[OT] Statisics - was RE: [GZG] Re: laser classes

Not wholly applicable.	We aren't rolling thousands of dice and trying
to determine how often a set of 9 6's comes up.  The question is what is
the probability that exactly 9 6's and a 4 or 5 will show up if you roll
10 dice.

Be the FT rules definition, the first 9 have to be 6's and the last one
a 4 or 5 so any combination that doesn't fit that criteria is excluded
by the definition of the problem.

If you change to problem to being 30 class 1 dice were rolled and 9 6's
were scored, that is a completely different animal.

--Binhan

-----Original Message-----
From: gzg-l-bounces@lists.csua.berkeley.edu
[mailto:gzg-l-bounces@lists.csua.berkeley.edu] On Behalf Of Daryl Lonnon
Sent: Monday, October 24, 2005 2:40 PM
To: gzg-l@lists.csua.berkeley.edu
Subject: [GZG] Re: laser classes

I respectfully disagree.

Tell you what, we each roll some dice 10,000 times (rerolling 6's). 
I'll roll mine in
chunks of 4, you'll roll yours as individuals (so I'm only going to
roll 2,500 times).
Everytime we roll EXACTLY 4 6's, we pay the other guy a dollar.  I'll
guarantee you
I'll end up on top.  Everytime we do it, I'll end up paying you about
50 dollars, you'll
end up paying me about 300.

The mistake you're making is that you're assuming independence between
dice rolls.
There IS dependence between set of rolls (within a set of rolls it's
independent).
Where I'm defining a set as a fistful of dice (4 in this case versus
1).  The dependence
between sets of rolls is that you don't get to roll that dice again if
you don't roll a 6!

Or to use your example from below (hopefully I get all the numbers
correct):

Probability of rolling EXACTLY 4 6's when rerolling 6's
(everthing in a parathensis can be viewed as one "roll (of potentially
 multiple dice)")

1 dice:
1st   2nd   3rd   4th	5th
(1/6) * (1/6) * (1/6) * (1/6) * (5/6) = 0.00064

2 dice:
(1/6 * 1/6) * (1/6 * 1/6) * (5/6 * 5/6) +
(1/6 * 1/6) * 2(1/6 * 5/6) * (1/6) * (5/6) +
2(1/6 * 5/6) * (1/6) * (1/6) * (1/6) * (5/6) =
5 * (1/6)^4 * (5/6)^2 = 0.002679

3 dice
(1/6 * 1/6 * 1/6) * 3 (5/6 * 5/6 * 1/6) * (5/6) +
3 (1/6 * 1/6 * 5/6) * (1/6 * 1/6) * (5/6 * 5/6) +
3 (1/6 * 1/6 * 5/6) * 2 (1/6 * 5/6) * (1/6) * (5/6)
3 (1/6 * 5/6 * 5/6) * (1/6) * (1/6) * (1/6) * (5/6) =
20 * (1/6)^4 * (5/6)^3 = 0.0089

4 dice
(1/6 * 1/6 * 1/6 * 1/6) * (5/6 * 5/6 * 5/6 * 5/6) +
4 (1/6 * 1/6 * 1/6 * 5/6) * 3 (1/6 * 5/6 * 5/6) * 5/6 +
6 (1/6 * 1/6 * 5/6 * 5/6) * (1/6 * 1/6) * (5/6 * 5/6) +
6 (1/6 * 1/6 * 5/6 * 5/6) * 2 (1/6 * 5/6) * (1/6) * (5/6) +
4 (1/6 * 5/6 * 5/6 * 5/6) * (1/6) * (1/6) * (1/6) * (5/6) =
35 * (1/6)^4 * (5/6)^4 = 0.013

Hopefully this makes some sense,

Daryl

On 10/24/05, B Lin <lin@rxkinetix.com> wrote:
> As previously covered - it doesn't matter if you roll 1 die twice or
two
> dice once each, the odds are the same.  By transitive, it doesn't
matter
> if you roll one die 9 times, or nine dice once each or 3 dice and 6
dice
> together, the odds are the same.
>
>
>
> Short example:
>
>
>
> Chance to roll four 6's:
>
>
>
> Single die - 1 in 6 6x6x6x6 = 1 in 1296
>
> Two dice, twice each = 2 sets of 1 in 36 = 36 x 36 = 1296
>
> Two dice, 1 on A, 3 on B = 1 in 6 x 1 in 216 = 1296
>
> Two dice, 3 on A, 1 on B = 1 in 216 x 1 in6 = 1296
>
> Three dice, one on A, one on B, two on C = 6x6x36 = 1296
>
> Four dice, once each = 6x6x6x6 = 1296
>
>
>
> --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 1:18 PM
> To: gzg-l@lists.csua.berkeley.edu
> 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 <mailto: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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> > Gzg-l@lists.csua.berkeley.edu
> > http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
> >
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