The count will start at 1300 on Saturday 16 June at our office, which is Langdale House, 11 Marshalea Road, London SE1 1EN. We have a limited numbers of places for supporters to attend so email elections@openrightsgroup.org if you want to join us. Below is a writeup of the intended count method.
The election will be by Single Transferable Vote using Meek's method for counting and distributing votes.
In simple terms: voters rank (some or all of the) candidates in order of preference. Candidates' first preferences are added up. If any candidate gets just more than 1/3 of the votes  “the quota”  they are elected. If two candidates have not been elected, the candidate with the lowest number of votes is eliminated and their votes redistributed.
There are two key characteristics of Meek's method:
for every round after someone has been eliminated, it is as if they no longer existed and their name was not on the ballot papers
if one person has won, but the election is carrying on because there is no second winner yet, the winner's surplus votes above the current quota are proportionately redistributed to the other candidates.
The ORG election rules require that there is an additional candidate called “Reopen Nominations” (or affectionately RON) who is never eliminated and whose votes are never redistributed. RON may win one or two of the seats. Each seat won by RON will be thrown open to another election run using the same rules.
Example election
There are two candidates so the “quota” (immediate win) is 1/3 of votes in a round, plus a very small value. In these election we will be rounding up to the nearest integer for clarity. [If we used exactly 1/3 then it would be just possible to have all the votes evenly split between three candidates, adding a very small value to the 1/3 avoids this happening].
Any Australian or New Zealander should be able to understand the example below. It's a bit more complicated than an Australian House election, but far simpler than an Australian Senate election.
For our example, we'll assume five candidates with a spread of support and 215 ballot papers. (It's much less complicated if there are only a couple of popular candidates or fewer votes.)
FIRST ROUND:
There were 5 spoiled ballot papers (eg people who put crosses against multiple candidates). They are excluded from the count, so we proceed as if there were only 210 votes cast. I'll ignore the possibility of spoiled ballots in later rounds, but if someone has only marked one or two candidates who are eliminated, their ballot will not count from then on and will be excluded for all purposes such as working out the quota.
We count the first preference votes
Candidate 
Votes 
A 
60 
B 
50 
C 
40 
D 
30 
E 
20 
RON 
10 
Quota = 210/3 + e = 71. Noone's achieved this yet, so we eliminate the lowest ranked candidate other than RON.
E is eliminated. E's 20 votes get distributed according to the next preference.
SECOND ROUND:
Candidate 
1^{st} round 
Distr. 
2^{nd} Round 
A 
60 
10 
70 
B 
50 

50 
C 
40 

40 
D 
30 

30 
E 
20 
(20) 

RON 
10 
10 
20 
Quota = 71. Noone's achieved this yet.
D is eliminated. D's 30 votes get distributed according to the next preference.
THIRD ROUND:
Candidate 
1^{st} round 
Distr. 
2^{nd} Round 
Distr. 
3^{rd} Round 
A 
60 
10 
70 
10 
80 
B 
50 

50 

50 
C 
40 

40 

40 
D 
30 

30 


E 
20 
(20) 

(30) 

RON 
10 
10 
20 
20 
40 
A has hit quota and is elected.
A has more votes than are needed to reach quota (9 more) so we need to redistribute A's excess votes proportionately. To do this we make another table showing A's second preferences. Only candidates still in the running count (remember anyone who was eliminated count as if they didn't exist), so A's votes with preferences for D or E fall through to B, C or RON.
Distribution of second choices for A 

Candidate 
Votes 
B 
30 
C 
30 
RON 
20 
A will only be allowed to keep 71 (the quota we worked out earlier) of its votes, the remaining 9 will be redistributed, so we calculate a “keep value” (the proportion of votes kept) and a “distribute value” (the proportion of votes passed on).
Keep value = 71/80 = 0.8875
so Distribute = 1  Keep = 0.1125
Using the distribute value of 0.1125 we can then work out how many votes to pass on to B, C and RON.
Distribution of second choices for A 

Candidate 
Votes 
x 0.1125 
B 
30 
3.375 
C 
30 
3.375 
RON 
20 
2.25 
(We keep A's distributed votes in a separate set of sorted piles from
the main votes, and just add to the totals for each round when
calculating whether a candidate has achieved quota and thus second
place. Three significant figures are sufficient for 210 votes.)
So our third round totals after A's votes have been distributed are:
Candidate 
1^{st} round 
Distr. 
2^{nd} Round 
Distr. 
3^{rd} Round 






start 
(from A) 
end 
A 
60 
10 
70 
10 
80 
(9) 
71 
B 
50 

50 

50 
3.375 
53.375 
C 
40 

40 

40 
3.375 
43.375 
D 
30 

30 
(30) 



E 
20 
(20) 





RON 
10 
10 
20 
20 
40 
2.25 
42.25 
Quota = 71. Noone's won the second place yet, so C's 40 votes get redistributed.
FOURTH ROUND:

1^{st} round 
Distr. 
2^{nd} Round 
Distr. 
3^{rd} Round 
Distr. 
4^{th} Round 






start 
from A 
end 

start 
from A 
end 
A 
60 
10 
70 
10 
80 
(9) 
71 
10 
90 
(19) 
71 
B 
50 

50 

50 
3.375 
53.375 
20 
70 
8.44 
78.44 
C 
40 

40 

40 
3.375 
43.375 
(40) 



D 
30 

30 
(30) 







E 
20 
(20) 









RON 
10 
10 
20 
20 
40 
2.25 
42.25 
10 
50 
10.55 
60.55 
A +10 = 90
B +20 = 70
RON +10 = 50
The above table includes the redistribution of votes from A.
A's votes' distribute = 1  (71/90) = 0.211
Distribution of second choices for A 

Candidate 
Votes 
x 0.211 
B 
40 
8.44 
RON 
50 
10.55 
Quota = 71. B is over quota, so B wins second place.