As previously covered – it doesn’t
matter if you roll 1 die twice or two dice once each, the odds are the
same. By transitive, it doesn’t matter if you roll one die 9 times,
or nine dice once each or 3 dice and 6 dice together, the odds are the same.
Short example:
Chance to roll four 6’s:
Single die – 1 in 6 6x6x6x6 = 1 in
1296
Two dice, twice each = 2 sets of 1 in 36 =
36 x 36 = 1296
Two dice, 1 on A, 3 on B = 1 in 6 x 1 in
216 = 1296
Two dice, 3 on A, 1 on B = 1 in 216 x 1
in6 = 1296
Three dice, one on A, one on B, two on C =
6x6x36 = 1296
Four dice, once each = 6x6x6x6 = 1296
--Binhan
From:
gzg-l-bounces@xxxxxxxxxxxxxxxxxxxxxxx [mailto:gzg-l-bounces@xxxxxxxxxxxxxxxxxxxxxxx]
On Behalf Of Roger Books
Sent: Monday, October 24, 2005
1:18 PM
To: gzg-l@xxxxxxxxxxxxxxxxxxxxxxx
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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