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Actually, following from what Roger say, where
you're talking discrete, unrelated, events (eg. the values on 2 dice) that need
to happen for an event to occur, I think you're better off working out the
probability of it not occuring, especially if the possibility of later events
(eg. subsequent dice rolls) is dependant on earlier, and finaly subtracting the
chance of it not happening from 1 to get the chance of it occuring.
Eg. The odds of rolling 6 on a (perfect) die is
1/6th. The odds of rolling a 6 if you roll 6 dice is not, even
theoretically 1/6th x 6, as that = 1, which is a certainty, and I've gone
more than 6 rolls between 6's, especially when that's what I needed. The
odds of rolling a 6 with 6 dice = 1 - (5/6ths^6) (raised to the power of 6).
This actually comes out as 1-0.335 or .665 or 66%.
I'm fairly sure that's what's need here, but it's
late, I'm knackered and I can't remember the precise details. Is it
4 or 5 = 1 point, 6 = 2 points and reroll, with the
same results?
Have to say that my stats is only basic, so it may
be beyond me anyway (short of taking the easy way out - writting a piece of VBA
to explore the full possibilities of all the rolls).
CJ
----- Original Message -----
Sent: Monday, October 24, 2005 8:18
PM
Subject: Re: [GZG] laser classes
No, we are calculating the chance of rolling 19 points of
damage on two dice. To do a statistical analysis on this you have to
account for the cases where one die does not roll up 6s all the way and you
have extra die rolls with the other.
Roger
On 10/24/05, B Lin
<lin@xxxxxxxxxxxxx> wrote:
Still moot, if Die
A is a 3, then it's a miss. We are only calculating the probability of
nine 6's and a 4 or 5. If 50 dice were rolled previous to this run,
but none hit, it doesn't change the calculation of the 10 rolls that matter
(i.e. dice aren't linked).
In addition, the
statement was that he suffered 19 points from 2 class 1 batteries in one
turn, so the assumption was that the two batteries contributed to the
damage, as opposed to one doing 19 and the other none. It could be
that one caused 4 points (2 rolls) and the other 15 points (8 rolls),
but the odds calculation doesn't change as long as the total dice
rolled to get that result equals ten.
There is a minor
caveat in this; due to the way the FT rule is written, the 6's have to come
first, then the single point roll, as you only get a re-roll by rolling a
six. So to recreate the actual event you have to calculate in the order of
the 9 6's first then the last die. Statistically it doesn't matter
which order the dice were rolled because the odds remain the
same.
--Binhan
But it could be 11 dice.
3 on die A
6 6 6
6 6 6 6 6 6 4:5 om Die B.
Watch out for extra
permutations.
Roger
If you break down the last two
rolls (the 9th six and a single point hit
(4 or 5) the odds are still the
same whether you roll one die or two.
Example: single die, chance of
a 6 (1 in 6), chance of a 4 or 5 (2 in 6
or 1 in 3) so chance of rolling
a 6 plus a 4 or 5 equals 1 in 18(6 x 3)
or rolling the other way (4 or 5
first, then a 6)= 1 in 18 (3 x 6). For
a total of 2 in 18 or 1 in
9.
Rolling two dice - chance of rolling a combination of a 6 with a 5
or 4
on the other die (11 chances out of 36 have a 6 (11/36) but only 4
of
those have a 4 or 5 in them (4 of 46) final odds = 1 in 9)
6:6
(5:6)(4:6) 3:6 2:6 1:6
(6:5)
5:5 4:5 3:5 2:5 1:5
(6:4)
5:4 4:4 3:4 2:4 1:4
6:3 5:3 4:3 3:3 2:3 1:3
6:2 5:2 4:2 3:2 2:2 1:2
6:1 5:1 4:1 3:1 2:1 1:1
What
the calculation is 9 6's and a 4 or 5 on a single die for a total
of 10
dice. It doesn't matter statistically which die (#1-10) rolls the
non-six, so using ten dice, one die or any number of dice in
between
that add up to ten doesn't matter for the
calculation.
--Binhan
-----Original Message-----
From: gzg-l-bounces@xxxxxxxxxxxxxxxxxxxxxxx
[mailto:gzg-l-bounces@xxxxxxxxxxxxxxxxxxxxxxx] On Behalf Of
McCarthy,
Tom
Sent: Monday, October 24, 2005 11:20 AM
To: gzg-l@xxxxxxxxxxxxxxxxxxxxxxx
Subject: RE: [GZG] laser
classes
I just meant that two dice lets you have a bad roll in the
mix.
2 dice, for example, let's you roll: 6,6; 6,6; 6,2; 6; 6; 6; 6;
4 or
6,6; 6,6; 6,1; 6; 6; 6; 6; 4 or 6,6; 6,6; 6,4; 6; 6; 6; 6; 1 and
still
reach 19.
Of course, if you are firm in your belief that you
are as likely to
reach 19 points of damage with 1 die as you are with
multiple dice, then
I really have no argument to counter
that.
> -----Original Message-----
> From:
gzg-l-bounces+tom.mccarthy=xwave.com@xxxxxxxxxxxxxxxxxxxxxxx
> [mailto: gzg-l-bounces+tom.mccarthy=xwave.com@xxxxxxxxxxxxxxxxxxxxxxx
]
On
> Behalf Of B Lin
> Sent: Monday, October 24, 2005
1:11 PM
> To: gzg-l@xxxxxxxxxxxxxxxxxxxxxxx
> Subject: RE: [GZG]
laser classes
>
> It actually doesn't make any difference
whether you use one die or two
> or ten -
> For example:
>
1 die - chance to roll a 6 = 1 in 6, chance to roll two sixes 1 in
36
(1
> in 6 x 1 in 6)
> 2 dice - chance to roll two sixes, 1
in 36.
>
> You can either roll a single die ten times or roll
ten dice once each
> and the odds are exactly the
same. Remember dice have no memory and
are
> not linked
to each other (in theory) so one die's result is not
affected
> by
the result of a previous roll, a future roll or neighboring die's
>
roll.
>
> --Binhan
>
> -----Original
Message-----
> From: gzg-l-bounces@xxxxxxxxxxxxxxxxxxxxxxx
> [mailto:
gzg-l-bounces@xxxxxxxxxxxxxxxxxxxxxxx] On Behalf Of McCarthy,
>
Tom
> Sent: Monday, October 24, 2005 10:48 AM
> To: gzg-l@xxxxxxxxxxxxxxxxxxxxxxx
> Subject: RE: [GZG]
laser classes
>
> For 6 to the 9th, I get
10,077,696. That makes the odds of getting
> exactly 19
points on one die to be 1 in 30,233,088 or so. On
two
dice,
> it's more likely, but still pretty unlikely.
>
>
>
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