, a ballot that contains votes for too many candidates for a given office is counted as one illegal vote cast for that office, because it is not possible for the tellers to determine which candidate(s) the voter prefers."

In this case,

• the ballot contains too many votes for one candidate rather than votes for too many candidates (1st bold portion),
  and
• (2nd bold portion) it is obvious which candidate the voter prefers.

So, I would count it as one vote for candidate B.

[Dan Honemann]

I vote with Messrs. Novosielski, Brown, and Kapur, making 4 votes for A, so A is elected.

[Rob Elsman]

On 8/25/2022 at 8:13 PM, Rob Elsman said:

Since four votes are necessary to be elected, only A has been elected.  B's two or three votes are insufficient.  C's two votes are also insufficient.

I must be mincemeat pie.  😁

[Richard Brown]

On 8/26/2022 at 12:48 PM, Rob Elsman said:

I must be mincemeat pie.  😁

I know the feeling! 😢

[Richard Brown]

On 8/25/2022 at 8:13 PM, Rob Elsman said:

Since four votes are necessary to be elected, only A has been elected.  B's two or three votes are insufficient.  C's two votes are also insufficient.

We agree on candidates A and C, but I think it’s important that we nail down whether B received two or three votes. I say he received three votes because the two votes for him that were improperly cast on one ballot indicate a clear intention to vote for B and no one else. I think that should count as one vote for B.

[Rob Elsman]

The confusing way it was written is ambiguous to me.  I would have no basis to agree or disagree with anyone's interpretation.  User "puzzling" would have to tell us more clearly what was intended, and I would certainly accept whatever enlightenment we received as authoritative.  However, as to the original question about which persons might have won, neither two or three votes was sufficient, and B certainly was not elected.  I suspect that we all agree on this, even that Honemann guy.  😁

[puzzling]

On 8/26/2022 at 7:09 PM, Rob Elsman said:

The confusing way it was written is ambiguous to me.  I would have no basis to agree or disagree with anyone's interpretation.  User "puzzling" would have to tell us more clearly what was intended, and I would certainly accept whatever enlightenment we received as authoritative.  However, as to the original question about which persons might have won, neither two or three votes was sufficient, and B certainly was not elected.  I suspect that we all agree on this, even that Honemann guy.  😁

I was puzzling if candidate B was also elected (so there was no need for a second  ballot)  but I seem to be overruled in that.

I thought maybe we can conclude that the two members voting for A only, should be satisfied (all their candidates sre selected) so we can remove these ballots from the stack and treat the remaining balls as the second round. But I see that all others think these members still should be given the opportunity to vote for another in a new round) 

 

the member voting twice for B was just a late addition,  I thought everyone would know that voting twice for the same candidate (in one ballot) is counted as voting once for that candidate.

I guess it is easiertonunderstand if you count ballot slips rathet than votes.

The problems of what to do if some member votes twice for the same candidate (on the same ballot) and ballots only supporting  one candidate  simply disappear.

 

[Rob Elsman]

Oddly enough, the latest information is as confusing and ambiguous as the first.  I still do not understand whether

(A) four voters cast three legal votes for B; or,

(B) three voters cast two legal votes for B.

Is it (A) or (B)?

[Richard Brown]

On 8/26/2022 at 1:37 PM, puzzling said:

the member voting twice for B was just a late addition,  I thought everyone would know that voting twice for the same candidate (in one ballot) is counted as voting once for that candidate.

I agree that one person voting twice for the same candidate (Candidate B ) on one ballot is counted as voting once for that candidate, but apparently not everyone agrees with that. I really do think it’s important that we nail down how many votes candidate B actually received to know how to properly attribute and count votes.