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

On 10/24/05, B Lin <lin@rxkinetix.com> wrote:
> Not wholly applicable.  We aren't rolling thousands of dice and trying
> to determine how often a set of 9 6's comes up.

Actually you are, you just don't think you are :-). One of
the easiest ways of calculating the probability of an event
is to run the event 1000's of times and count the number
of times it occurs and then divide by the number of times
you tried.  It's called a Monte Carlo simulation.  It's often
(with the use of a computer) significantly simplier than
calculating a probability the old fashion way. It's not 100%
accurate, but it's close enough.  So I'd say, when you're
trying to calculate the probability of event occuring, it's
effectively the same thing as attempting to do the event
an infinite number of times (or some really large number)
and counting the number of times it occured (and dividing
by the number of attempts).

The simulation I described, was closely related to the
original problem (the odds of rolling 19 points of damage
on 2 class-1 beams).

> 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.

Well that's 10 * (1/6)^9 * (2/6) * 2 ... but that's not the same
as the odds of you rolling 9 6's in a row followed by a single
4 or 5 (which is (1/6)^10 * 2) which is also not the same as
rolling 2 beam dice of damage and getting 19 pts of damage.

> 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.

I'm starting to think you didn't actually read what I wrote.

Daryl

> --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.
> > >
> > >
> > > _______________________________________________
> >
> >
> >
> > > Gzg-l mailing list
> > > Gzg-l@lists.csua.berkeley.edu
> > > http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
> > >
> > > _______________________________________________
> > > Gzg-l mailing list
> > > Gzg-l@lists.csua.berkeley.edu
> > > http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
> >
> >
> > _______________________________________________
> > Gzg-l mailing list
> > Gzg-l@lists.csua.berkeley.edu
> > http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
> >
> > _______________________________________________
> > Gzg-l mailing list
> > Gzg-l@lists.csua.berkeley.edu
> > http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
> >
> >
> >
> >
> > _______________________________________________
> > Gzg-l mailing list
> > Gzg-l@lists.csua.berkeley.edu
> > http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
> >
> >
> >
> >
> >
>
> _______________________________________________
> Gzg-l mailing list
> Gzg-l@lists.csua.berkeley.edu
> http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
>
> _______________________________________________
> Gzg-l mailing list
> Gzg-l@lists.csua.berkeley.edu
> http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
>

_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l