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# RE: [GZG] laser classes

From: "B Lin" <lin@r...>
Date: Mon, 24 Oct 2005 12:41:53 -0600
Subject: RE: [GZG] laser classes
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http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lStill 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

________________________________

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
<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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>
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