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

From: Tony Christney <tchristney@t...>
Date: Tue, 25 Oct 2005 12:23:18 -0700
Subject: Re: Statistics - was RE: [GZG] Re: laser classes
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Hi,

I think that the only mistake that Binhan is making is that there
is an additional restriction on the two beam vs. one beam problem.

There is an additional die that must miss for the two beam problem.

For the one beam problem, the die rolls must follow this sequence:
6,6,6,6,6,6,6,6,6,{4,5}

For the two beam problem the die rolls must follow this sequence:
6,6,6,6,6,6,6,6,6,{4,5},{1,2,3}

Note that there is an extra roll for the two beam case. Having said
that, I don't think that it changes the odds very much.

Other than that, what Binhan is saying about the distribution of
dice for the two problems being irrelevant is true.

Cheers,
Tony.

On 25-Oct-05, at 10:39 AM, B Lin 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] 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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