[EM] IRV for president: 2 votes overpower 105 million
Craig Carey
research at ijs.co.nz
Mon Jul 14 00:34:04 PDT 2003
Mr Keshet of the wikipedia.org website gave grounds saying he could not unstand
the lot and dashed off.
I have produced something that can be put into advocacy material.
Here I show that the Alternative Vote will almost negate (i.e. wrongly lose)
the votes of 105 million Americans (after the numbers are multiplied out.
The superior power eradicating approx 101 millions is 2 voters.
Simple enough since they were using a special case of the STV ballot paper.
In Takoma Park, Maryland, (77deg0min West, 38deg59min North), in USA's North
East, AV was named it the Instant Runoff Vote. It is long way from Washington.
I now take the view that the CVD should trademark (whatever), the brand name
IRV. It could be like opening a solar sail to catch Green puffs and slow
the CEO's descent.
The following text is a revised part of my submission to the wikipedia.org.
The text is also here for a while: http://www.ijs.co.nz/irv-wrong-winners.htm
This is the final 4 candidate example (except for changes in the 2 totals).
Beyond the following is the very simple example where 'IRV' negates or loses
43.75% of the votes for the president.
_________[start]_____________________________________________________________
The example below implies that if the President of the United States of
America was elected with the Alternative Vote (known under the popular
name, IRV, coined by Maryland's Centre of Voting and Democracy), then 2
Alternative Vote papers will neutralize and be victorious over about 100
million Americans that are set against thise two other individuals.
[Currently the population of USA is "280,562,489 (July 2002 est.)" says
the CIA Factbook. That times 0.375 equals 105.2 million].
· · · · · · · · · Votes · · · · · · · · · · · · ·Percentage
· ·+----------------------------------+ ·-------------------------
· ·|· A · · 50,000,001 · 199,999,999· | · 20.0000004· 49.99999975
· ·|· BA· · 49,999,999 · 50,000,001· | · 19.9999996· 12.50000025
· ·|· CB · 100,000,000 · 100,000,000· | · 40· · · · · 25
· ·|· DBA · 50,000,000 · 50,000,000· | · 20· · · · · 12.5
· ·+----------------------------------+ ·-------------------------
·Winner: · · · · · · A · · · · · · B
·Total: · ·250,000,000 · 400,000,000
·
The calculations leading to the winners A and B:
·
· ·+-------------------------------------+
· ·| A· · 100,000,000 | A · ·199,999,999 |
· ·| C· · 100,000,000 | BA · 100,000,001 |
· ·| DA · ·50,000,000 | CB · 100,000,000 |
· ·+-------------------------------------+
· ·| A· · 150,000,000 | A · ·199,999,999 |
· ·| C· · 100,000,000 | B* · 200,000,001 |
· ·+-------------------------------------+
Notes: (1): 199,999,999 - 50,000,001 = 149,999,998.
(2): (149,999,998-2)/400,000,000 = 37.499999%.
If an improvement of AV (a variant) were to be checked, then there could
be sound suspicion about whether it is actually better, if the 37.5%
figure didn't reduce enough. It may shift a bit so a search (e.g. using
"for loops") would find a worst case example of the variant.
The method of the city of London allows only two preferences on papers.
The rule of the previous paragraph implies that 3 preferences would be
too many. A better idea is to put a 1/3 quota in somehow.
AV is conjectured to be not corrupt by the author, i.e. here the (BA)
papers must have an FPTP power around 1.0, and hence the 150 (A) papers
have a power under about 1/74,999,999 (it could be negative). As would
happen in real STV elections, the method seems to be transferring some
of the votes to a wrong candidate. To get the power of the papers up
again, some of the voters would do the same themselves. They lobby their
political party to runs simulations for them to make up for the loss of
the right to vote. Exactly which wrong candidate the 150 million ought
start approving of, is made less interesting by the lack of knowledge on
whether the Alternative Vote got it wrong in the first election, the
second, or both.
_________[end]_______________________________________________________________
Without the middle dots:
Votes Percentage
+----------------------------------+ -------------------------
| A 50,000,001 199,999,999 | 20.0000004 49.99999975
| BA 49,999,999 50,000,001 | 19.9999996 12.50000025
| CB 100,000,000 100,000,000 | 40 25
| DBA 50,000,000 50,000,000 | 20 12.5
+----------------------------------+ -------------------------
Winner: A B
Total: 250,000,000 400,000,000
The power of the '(A)' papers is about
REDLOG found the papers and 'for loops' optimized the numbers.
How many papers to use in an example that is said to be about the
presidential election ?. I.e. past or projected countable papers.
-------------
Here is the REDLOG program that found the numbers for the 4 papers.
This is here in case someone wants to change the totals and not
have someone later claim to have a better example for those totals.
tot2:=400000000; tot1:=0.625*tot2; s:=tot1/5; r:=30; r1:=r+1;
k1:=0;k2:=0;k3:=0;k4:=0;k5:=0;
precision 13; % number of decimal digits if "on rounding"
on fullprec; % off implies don't print trailing zeros
off rounded;
for ba1:=s-2-r:s-2+r do begin
ba2 := ba1 + 2;
for a1:=ba1+1:ba1+r1 do
for dba:=ba1+1:min(ba1+r1,ba2-1) do begin
for cb:=ba2+dba-1 step -1 until ba2+dba-r1 do begin
tot1 := a1+ba1+cb+dba;
if tot1 = 249999999 + 1 then begin % 249999999 is about optimal
%write "cb=",cb," dba=",dba," ba2=",ba2," tot1=",tot1;
for a2:=ba2+dba+cb-1 step -1 until ba2+dba+cb-r1 do
% if cb = tot1 - a1-ba1-dba then ...
if tot2 = a2+ba2+cb+dba then begin
if ba1<cb then k1:=1; if dba < cb then k2:=1; if cb < a1+ba1+dba then k3:=1;
if dba < cb then k4:=1; if cb < a2 then k5:=1;
if ba1<cb and (dba < a1+ba1 and dba < cb) and (cb < a1+ba1+dba)
and dba < a2 and dba < cb and a1 < a2 and cb < a2 then begin
m1 := 1000*((a2-a1)-1.999*(ba2-ba1))/(0.6*tot1) - 999;
m2 := 1000*((a2-a1)/(ba2-ba1)) / (0.15*tot1) - 1999;
m3 := 1 + 0.0000001 / (1 + abs(tot1*((cb/dba) - 2)));
on rounded;
write "@ ",m1+m2+m3,": ",m1," ",m2," ",m3," co1";
off rounded;
write " ba1 ",ba1," a1 ",a1," dba ",dba," ba2 ",ba2," cb ",cb," a2 ",a2,
" ",ba2+dba-cb," ",r;
end; end; end; end; end; end; end;
write {k1,k2,k3,k4,k5};
a1 < a2 # wide
ba1 < ba2 # close, redundant
-----------
ba1 < min(a1,dba) # 1close
ba1 < cb # 1wide
dba < ba2 # 2close
dba < min(a2,cb) # 2wide
-----------
dba < min(a1+ba1,cb) # 1wide
cb < ba2+dba # 2close
cb < a2 # 2wide
-----------
cb < a1+ba1+dba # 1wide
a2 < ba2+cb+dba # 2close
Craig Carey
Auckland, New Zealand
http://groups.yahoo.com/group/politicians-and-polytopes/
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