[GZG] [OT] statistics, was: laser classes

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Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lThat is what is
used when your chances are increasing. For instance if I
wanted to know my chance of rolling at least one six on two dice it
would
be:

1-( 5/6 * 5/6 ).

If I wanted to know the chance of rolling two sixes on two dice it's:

1/6 * 1/6

Roger Books

On 10/24/05, R Campbell-Jones <rcj@dircon.co.uk> wrote:
>
> Actually, 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 <roger.books@gmail.com>
> *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
> >
> >   ------------------------------
> >
> > *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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> > >
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> >
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