============================== CFJ 1577 ==============================
The Registrar has not legally resolved the Rebellion that occurred
on December 15th, 2005.
Called by Sherlock: 17 Dec 2005 13:27:57 GMT
Assigned to Quazie: 01 Jan 2006 01:31:43 GMT
Quazie recused: 23 Jan 2006 00:00:00 GMT
Assigned to OscarMeyr: 13 Feb 2006 08:45:52 GMT
Judged FALSE by OscarMeyr: 13 Feb 2006 20:03:35 GMT
Registrar OscarMeyr used a roll of 1 - 12 to determine the
success of the Rebellion. However, per eir own admission,
there were only 11 registered players at the time of Rebellion.
Rule 1079 (Definition of "Random") is fairly clear that the
choice has to made between the "possible outcomes":
(a) When a Rule requires a random choice to be made, then
the choice shall be made using whatever probability
distribution among the possible outcomes the Rules
provide for making that choice. If the Rules do not
specify a probability distribution, then a uniform
probability distribution shall be used.
By Agoran rules, 12 was not a possible outcome of Rebellion,
so it should not have been included as a possible roll in the
determination, regardless of whether the Registrar would have
ignored it as a result.
Judge OscarMeyr's Arguments:
The result of my random determination (selecting a random number from 1
to 12, and trying again on a result of 12) to resolve the rebellion was
mathematically equivalent to selecting a random number from 1 to 11.
Goethe posted eir proof of this in
Judge OscarMeyr's Evidence:
Judge's evidence, the aforementioned post by Goethe:
>> Would u mind submitting your math you menitioned doing as evidence
>> on CFJ 1577?
Gratuitous Evidence for CFJ 1577, by request:
Part 1. Math.
The algorithm used is: Roll 1d12, and re-roll on every 12 (forever
if needed), until a 1 through 11 comes up, terminating the process
with the last number.
What are the odds of a 1 coming up in the terminus?
Odds are 1/12 of rolling a 1 on the first roll, terminating.
Odds are 1/12 * 1/12 of rolling a 12 on first roll, then
a 1 on the second roll, terminating.
Odds are 1/12 * 1/12 * 1/12 of rolling 12-12-1, terminating.
Odds are (1/12)^4 of rolling 12-12-12-1, terminating.
Since the above sequences are mutually exclusive, the probabilities
can be added, giving the probability of rolling a 1 as:
SUM[k=1...inf](1/12)^k = (1/12) / (1-(1/12)) = 1/11
(proof ommitted, see equation 9 at
So there's a 1/11 chance of terminating the process with a 1 (or a
2, or 3 or ... or 11).
Of course, as Maud pointed out, if the only allowable outcomes are
1 to 11, and there's no reason to bias any one of those, so the math
isn't really needed to make the same argument.
Part 2. Why this doesn't matter
As I said before, my play of Not Your Turn
"took back" one Rebel Rabble, and changed the odds. So the odds
should have been 9/11 of failure, not 10/11. So OscarMeyr chose
from the wrong distribution. E pointed out that eir roll of "10"
would have been a failure under either distribution, so e shouldn't
be required to re-roll. This IMO is a more interesting reason to
consider this CFJ.