Here's my take
------ A -------------- B ---------------- C --------------- D
- 1 - Lab (14,869) --- Con (11,592) ----- Lib (5,742) ------- Grn (908)
- 2 - BNP (4,340) ----- BNP (444) -------- BNP (1,024) ------ Lab (1,215)
- 3 - Con (7,109) ----- Lib (3,401) ------ Lab (981) --------- Ind (62)
I have amended your table into voter groups to better represent reality (number of votes)
FPTP: Lab win
AV:
- Grn lose
- D's second vote would go to the Lab party. This now means Lab have an additional 1,215 votes from the D block which are added to Lab
*will not count for anything unless second choice party loses.
------ A ----------------- B --------------- C ---------- D
- 1 - Lab (16,084) --- Con (11,592) ---- Lib (5,742) ------- gone
- 2 - BNP (4,340) ----- BNP (444) ------- BNP (1,024) ------ t/o
- 3 - Con (7,109) --- Lib (3,401) ------- Lab (981) ----- Ind (62)*
Now the
Lib has fewest votes so they are eliminated and block C has their 2nd vote distributed
------ A ---------------- B ---------------- C ------------ D
- 1 - Lab (16,084) --- Con (11,592) -------gone --------- gone
- 2 - BNP (4,340) ------- BNP (444) ------ BNP (1,024) --- t/o
- 3 - Con (7,109) --- Lib (3,401) ------- Lab (981)* ----- Ind (62)*
- The BNP now have 1024 votes
- This is not enough so the BNP are eliminated and so block C's third vote will be counted.
- Lab gain 981 third choice votes from block C's third choice
------ A ---------------- B ---------------- C ------------ D
- 1 - Lab (17,065) --- Con (11,592)---------gone -------- gone
- 2 - BNP (4,340) ------ BNP (444) -------- gone --------- t/o
- 3 - Con (7,109) ------ Lib (3,401) ------- t/o ------- Ind (62)*
- The Con party has the fewest votes.
- The BNP have been eliminated as have the Lib
- Ind's votes were not counted as second preferences used
- Therefore Lab wins
This is my interpretation of how AV works, it could well be wide of the mark though. AV certainly isn't as simple as FPTP!
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Your interpretation is wrong. In the initial position (which, in fact, must be at least the third round since the BNP and Ind candidates must have been eliminated earlier) you have more votes for the group D second choice than their first; this is impossible. When the Lib Dem candidate is eliminated the votes can only be redistributed to remaining candidates - Lab or Con.
To illustrate, let us assume a four way contest between candidates Arthur, Beverley, Carol and Duncan.
Arthur is the first choice of 10000 voters, Beverley of 9000.
There are 3000 people who voted for Carol as their first choice.
700 of them had Arthur as second choice, 1500 Beverley, 700 Duncan and 100 had no second choice. Of the 700 who voted C/D 500 had Beverley third and 200 Arthur.
Duncan is the first choice for 1500 voters, of whom 500 have Carol second and Beverley third, 400 Carol then Arthur, 400 Beverley then Carol and 200 Arthur then Carol.
First round: A 10000, B 9000, C 3000, D 1500
No one has a majority of the votes so Duncan is eliminated.
Second round: A 10200, B 9400, C 3900
No one has a majority so Carol is eliminated.
Third round:
A 10000 + 200 (second choice after D) + 700 (2nd after C) + 400 (3rd after D and C) + 200 (3rd after C and D) = 11,500
B 9000 + 400 (2nd after D) + 1500 (2nd after C) + 500 (3rd after D and C) + 500 (3rd after C and D) = 11,900
Beverley is elected
This looks far more complicated than it really is, largely because there are n! groups of voters for n candidates - in this case 4 factorial (4! = 4x3x2x1 = 24) though this falls rapidly after each round - in the second round there are only 3x2x1 = 6 groups voting ABC, ACB, BAc, BCA, CAB or CBA, and in the final round only two AB or BA.
It is somewhat inaccurate to refer to second or third choices in multiple rounds what counts in, say, the fourth round of voting is your highest ranked remaining candidate - this could be your fourth choice but it makes no difference in what order your first three were eliminated, your vote must be applied to someone who is still in with a chance of winning.