Prev: RE: [GZG] Some easy Dirtside questions Next: Re: [GZG] laser classes

RE: [GZG] laser classes

From: "B Lin" <lin@r...>
Date: Mon, 24 Oct 2005 14:20:52 -0600
Subject: RE: [GZG] laser classes

_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lActually you don't.
 You only have to look at the cases that will
produce the correct result, for instance we don't look at cases with 3
batteries, or situations that use 20 die rolls or include the X number
of die rolls that occurred before the streak.  The case in question is 9
sixes and a 4 or 5 for ten dice total.	We ignore the two rolls after
these ten because they result in misses (one miss for each battery,
otherwise they would either score more damage and possibly receive an
additional die-roll which changes the problem.) and so fall outside of
the question.



So in your case listed below, yes, it's a possible case but we would
discount the 3 on Die A since it's a miss, so you are back to 2 sixes,
plus seven sixes and a four which adds up to the required total of 9
sixes and a 4 or 5.  Another possibility is three 6's and  a five on Die
A and then six 6's on Die B.



Because of the re-roll rule, there is only one possible combination to
get this result - a series of 9 sixes (split anyway with at least one
die to each battery) and then a roll of 4 or 5.  It doesn't matter which
die, A or B rolls the 4 or 5, but the 4 or 5 has to happen at the end of
the streak.  This ordering is forced by the rules definition, not by
statistical analysis. From statstical point of view the 4 or 5 could
occur with either die at any position since the dice aren't linked
statistically to each other and the overall odds aren't affected by the
order of the dice.



--Binhan

________________________________

From: gzg-l-bounces@lists.csua.berkeley.edu
[mailto:gzg-l-bounces@lists.csua.berkeley.edu] On Behalf Of Roger Books
Sent: Monday, October 24, 2005 2:00 PM
To: gzg-l@lists.csua.berkeley.edu
Subject: Re: [GZG] laser classes



So you are saying if I roll two 6s and a 3 on one die and a
7 6s and a 4 this combination doesn't count as a possible
combination?

You are starting from the assumption that you can roll a handful of dice
and get the correct number of combinations.  You need to include all
possible 
combinations.

Roger 

On 10/24/05, B Lin <lin@rxkinetix.com> wrote:

The dice are only linked since we are looking at a specific incident and
by the rule that a 6 has to be rolled before you get a re-roll.  For all
we know they rolled 22 dice for Class 1 beams that turn and only 2 got a

re-roll but went on to score an additional 15 points of damage.  The
point here is that all those events (all the non-rerolls) have no effect
on the remaining 15 points of damage.  By default, any roll that doesn't

produce damage is not included in the calculation (for instance there
really had to be 12 rolls not 10 since at some point they would have had
to roll a miss for each beam battery to end the streak). If the
statement had been 19 points of damage from 2 class 1 batteries over the

game, then the calculation changes, since you then have to include all
the die rolls of all the Class 1 batteries fired during the game, but
the only way to get 19 points of damage from two Class 1 beam batteries 
in a single turn is to roll 9 6's and a 4 or 5. (Had the damage total
been an even number, then there would have been two possible damage
combinations; all 6's or all 6's with the last two being 4 or 5's).

--Binhan

-----Original Message-----
From: gzg-l-bounces@lists.csua.berkeley.edu
[mailto:gzg-l-bounces@lists.csua.berkeley.edu ] On Behalf Of Doug Evans
Sent: Monday, October 24, 2005 1:02 PM
To: gzg-l@lists.csua.berkeley.edu
Subject: RE: [GZG] laser classes

For this discussion, these dice are linked, though. 

Die A could be 3, and die B could be 6; however, die B could be 3, and
die
A could be 6.  Or both could be 6.

If both are six, then you move on to the next roll with both dice.
However,
as either of the previous dice could have been six even if the other 
wasn't, you have a slightly greater chance of starting a run of sixes
with
two dice than with one. And that slightly better chance is found in each
case of rolling two dice.

I think.

Gawd, my brain hurts, and I have to go to a staff meeting. 

I think I gave up before, but this time I REALLY mean it.

The_Beast

_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu 
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l

_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l



Prev: RE: [GZG] Some easy Dirtside questions Next: Re: [GZG] laser classes