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

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Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lActually no. You
are extrapolating backwards from the calculated result
in a different direction than the original question.

The odds (1 in 30,233,088) are for getting to the final total of 19
points.  Each step has a smaller probability, but when combined, reach
the final ratio. For each die roll, the chance to get a six is 1 in 6.
So in your case, whether you roll a single die twice, or two dice at the
same time, each individual die has the same chance to get a 6.



An example is listed below:



If you are proceeding from the beginning with two class one batteries.
This means you get two dice to roll.  To get to 19 points of damage you
need to get a series of re-rolls.



For this example I will assume that the rolls end up even between the
two batteries, but the case works just as well for any combination of
hits (4,5 or 6) between the two batteries that add up to ten dice total.



First roll - Both dice have to get a six, otherwise you don't get a
re-roll - chance of getting 2 6's on 2 dice, 1 in 36

Second roll, Both dice have to get a six otherwise you don't get a
re-roll - chance of getting 2 6's on 2 dice, 1 in 36

Third roll, Both dice have to get a six otherwise you don't get a
re-roll - chance of getting 2 6's on 2 dice, 1 in 36

Fourth roll, Both dice have to get a six otherwise you don't get a
re-roll - chance of getting 2 6's on 2 dice, 1 in 36

Fifth roll, One die gets a 6, the other die gets a 4 or 5 with no
re-roll, (1 in 6 and 1 in 3), overall 1 in 18



Technically there is a sixth roll, but by definition of the problem you
don't count the rolled miss.  If the roll wasn't a miss, then the point
total would be higher and that's a different problem.



This is no different than rolling a single battery 9 times for a six and
then once for a 4 or 5.



You are asking a different question: Does using more batteries produce
an increased chance of causing 19 points, and the answer is yes.  The
above analysis was for a specific event and limited itself to fixed
end-point - 19 points in 10 dice.  You are asking if given an unlimited
pool of dice, would more dice have a greater chance of scoring 19 points
than fewer and the answer is yes. If you limit the dice pool (10 dice)
the answer is no.



--Binhan



________________________________

From: gzg-l-bounces@lists.csua.berkeley.edu
[mailto:gzg-l-bounces@lists.csua.berkeley.edu] On Behalf Of Roger Books
Sent: Tuesday, October 25, 2005 11:55 AM
To: gzg-l@lists.csua.berkeley.edu
Subject: Re: Statistics - was RE: [GZG] Re: laser classes



Then I, sir, am going to start rolling my dice one beam at a time.

I have the same chance coming up with 19 on one beam as I do on two.

Therefore if I roll the dice per beam I have a 1 in 1.6 million(?)
chance of getting 19 with the first roll.  With the second die I have a
1 in 1.6 million.  That means if I roll my dice seperately I have a 2 in
1.6 million and have just doubled my odds.  

Of course I could come up with 38, but that is fine also.

The 1.6 million is what I believe I heard.  If it is wrong sub in the
correct value.

Roger Books

On 10/25/05, B Lin <lin@rxkinetix.com> wrote:

Again you are confusing the question.  It doesn't matter how many dice
you roll ot what grouping you use to roll as long as the total number of
dice rolled is the same.  You keep insisting that more dice produce
better odds, and that is true if their total is more. i.e. rolling two
dice has almost twice the odds of getting a 6 compared to rolling one
die.  But comparing two single dice to a pair of dice the odds of
rolling a 6 are exactly the same. You are confusing a unit of one being 
equivalent to a unit of 2 or 4 and they are not.  The totals need to be
equivalent (i.e. both sets total 9 dice) for you to directly compare
them.

Throughout the argument I have constantly stressed the point that the 
number of dice be equivalent and it didn't matter what grouping you
rolled them as long as the total # was equal.  You keep insisting that a
set of 4 (or 2) is better than 1 and I don't disagree, but 4 isn't equal

to 1.  The correct comparison is a set of 4 dice against a set of 4
single dice - the total number of dice are equivalent.	Your argument
seems to imply that the odds of getting 9 6's is different if you roll
the dice in pairs or in fours vs. rolling each die singly, and that is
not true.

The question was not if the odds of rolling 19 points of damage on one
die is the same as rolling 19 points of damage using 2 dice.  The 
question was what were the odds of getting 19 points of damage from two
Class One batteries.  Working backwards, the only way to get 19 points
of damage using the re-roll rule is that 9 6's and a 4 or 5 had to be 
rolled.  This delineated the problem right there - we are only
considering cases where 9 6's and a 4 or 5 are rolled, any other
combination doesn't match the 2 battery limit or the 19 point limit.

At this point the problem has dealt with the number of batteries and 
re-roll issues and they are no longer relevant.  The question is now,
what are the odds of rolling the exact sequences 9,9,9,9,9,9,9,9,9,5 or
9,9,9,9,9,9,9,9,9,4 on 10 dice.

For that there is a single answer as previously shown in other posts. 

--Binhan

-----Original Message-----
From: gzg-l-bounces+lin=rxkinetix.com@lists.csua.berkeley.edu
[mailto: gzg-l-bounces+lin=rxkinetix.com@lists.csua.berkeley.edu
<mailto:gzg-l-bounces+lin=rxkinetix.com@lists.csua.berkeley.edu> ] On
Behalf Of Daryl Lonnon
Sent: Tuesday, October 25, 2005 10:53 AM
To: gzg-l@lists.csua.berkeley.edu 
Subject: Re: Statistics - was RE: [GZG] Re: laser classes

Oh well ..

Obviously someone is not understanding someone (could very
well be me).

If your contention is that it's the same odds to roll 19 pts of 
damage on 1 beam die as it is on 2 beam dice, then we'll
just have to agree to disagree.

Daryl

p.s. If you wish to continue this discussion we can do it
over e-mail and not annoy the other list members. 

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