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

The reason it's confusing is that you haven't stated the problem
clearly. You have taken the re-roll portion out of context and
re-applied it. For instance, your example, for some reason, adds more
dice rolls past 4, they should be compared directly to 4 dice rolls- it
doesn't matter if the 5th or subsequent rolls are a 1 or a 6, only the
that the first 4 were 6's (by the assumed definition of "EXACTLY 4 6's"
and not 4 6's and a couple of other numbers...). You've added points
that are irrelevant to generate your numbers so they don't compare
directly with each other. Along those lines, your example has you
rolling many more times in your 4 dice example than the single die
example which produces an artificially high probability. To make it
comparable you have to compare rolling a set of 4 dice once (4 rolls
total) against rolling 4 single dice (4 rolls total).  Your example is
rolling a single die and re-rolling only sixes, vs. rolling a set of 4
dice and re-rolling 6's. It's a clear case where you are comparing 1
against 4.  

The chance to roll 4 sixes with a die rolled 4 times is 1 in 1296. It
doesn't matter if you take one die, roll it 4 times or take 4 dice, roll
them once each or 2 dice and roll one 3 times and one once, you still
roll a total of 4 dice that are independent of each other and produce
the exact same number of permutations - 1296. Of those permutations only
one has 4 sixes in it. Therefore the odds of rolling 4 sixes and EXACTLY
4 6's is 1 in 1296 no matter which order or grouping of dice you use. 

The dice are only linked because the FT rule says that to get a re-roll
you have to roll a 6, this narrows the search to only dice rolls that
contain a 6, except for the last one, which is looking for a 4 or 5.
This means to produce 19 points of damage you need 9 rolls of 6 followed
by a roll of 4 or 5.  The problem is carefully delineated and already
has taken the re-rolls into account.  There are exactly 2 possibilities
that follow this definition 9,9,9,9,9,9,9,9,9,5 and 9,9,9,9,9,9,9,9,9,4
out of 60,466,176 permutations. 

By defining the problem as 9 6's we have already taken the re-roll in
account, in addition, the exact number of re-rolls was already
determined (9). Your example seems to imply that if you get a six you
are allowed an additional roll even past the stated number of rolls (4).

You don't have to consider multiple batteries for the following reasons-

Part 1)The total is an odd number of points (19).  You can only get a
re-roll on a 6 and that produces 2 points (an even number). So to
generate an odd number of points you need an odd point roll (4 or 5).
Two odd rolls make an even and three would not be possible since you
wouldn't get the reqired number of re-rolls. Therefore only 1 die can
roll a 4 or 5 to produce a single or odd number of points.  

Part 2) Batteries are batteries - a class 1 beam battery gets one die,
no special modifiers or rules.	Same batteries are interchangeable (i.e.
a class 1 is the exact same as any other class 1) each produces 1 die of
fire that has the same to hit chance as any other Class 1.

Part 3) If batteries are the same then it doesn't matter which die roll
is associated with which battery as they give the same result - a 6 is a
6, and a 6 by a class 1 battery is the same as a 6 by a class 1 battery.

Part 4) if the 6's are equivalent then it doesn't matter if they come
from 1 battery or 2. And it doesn't matter which battery rolls the 4 or
5.

Final - All the rolls can be treated as having come from a single Class
1 battery.

If there were three or more batteries involved, then the calculation
would then have to include partial hits from multiple batteries and
would be much more complicated.

--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 3:35 PM
To: gzg-l@lists.csua.berkeley.edu
Subject: 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
>

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