Re: [GZG] laser classes

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Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lActually, following
from what Roger say, where you're talking discrete, unrelated, events
(eg. the values on 2 dice) that need to happen for an event to occur, I
think you're better off working out the probability of it not occuring,
especially if the possibility of later events (eg. subsequent dice
rolls) is dependant on earlier, and finaly subtracting the chance of it
not happening from 1 to get the chance of it occuring.

Eg. The odds of rolling 6 on a (perfect) die is 1/6th. The odds of
rolling a 6 if you roll 6 dice is not, even theoretically 1/6th x 6, as
that = 1, which is a certainty, and I've gone more than 6 rolls between
6's, especially when that's what I needed. The odds of rolling a 6 with
6 dice = 1 - (5/6ths^6) (raised to the power of 6). This actually comes
out as 1-0.335 or .665 or 66%.

I'm fairly sure that's what's need here, but it's late, I'm knackered
and I can't remember the precise details. Is it

4 or 5 = 1 point, 6 = 2 points and reroll, with the same results?

Have to say that my stats is only basic, so it may be beyond me anyway
(short of taking the easy way out - writting a piece of VBA to explore
the full possibilities of all the rolls).

CJ
  ----- Original Message ----- 
  From: Roger Books 
  To: gzg-l@lists.csua.berkeley.edu 
  Sent: Monday, October 24, 2005 8:18 PM
  Subject: Re: [GZG] laser classes

  No, we are calculating the chance of rolling 19 points of damage on
two dice.  To do a statistical analysis on this you have to account for
the cases where one die does not roll up 6s all the way and you have
extra die rolls with the other.

  Roger

  On 10/24/05, B Lin <lin@rxkinetix.com> wrote:
    Still moot, if Die A is a 3, then it's a miss.  We are only
calculating the probability of nine 6's and a 4 or 5.  If 50 dice were
rolled previous to this run, but none hit, it doesn't change the
calculation of the 10 rolls that matter (i.e. dice aren't linked).

    In addition, the statement was that he suffered 19 points from 2
class 1 batteries in one turn, so the assumption was that the two
batteries contributed to the damage, as opposed to one doing 19 and the
other none.  It could be that one caused 4 points (2 rolls) and the
other 15 points (8 rolls),  but the odds calculation doesn't change as
long as the total dice rolled to get that result equals ten.

    There is a minor caveat in this; due to the way the FT rule is
written, the 6's have to come first, then the single point roll, as you
only get a re-roll by rolling a six. So to recreate the actual event you
have to calculate in the order of the 9 6's first then the last die. 
Statistically it doesn't matter which order the dice were rolled because
the odds remain the same.

    --Binhan

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    From: gzg-l-bounces+lin=rxkinetix.com@lists.csua.berkeley.edu
[mailto:gzg-l-bounces+lin=rxkinetix.com@lists.csua.berkeley.edu] On
Behalf Of Roger Books
    Sent: Monday, October 24, 2005 12:31 PM
    To: gzg-l@lists.csua.berkeley.edu
    Subject: Re: [GZG] laser classes

    But it could be 11 dice.

    3 on die A
    6 6 6 6 6 6 6 6 6 4:5 om Die B.

    Watch out for extra permutations.

    Roger

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

    If you break down the last two rolls (the 9th six and a single point
hit
    (4 or 5) the odds are still the same whether you roll one die or
two.

    Example: single die, chance of a 6 (1 in 6), chance of a 4 or 5 (2
in 6 
    or 1 in 3) so chance of rolling a 6 plus a 4 or 5 equals 1 in 18(6 x
3)
    or rolling the other way (4 or 5 first, then a 6)= 1 in 18 (3 x 6).
For
    a total of 2 in 18 or 1 in 9.

    Rolling two dice - chance of rolling a combination of a 6 with a 5
or 4 
    on the other die (11 chances out of 36 have a 6 (11/36) but only 4
of
    those have a 4 or 5 in them (4 of 46) final odds = 1 in 9)

    6:6 (5:6)(4:6)  3:6  2:6  1:6
    (6:5) 5:5  4:5  3:5  2:5  1:5
    (6:4) 5:4  4:4  3:4  2:4  1:4 
    6:3  5:3  4:3  3:3	2:3  1:3
    6:2  5:2  4:2  3:2	2:2  1:2
    6:1  5:1  4:1  3:1	2:1  1:1

    What the calculation is 9 6's and a 4 or 5 on a single die for a
total
    of 10 dice. It doesn't matter statistically which die (#1-10) rolls
the 
    non-six, so using ten dice, one die or any number of dice in between
    that add up to ten doesn't matter for the calculation.

    --Binhan

    -----Original Message-----
    From: gzg-l-bounces@lists.csua.berkeley.edu
    [mailto:gzg-l-bounces@lists.csua.berkeley.edu] On Behalf Of
McCarthy,
    Tom
    Sent: Monday, October 24, 2005 11:20 AM 
    To: gzg-l@lists.csua.berkeley.edu
    Subject: RE: [GZG] laser classes

    I just meant that two dice lets you have a bad roll in the mix.

    2 dice, for example, let's you roll: 6,6; 6,6; 6,2; 6; 6; 6; 6; 4 or

    6,6; 6,6; 6,1; 6; 6; 6; 6; 4 or 6,6; 6,6; 6,4; 6; 6; 6; 6; 1 and
still
    reach 19.

    Of course, if you are firm in your belief that you are as likely to
    reach 19 points of damage with 1 die as you are with multiple dice,
then 
    I really have no argument to counter that.

    > -----Original Message-----
    > From: gzg-l-bounces+tom.mccarthy=xwave.com@lists.csua.berkeley.edu
    > [mailto:
gzg-l-bounces+tom.mccarthy=xwave.com@lists.csua.berkeley.edu ]
    On
    > Behalf Of B Lin
    > Sent: Monday, October 24, 2005 1:11 PM
    > To: gzg-l@lists.csua.berkeley.edu
    > Subject: RE: [GZG] laser classes
    >
    > It actually doesn't make any difference whether you use one die or
two
    > or ten -
    > For example:
    > 1 die - chance to roll a 6 = 1 in 6, chance to roll two sixes 1 in
36
    (1
    > in 6 x 1 in 6)
    > 2 dice - chance to roll two sixes, 1 in 36.
    >
    > You can either roll a single die ten times or roll ten dice once
each 
    > and the odds are exactly the same.  Remember dice have no memory
and
    are
    > not linked to each other (in theory) so one die's result is not
    affected
    > by the result of a previous roll, a future roll or neighboring
die's 
    > roll.
    >
    > --Binhan
    >
    > -----Original Message-----
    > From: gzg-l-bounces@lists.csua.berkeley.edu
    > [mailto: gzg-l-bounces@lists.csua.berkeley.edu] On Behalf Of
McCarthy,
    > Tom
    > Sent: Monday, October 24, 2005 10:48 AM
    > To: gzg-l@lists.csua.berkeley.edu
    > Subject: RE: [GZG] laser classes
    >
    > For 6 to the 9th, I get 10,077,696.  That makes the odds of
getting
    > exactly 19 points on one die to be 1 in 30,233,088 or so.  On two
    dice,
    > it's more likely, but still pretty unlikely. 
    >
    >
    > _______________________________________________

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