RE: [GZG] laser classes

Correct with some minor corrections.

Your case #1 where Die 1 rolls a miss right off the bat is eliminated
since that would result in a SINGLE beam doing 19 points of damage,
since you multiply by two (die 1 results and die 2 results) you only
have 18 cases left.  By implication, Case #10 is unlikely since I don't
think he would have included the second beam to the total if it only
provided a single point.  Again subtract two cases from the total
leaving 16.

Each case has a 1 in 60,466,176 but remember that there are two
possibilities for scoring a single point (rolling a 4 or 5) so the odds
are actually  1 in 30,233,088.	The cases are additive so with 16 cases
odds are 16 in 30,233,088 so the final count is 1 in 1,889,568.

--Binhan

-----Original Message-----
From: gzg-l-bounces@lists.csua.berkeley.edu
[mailto:gzg-l-bounces@lists.csua.berkeley.edu] On Behalf Of Brian B
Sent: Wednesday, October 26, 2005 8:28 AM
To: gzg-l@lists.csua.berkeley.edu
Subject: Re: [GZG] laser classes

On 10/25/05, B Lin <lin@rxkinetix.com> wrote:
> The straight beam damage (not against shields) is 1 point of damage
for
> a roll of a 4 or 5 and two points of damage and a re-roll for a 6.
>
> --Binhan

Thanks for clarifying.	The reason I asked was because I wanted to
wrap my own head around this, and was looking back on James' original
post, because it seemed to me that what he originally said got lost in
the shuffle.  Maybe looking at it again will clarify things:

" i have a problem with 2 class 1 laser's doing a total of 19pts of
damage in one turn"

So we have two dice being rolled, and one of those dice does X amount
of damage, and the other Y, and the total damage added up equals 19. 
Setting aside all of the sixes, in order to reach an odd number of
damage, one of the two dice had to end its rolls with a 4 or 5, and
the other die ended its rolls with a 1, 2, or 3, for 0 damage.	If
they had both ended with a 4 or 5, or both with 3 or less, the total
damage would have been equal.

>From there I realized that Tom B was on the right track with adding up
all the combinations that equal 19.  Since one of the dice ends with a
4 or 5, that leaves 18 points done by 6's, or a total of 9 6's in a
row.  That means the odds of rolling 19 points of damage is equal to
the aodds of one die rolling X 6's in a row, followed by a 4 or 5, and
the other rolling 9-x 6's in a row, followed by a 1, 2, or 3.  And
vice versa.  So:

The odds of rolling 1, 2, or 3 (Result A) are 1 in 2.
The odds of rolling a 4 or 5 (Result B) are 1 in 3.
The odds of rolling a 6 (Result C) are 1 in 6.

So the following roll combinations result in 19 points of damage:

Die 1					    Die 2
1.A					       CCCCCCCCCB
2.CA						CCCCCCCCB
3.CCA					       CCCCCCCB
4.CCCA					      CCCCCCB
5.CCCCA 				     CCCCCB
6.CCCCCA				     CCCCB
7.CCCCCCA				      CCCB
8.CCCCCCCA				      CCB
9.CCCCCCCCA				      CB
10.CCCCCCCCCA				       B

Noww, I could go on to calculate:
CCCCCCCCCB				    A
Etc., but since a=b is the same as b=a, I'll just double the numbers
derived from above.

1. that's 1 in 2 multiplied by 1 in 6^9x3, or 1 2x3x6^9 or 1 in 6x6^9
or 1 in 6^10.
I think I see where this is going.
2 thats 1 in 6x2 multiplied by 1 6^8x3 which is.... yes, 1 in 6^10....
each of the 20 results have the same odds.  that means the odds ar
1x20 in (6^10)x20, or 1 in 6^10.  SO the odds of rolling 19 points of
damage when firing two beam 1's is 1 in 60, 466,176.

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