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

On 10/25/05, Tony Christney <tchristney@telus.net> wrote:
> 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.

It makes it 10 times more likely. So the sequences look like:

{4,5},6,6,6,6,6,6,6,6,6,{1,2,3} --|
6,{4,5},6,6,6,6,6,6,6,6,{1,2,3}   |
6,6,{4,5},6,6,6,6,6,6,6,{1,2,3}   |- occurs 10 times
....
6,6,6,6,6,6,6,6,6,{4,5},{1,2,3}---|

{1,2,3},6,6,6,6,6,6,6,6,6,{4,5}---|
6,{1,2,3},6,6,6,6,6,6,6,6,{4,5}   |- occurs 10 times
...
6,6,6,6,6,6,6,6,6,{1,2,3},{4,5}---|

The odds of each one of these strings in the same:
(1/6)^9*(1/2)*(1/3) * 20 =
Which takes the odds from 1 in 30 million to 1 in 3 million
(roughly).

Oddly enough the difference in odds of rolling any odd
number of damage points (between 1 die and 2 dice)
in full thrust is = (dmg_pts+1)/2

So you're twice as likely to get 3 pts damage with 2 dice
as you are with 1.  Your 3 times more likely to get 5 damage
points, Your 4 times more likely to get 7 damage points (and
so on and so on).

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

Daryl

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