Re: OA Elections (math & my, my)
Ted Burton (tedburtn@HALCYON.COM)
Fri, 2 Jun 1995 22:11:48 -0600
Fred Rogers <email@example.com> wrote:
>On Wed, 31 May 1995 10:29:05 -0400, Fred Weber <Hoopcoach7@AOL.COM> writes:
>> I too have seen the problems with OA elections.....and I have a
>> comment....First how can 4 out of five get elected....I thought the rule was
>> 50% of those eligible....
>This one I can answer. The rules state that any scout who appears on
>50% of the ballots is elected. With five eligible, each ballot can
>contain up to 3 names. Thus, if the votes split right, all five could
After reading the recent dialogue in which I participated somewhat
doctrinarily (is that a word?), most especially Fred's latest, I broke down
and put the election process on a spreadsheet. What emerges is that, yes,
the election results can indeed generate the election of all those
eligible. It is equally true the same process can be deadly to all under
certain combinations of number of voters/number of eligibles voted-for.
This results from the fact that if the votes are split evenly across all
candidates, then most ratios of all voters to all candidates will result
in all being elected. However other ratios of all votes to all candidates
would result in none being elected.
My apologies to Stan for being in the mind set that only half can be
elected; rather it is that only half may be on any one voter's ballot.
The problem of the SM with 44 eligible, and no one being elected, is not as
perplexing as it might seem. If he had 44 voters, distributing the maximum
number of votes evenly among all 44 candidates, all would have been elected
by the exact number of votes it takes. If he had 200 voters, distributing
the maximum number of votes evenly among all 44 candidates, all would have
been elected by the exact number of votes it takes. Under either of the
situations, voters voting other than evenly would have elected some of the
44. His problem is short ballots.
The real complexity is introduced if some voters vote for fewer than the
maximum number allowed. If fewer than the maximum number of votes are
actually cast per ballot, and the ballots are in fact evenly split among
all candidates, then no one will be elected; those 'no' votes of blank or
less-than-full ballots count. The more partial ballots you have, the more
likely it will be that candidates will not be elected. If you have 25
scouts voting, and 4 eligible scouts (meaning a maximum of 2 candidates
voted on any one ballot), and ten of the scouts vote only for 1 candidate
(rather than 2), and if the votes are evenly distributed among the 4, each
will get 10 votes of the 13 needed. If there are 44 eligible, and 75
voting, but 60 of those voting vote only for 11 candidates, not 22, the
average votes per candidate falls to 22 or 23, well short of the 38
The moral of the story under the present rules is to elect once a year so
that your number of candidates does not grow to the point that by the
nature of things it is unrealistic for all scouts to figure out who to vote
for, especially in larger troops where partial ballots are more likely. The
Scoutmaster of the troop large enough for people not to be fully familiar
with each other even when around a long time, must be strict with his own
use of his own 'veto', that is, use it unmercifully. Also, he might point
out the actual number of votes it is going to take to get elected, and
encourage people to focus on the best of the best, and not to think that
'oh well Harry is surely going to make it so I will vote for someone else.'
If one wishes to think of our SM with 44 eligible scouts, a rule change
could be useful. A way to get around the problem of a large number of
potentially unfamiliar candidates (partrols unfamiliar with the real goings
on in other patrols) would be to require a majority of the
_average_of_all_votes cast_, rather than of the ballots turned in. That is,
instead of being on half the ballots, having more votes than the average
computed by dividing all votes by the number of all persons permitted to be
on a ballot. What that in effect means is eliminating the rule that a blank
ballot is counted, and eliminating the rule that a partial ballot is a 'no'
as to all other candidates. One would total all votes cast, divide by the
number of candidates permitted to be on any one ballot, and see who
equalled half or more of that in votes. The concept of the no vote can be
preserved by permitting one to turn in a ballot that expressly says "no",
that is, that no one in the view of the voter deserves election, and
counting the no. If one has 4 candidates, Tom, Dick, Harry, and Joe, and is
permitted to vote for 2: a voter could vote Tom and Dick; or Tom and no; or
no, no. If doing that, then the "no's" could be counted in as votes cast.
However, blank or party blank ballots would be disregarded. This would fix
the problem of the SM with 44, unless those 44 really should all have been
By the way, if we hold no election because we fear Tom will not get
elected, well, then, it becomes certain that Tom will not be elected, and
neither will Dick or Harry.
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