_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lwhat a week for my
nac fleet, i have a problem with 2 class 1 laser's doing a total of 19pts of
damage in one turn, thanks to rerolls, so with my group I'm bringing in the
rule class 1's get one reroll, class 2's get two and class 3's get three etc
etc etc. I know what your thinking bad sportsmanship but it works both way's,
my nsl fleet with all there lasers are that bad thanks to my rerolls that the
fse player took off all his missile launcher's and gave them armour and
weapons upgrades and screens, so at the moment I'm wondering, equal pt's
battles don't seem to work well with some fleet's, of course I've taken into
account luck with rolling, but the worst fleet I play against is the U.N.S.C
man I blame bab5 for the graser's they tear me up badly, and it takes that
long to get into class three range with movement 4 and 2, and there missiles
are dam right lethal, so finally I get to my pt,I know that somefleet's are
meant to be state of the art, and that some are meant to be just in a bad
state, but are we risking reason why we play full thrust, so as we can power
game the fleets eg, my fleet has class 3 laser's, so he buy's a fleet with
class 4's, it just seem's to me any fleet played right of course, should be
able to take on any other fleet with the same chance of winning. well that's
my whinge over were about to test the ISLAMIC FED fleet we just downloaded, so
well put it up against all four major factions and the unsc, will do a rule
for suicide ships eg, were gunna die so lets ram lads.
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lSounds like an
extreme reaction to a very, very rare (and very extreme) event.
I've often wondered how FSE players felt going from FT2.0 to Fleet Book 1, as
their fleet went from primarily beams (same as everyone else) to salvo
missiles that it takes some time to learn how to use well (and which some
fleets, as published, excel at defending against).
For the local FSE player, on a 'small' table, very, very happy; on a 'large'
table, or a campaign system that enforces supply support of missles, the world
is SOOOOO unfair.
However, their missle change was accompanied with a transition to very fast
eggshells. That REALLY complicates the evaluations.
The_Beast
Tom wrote on 10/24/2005 08:39:03 AM:
...
> Iâve often wondered how FSE players felt going from FT2.0
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lMuch depends on the
other players strategy. Life can be difficult if your opponent surrounds her
big ships with escorts, moves at high speed and dodges. I've head games where
every missile hit ships with few PDS's and I've still lost. OTH Slow ESU type
ships tend to be crunchy.
Roger
> On 10/24/05, McCarthy, Tom <Tom.McCarthy@xwave.com> wrote:
> Slow ESU type ships tend to be crunchy.
*sob* My Komarov, my Komarov...
The_Beast
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lJust surround it
with sacrificial destroyers.
Roger
> On 10/24/05, Doug Evans <devans@nebraska.edu> wrote:
I had a variant Beijing/B! I was safe! ;->=
The_Beast
> Just surround it with sacrificial destroyers.
> On 10/24/05, Doug Evans <devans@nebraska.edu> wrote:
Oh, James, JUST to show you I wasn't ONLY wandering off topic, I have to agree
that your's is a very extreme example. However, it's true there's a
pretty wide variety of outcomes based on strings of re-rolls.
I am curious: are you restricting a single re-roll per class level of
weapon per turn? That is a bit fiddly. 'My class two had box cars, so I
have to remember that if either of the re-rolls are sixes, no further
re-rolls.'
'Ok, but I only had two sixes on my class three, so if one or both are
sixes, I can only get one more re-roll.'
On the other, other hand, re-rolls exist as a form of catastrophic
damage beyone the threshold checks, and are quite comfortable to me.
The_Beast
Oh, James, just to be clear, as SOMEONE will suggest you may be doing damage
somehow incorrectly, the pair of class one's rolled boxcars at least four
times in a row, then a five or six with a four, or a single six followed by a
four, right? Or, fewer boxcars, and a heck of a lot of single sixes in a row,
with a single four somewhere.
I've rolled as bad, but I thought it still extreme.
The_Beast
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lOn 10/24/05, Doug
> Evans <devans@nebraska.edu> wrote:
Upon reading James' post, two words came to mind:
Teske Field
I need say no more. ;-)
Mk
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lDefinitely a rare
event, but the rules themselves aren't broken. It would be like winning
Powerball (odds of 1 in 150 million) and saying that everyone should buy a
ticket because it's so easy to win. Streaks of luck happen, but it doesn't
mean the rules are broken. Otherwise casinos would not be in business long.
You need to examine the statistics of the situation - 2 class 1 beams
rolling about 16-17 6's in a row, the theoretical odds are 1 in 1.69 to
the 13th power or 16,900,000,000,000 - far more than the odds of winning
Powerball or the Earth getting wiped out by an asteroid. The more likely
scenario is that your opponent wasn't using purely random dice (dice, even
correctly manufactured dice are known to have biases, which
can be easily proven by rolling them 100+ times and recording how many
times each number comes up) or using techniques (low hand rolling, practiced
dice rotation etc) to increase the chances of rolling sixes. For instance if
the dice are biased or rolling techniques used that increase the chance of
rolling a six to 1 in 3, then the odds drop to 1
in 13,046,721 - a vast decrease in chances.
You need to examine all the factors that caused the aberrant situation, rather
than just picking what seems convenient. Possible solutions
before a rule change - require people to use standard dice or allow
everyone to use the same set of dice (therefore giving everyone the same
chances), eliminate dice rolling techniques by requiring the use of dice
rolling towers or by mandating people roll dice into a box a couple of feet
away a la Las Vegas so that they can not just drop or easily roll the dice in
a controlled manner.
The re-roll rule is in place to simulate the extremely low probability
outcomes ( a la Star Wars, a single Proton torpedo shot destroying a mega
(Beva?) ton space station) that people like to occasionally see. Obviously it
sucks when you are on the receiving end, but it does add flavor, otherwise a
giant ship like a DN is never going to lose to the smallest classes of ship
like DE's, DD's and frigates.
--Binhan
________________________________
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 james mitchell
Sent: Sunday, October 23, 2005 2:19 PM
To: gzg-l@lists.csua.berkeley.edu
Subject: [GZG] laser classes
what a week for my nac fleet, i have a problem with 2 class 1 laser's doing a
total of 19pts of damage in one turn, thanks to rerolls, so with my group I'm
bringing in the rule class 1's get one reroll, class 2's get two and class 3's
get three etc etc etc. I know what your thinking bad sportsmanship but it
works both way's, my nsl fleet with all there lasers are that bad thanks to my
rerolls that the fse player took off all his missile launcher's and gave them
armour and weapons upgrades and screens, so at the moment I'm wondering, equal
pt's battles don't seem to work well with some fleet's, of course I've taken
into account luck with rolling, but the worst fleet I play against is the
U.N.S.C man I blame bab5 for the graser's they tear me up badly, and it takes
that long to get into class three range with movement 4 and 2, and there
missiles are dam right lethal, so finally I get to my pt,I know that
somefleet's are meant to be state of the art, and that some are meant to be
just in a bad state, but are we risking reason why we play full thrust, so as
we can power game the fleets eg, my fleet has class 3 laser's, so he buy's a
fleet with class 4's, it just seem's to me any fleet played right of course,
should be able to take on any other fleet with the same chance of winning.
well that's my whinge over were about to test the ISLAMIC FED fleet we just
downloaded, so well put it up against all four major factions and the unsc,
will do a rule for suicide ships eg, were gunna die so lets ram lads.
bye for now james,
p.s just spoke to god, he said the hurricanes are our work, hope everyone's
ok.
Whoops, make that 'been rolled AGAINST as bad,...'
I've only rolled that against opponents in a dream...
Wait, three triple sixes in a row in Risk; that's about two thirds as good!
The_Beast
> I've rolled as bad, but I thought it still extreme.
Birhan, it isn't that many in a row; boxcars are only one in thirty six; to
the fourth power, that's only one in something like a million and a half.
Course, followed by the three more points score bumps it a bit, but
definitely easier than powerball. ;->=
Let me know if I got it all whack; statistics definitely pure voodoo in my
limited math.
The_Beast
Ok, it ain't the statistics, it's the weapon effect. Sixes are still two
points of damage, right? So, that's only eight in a row, right?
The_Beast
> Birhan, it isn't that many in a row; boxcars are only one in thirty
On Saturday, in a re-enactment of Trafalgar, I got a 6 for five shots
in a row, and a total of seven 6's out of seventeen shots. Not doing anything
flaky with the dice that I know of, although my son claims I have Evil Psychic
Dice control.
By the way, we had about 65 players in that game, one per ship. After
10.5 hours of play (26 turns), the Franco-Spanish side lost eight
strikes (including me) and one explosion; the Brits suffered four strikes,
including Victory and Royal Sovereign (which struck to me). Some time soon
I'll post about the GM's mechanics for running a game with that many players.
Oops, my mistake - I was figuring each point was a 6, but a 6 does two
points and a re-roll (assuming no shields), so to get 19 points, there
are only 8 6's and a 4 or 5. The chances of rolling 8 sixes in a row are 6 to
the 8th power or 6x6x6x6x6x6x6x6 = 1 in 12,877,056 then to roll
a 4 or 5 (2 chances out of 6 or 1 in 3) = 1 in 38,631,168 or roughly 1/4
the chance to win Powerball. Odds go up exponentially for each event, so twice
as many events is not twice the odds, but much much more.
Example: A single six = 1 in 6 Two sixes = 1 in 36 Three sixes = 1 in 216 Four
sixes = 1 in 1296
Note that going from 2 sixes to 4 sixes is not twice the odds. These assume
completely statistically random dice. If the dice are biased or you use
rolling techniques (i.e. setting the dice) then the odds of getting a 6
increase and the odds drop a lot. For instance if you can roll in such a way
to double the chance of a six to 1 in 3:
A single six = 1 in 3 Two sixes = 1 in 9 Three sixes = 1 in 27 Four sixes = 1
in 81
You then have a better chance of rolling three sixes than someone rolling
regular dice getting 2 sixes.
I'm not saying that long-shot events don't occur (see Powerball winners)
but they are statistically rare and usually if a bunch of rare events happen
it's probably because the odds are not what you think they are and something
else is affecting the result.
--Binhan
[quoted original message omitted]
Understood, Birhan, but I'm obviously screwing up something with my
calculator, cuz I only get 1,679,616, not over twelve million for six to the
eighth. Damn thing is, keeps coming out the same.
But, if winning the powerball is "(odds of 1 in 150 million)", isn't one
in thirty-nine millions more like four times the odds, not one fourth?
Really, I'm not good with figures, but I get 'em right once in a while.
;->=
Please note, my previous figuring included the assumption that fives give you
two hits but no rerolls. I REALLY need to play more often!
The_Beast
Oh, and nine sixes (18 points) plus a four or a five for 19. ;->=
So multiply whatever you come up with, for six to the eighth, by six times
three for the six plus either a four or a five.
The_Beast
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.
Nope you're right. My calculator was acting funny. 6 to the 8th is 1,679,616.
The math thing just isn't working for me today...;)
--Binhan
[quoted original message omitted]
> On two dice, it's more likely, but still pretty unlikely.
Uh, that's what I thought, but not really. Rolling four boxcars is the same as
rolling eight sixes in a roll. Six times eight.
Did have to smash the ole noggin' into the brick wall a few times before it
became clear.
However, 'pretty unlikely' DOES stand as obvious understatement. ;->=
The_Beast
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
[quoted original message omitted]
Ok, I hit my head on statistics again.
***
> On two dice, it's more likely, but still pretty unlikely.
Uh, that's what I thought, but not really. Rolling four boxcars is the same as
rolling eight sixes in a roll. Six times eight.
Did have to smash the ole noggin' into the brick wall a few times before it
became clear.
However, 'pretty unlikely' DOES stand as obvious understatement. ;->=
***
Isn't the first single six a higher odd than one in six?
The_Beast
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-----
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
[quoted original message omitted]
I can't help but think that Tom has a point, and not just because I brought
it up in a slightly different form ( ;->= ), but however you cut it,
it's a HUGE number, or a tiny chance.
I don't think enough games of FT have been played since the game's creation
for a reasonable chance of seeing it. James' opponent has immense braggin'
rights.
Chris nailed it right on the head.
The_Beast
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lBut 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
> On 10/24/05, B Lin <lin@rxkinetix.com> wrote:
For
> a total of 2 in 18 or 1 in 9.
[mailto:gzg-l-bounces+tom.mccarthy=xwave.com@lists.csua.berkeley.edu]
> On
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lStill 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
________________________________
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 Roger Books
Sent: Monday, October 24, 2005 12:31 PM
To: gzg-l@lists.csua.berkeley.edu
Subject: Re: [GZG] laser classes
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
> On 10/24/05, B Lin <lin@rxkinetix.com> wrote:
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
[quoted original message omitted]
> what a week for my nac fleet, i have a problem with 2 class 1 laser's
As has been stated, that was a 1 in 3 million chance of happening, and people
tend to remember the wild results and forget about all the average rolls.
> so at the moment I'm wondering , equal pt's
Not knowing your tactics, it is hard from someone on this list to know for
sure but often I have seen the real problem in this situation to be the
players not adapting their tactics to the fleets abilities. Also which
movement system are you using as they points are based on Cinematic not
Vector.
> but the worst fleet I play against is
Some of your terminology is not quite standard, like 'lasers' instead of beams
as FT is generic so they could be Lasers, Phasers, Particle Beams or whatever
your want. In this case you state "movement" and not "thrust" so that hints
you might be having ships only move 4 or 2 and not changing their speed by 4
or 2 which would make a big difference. Plus Cinematic or Vector? The the
big
grasers usually are single arc so are harder to use in Cinematic but have a
much easier time getting you into arc in Vector.
> so finally I get to my pt,I
That should be true given correct tactics, even playing ability and average
dice rolls. The NPV has some problems underpricing big ships so what kind of
mix of ships do you usually have in one of your fleets in a game? If he is
buying class 4s, that is a lot of mass where you can mount a whole lot more 2s
and 3s, just get up close and you should have more dice to roll than he does.
For this discussion, these dice are linked, though.
Die A could be 3, and die B could be 6; however, die B could be 3, and die A
could be 6. Or both could be 6.
If both are six, then you move on to the next roll with both dice. However, as
either of the previous dice could have been six even if the other wasn't, you
have a slightly greater chance of starting a run of sixes with two dice than
with one. And that slightly better chance is found in each case of rolling two
dice.
I think.
Gawd, my brain hurts, and I have to go to a staff meeting.
I think I gave up before, but this time I REALLY mean it.
The_Beast
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lNo, 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@rxkinetix.com> wrote:
Statistically
> it doesn't matter which order the dice were rolled because the odds
For
> a total of 2 in 18 or 1 in 9.
gzg-l-bounces+tom.mccarthy=xwave.com@lists.csua.berkeley.edu]
> On
The dice are only linked since we are looking at a specific incident and
by the rule that a 6 has to be rolled before you get a re-roll. For all
we know they rolled 22 dice for Class 1 beams that turn and only 2 got a
re-roll but went on to score an additional 15 points of damage. The
point here is that all those events (all the non-rerolls) have no effect
on the remaining 15 points of damage. By default, any roll that doesn't
produce damage is not included in the calculation (for instance there really
had to be 12 rolls not 10 since at some point they would have had to roll a
miss for each beam battery to end the streak). If the statement had been 19
points of damage from 2 class 1 batteries over the game, then the calculation
changes, since you then have to include all the die rolls of all the Class 1
batteries fired during the game, but the only way to get 19 points of damage
from two Class 1 beam batteries in a single turn is to roll 9 6's and a 4 or
5. (Had the damage total been an even number, then there would have been two
possible damage combinations; all 6's or all 6's with the last two being 4 or
5's).
--Binhan
[quoted original message omitted]
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lAs 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@lists.csua.berkeley.edu
[mailto:gzg-l-bounces@lists.csua.berkeley.edu] On Behalf Of Roger Books
Sent: Monday, October 24, 2005 1:18 PM
To: gzg-l@lists.csua.berkeley.edu
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@rxkinetix.com> 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
________________________________
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 Roger Books
Sent: Monday, October 24, 2005 12:31 PM
To: gzg-l@lists.csua.berkeley.edu
Subject: Re: [GZG] laser classes
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
On 10/24/05, B Lin < lin@rxkinetix.com <mailto:lin@rxkinetix.com> >
wrote:
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
[quoted original message omitted]
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lSo you are saying
if I roll two 6s and a 3 on one die and a 7 6s and a 4 this combination
doesn't count as a possible combination?
You are starting from the assumption that you can roll a handful of dice and
get the correct number of combinations. You need to include all possible
combinations.
Roger
> On 10/24/05, B Lin <lin@rxkinetix.com> wrote:
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lActually you don't.
You only have to look at the cases that will produce the correct result, for
instance we don't look at cases with 3 batteries, or situations that use 20
die rolls or include the X number of die rolls that occurred before the
streak. The case in question is 9 sixes and a 4 or 5 for ten dice total. We
ignore the two rolls after these ten because they result in misses (one miss
for each battery, otherwise they would either score more damage and possibly
receive an
additional die-roll which changes the problem.) and so fall outside of
the question.
So in your case listed below, yes, it's a possible case but we would discount
the 3 on Die A since it's a miss, so you are back to 2 sixes, plus seven sixes
and a four which adds up to the required total of 9 sixes and a 4 or 5.
Another possibility is three 6's and a five on Die A and then six 6's on Die
B.
Because of the re-roll rule, there is only one possible combination to
get this result - a series of 9 sixes (split anyway with at least one
die to each battery) and then a roll of 4 or 5. It doesn't matter which die, A
or B rolls the 4 or 5, but the 4 or 5 has to happen at the end of the streak.
This ordering is forced by the rules definition, not by statistical analysis.
From statstical point of view the 4 or 5 could occur with either die at any
position since the dice aren't linked statistically to each other and the
overall odds aren't affected by the order of the dice.
--Binhan
________________________________
From: gzg-l-bounces@lists.csua.berkeley.edu
[mailto:gzg-l-bounces@lists.csua.berkeley.edu] On Behalf Of Roger Books
Sent: Monday, October 24, 2005 2:00 PM
To: gzg-l@lists.csua.berkeley.edu
Subject: Re: [GZG] laser classes
So you are saying if I roll two 6s and a 3 on one die and a 7 6s and a 4 this
combination doesn't count as a possible combination?
You are starting from the assumption that you can roll a handful of dice and
get the correct number of combinations. You need to include all possible
combinations.
Roger
> On 10/24/05, B Lin <lin@rxkinetix.com> wrote:
The dice are only linked since we are looking at a specific incident and
by the rule that a 6 has to be rolled before you get a re-roll. For all
we know they rolled 22 dice for Class 1 beams that turn and only 2 got a
re-roll but went on to score an additional 15 points of damage. The
point here is that all those events (all the non-rerolls) have no effect
on the remaining 15 points of damage. By default, any roll that doesn't
produce damage is not included in the calculation (for instance there really
had to be 12 rolls not 10 since at some point they would have had to roll a
miss for each beam battery to end the streak). If the statement had been 19
points of damage from 2 class 1 batteries over the
game, then the calculation changes, since you then have to include all the die
rolls of all the Class 1 batteries fired during the game, but the only way to
get 19 points of damage from two Class 1 beam batteries in a single turn is to
roll 9 6's and a 4 or 5. (Had the damage total been an even number, then there
would have been two possible damage combinations; all 6's or all 6's with the
last two being 4 or 5's).
--Binhan
[quoted original message omitted]
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lActually, 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
[quoted original message omitted]
B Lin schrieb:
> It actually doesn't make any difference whether you use one die or two
> You can either roll a single die ten times or roll ten dice once each
Unless the odds per die are not the same for each die. Possible cuases have
already been discussed. If a die is unintentionally biased, the odds may be
different for different ones. It would also be harder to manipulate throwing
several dice than throwing a single one.
Greetings Karl Heinz
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lOf course, your
gaming group could always be harbouring a closet telekinetic...
CJ
[quoted original message omitted]
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-l
> sorry men I didn't explain it that well yes the to 2 class 1 beams
Is that rerolls per each die or for all the dice in a single beam battery?
Do you roll each beam separately? We just pile all the beam dice being fired
from one ship at another together and roll them all at once to save time so
you never know which beam that single die with all the rerolls came from.
Separating them out would take more time and I like rolling handfulls of dice.
This change will also penalize small escorts probably more than larger ships
as a high percentage of their weapons mass comes from Beam 1's and maybe a
Beam 2. Tbey are already overpriced when using NPV and this will just hurt
them more loosing rerolls that a larger ship would get to keep.
You know, I'm probably going to regret this, but I've been watching this
discussion for some time, and since I haven't played Full Thrust, and my books
were put sadly away when I gave up on finding local opponents, I have to ask:
What is the rule for rolling beam weapons
with regards to re-rolls and damage?
The straight beam damage (not against shields) is 1 point of damage for
a roll of a 4 or 5 and two points of damage and a re-roll for a 6.
--Binhan
[quoted original message omitted]
_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lsorry men I didn't
explain it that well yes the to 2 class 1 beams did a total of 19 between them
in one turn, and for some reason in our games, it's not unusual to reroll 3
times on the one die eg, 6.6 and 6, no there not special dice, and one of my
gamer's now works for the Australian department of statistic's and he can
never believe the dice rolling in our games, then he rattles of the chance of
that happening,but it doesn't work at the casino for some reason. but since it
seem's to be that regular an event in our games, I'll still be bringing in the
rule you cannot have more rerolls than the class of your beam weapon.
cheer james
[quoted original message omitted]
> On 10/25/05, B Lin <lin@rxkinetix.com> wrote:
Thanks for clarifying. The reason I asked was because I wanted to wrap my own
head around this, and was looking back on James' original post, because it
seemed to me that what he originally said got lost in the shuffle. Maybe
looking at it again will clarify things:
" i have a problem with 2 class 1 laser's doing a total of 19pts of damage in
one turn"
So we have two dice being rolled, and one of those dice does X amount of
damage, and the other Y, and the total damage added up equals 19. Setting
aside all of the sixes, in order to reach an odd number of damage, one of the
two dice had to end its rolls with a 4 or 5, and the other die ended its rolls
with a 1, 2, or 3, for 0 damage. If they had both ended with a 4 or 5, or both
with 3 or less, the total damage would have been equal.
> From there I realized that Tom B was on the right track with adding up
The odds of rolling 1, 2, or 3 (Result A) are 1 in 2. The odds of rolling a 4
or 5 (Result B) are 1 in 3. The odds of rolling a 6 (Result C) are 1 in 6.
So the following roll combinations result in 19 points of damage:
Die 1 Die 2 1.A CCCCCCCCCB 2.CA CCCCCCCCB 3.CCA CCCCCCCB 4.CCCA CCCCCCB
5.CCCCA CCCCCB 6.CCCCCA CCCCB 7.CCCCCCA CCCB 8.CCCCCCCA CCB 9.CCCCCCCCA CB
10.CCCCCCCCCA B
Noww, I could go on to calculate: CCCCCCCCCB A Etc., but since a=b is the same
as b=a, I'll just double the numbers derived from above.
1. that's 1 in 2 multiplied by 1 in 6^9x3, or 1 2x3x6^9 or 1 in 6x6^9 or 1 in
6^10. I think I see where this is going. 2 thats 1 in 6x2 multiplied by 1
6^8x3 which is.... yes, 1 in 6^10.... each of the 20 results have the same
odds. that means the odds ar 1x20 in (6^10)x20, or 1 in 6^10. SO the odds of
rolling 19 points of damage when firing two beam 1's is 1 in 60, 466,176.
Correct with some minor corrections.
Your case #1 where Die 1 rolls a miss right off the bat is eliminated since
that would result in a SINGLE beam doing 19 points of damage, since you
multiply by two (die 1 results and die 2 results) you only have 18 cases left.
By implication, Case #10 is unlikely since I don't think he would have
included the second beam to the total if it only provided a single point.
Again subtract two cases from the total leaving 16.
Each case has a 1 in 60,466,176 but remember that there are two possibilities
for scoring a single point (rolling a 4 or 5) so the odds are actually 1 in
30,233,088. The cases are additive so with 16 cases odds are 16 in 30,233,088
so the final count is 1 in 1,889,568.
--Binhan
[quoted original message omitted]
> On 10/26/05, B Lin <lin@rxkinetix.com> wrote:
WRONG. Despite the fact that that is a SINGLE beam doing all 19 points of
damage, it's STILL one of the possible outcomes of firing
two beams -- one does no damage, and the other does all the damage.
> since you multiply by two (die 1 results and die 2 results) you only
You're fiddling. I'm talking about the exact number of possible outcomes where
rolling the dice for two beams resulted in exactly 19 points of damage. Since
one of those beams had to do an odd number of points, those outcomes have to
include 18 by one and 1 by the other.
Again subtract two cases from the total
> leaving 16.
No. There are still 20. Not thast it's ultimately relevant, but there are 20
possible cases.
> Each case has a 1 in 60,466,176 but remember that there are two
I'm well aware of that. That's why the odds of scoring a single point are
calculated as 1 in 3, not 1 in 6.
so the odds
> are actually 1 in 30,233,088.
Wrong. As I said, I already took into account that EITHER a 4 OR a 5 can
result in 1. Read back through my post again.
The cases are additive so with 16 cases
> odds are 16 in 30,233,088 so the final count is 1 in 1,889,568.
Wrong a final time. Remember that when you add, you have to add to
BOTH sides -- so even using your incorrect 30,233,088 16 cases with
odds of 1 in 30,233,088 per case means odds odf 16 in 483,729,408 which comes
out to 1 in 30,233,088.
RETRACTION
> On 10/26/05, Brian B <brianbinor@gmail.com> wrote:
> The cases are additive so with 16 cases
I was wrong on this point. I stand by the rest, but on this point, B was
right.
> On 10/26/05, Brian B <brianbinor@gmail.com> wrote:
Which means that, sticking with my 20 cases and 60,466,088, the odds end up as
1 in 3,023,304.4
> --- Doug Evans <devans@nebraska.edu> wrote:
I've done that several times; I think I've gotten over 30 points
off a single beam die. Ask Indy. ^_- (Admittedly, a couple of
rerolls is *much* more common than a string like that....)
Of course, what doesn't get talked about here so much is the fact that, when
playing Epic all those years ago (which also needed high numbers on d6s to
hit, for those not in the know), I couldn't hit the broad side of a barn. So I
figure my dice rolling averaged out between games. <grin>
That said, it's not all *that* small a chance. You have to consider not just
the 'attack' dice, if you are looking for sequences like this, but also
initiative dice, threshold check rolls, and repair rolls. There are a *lot* of
dice getting rolled in any given game, and not all will stand out in people's
minds like the attack rolls will.
'Til later,