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Date: Tue, 25 Oct 2005 13:54:59 -0400

Subject: 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-lThen 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
1.6million. 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] 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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> Gzg-l@lists.csua.berkeley.edu
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>
> _______________________________________________
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> Gzg-l@lists.csua.berkeley.edu
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>
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