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.

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

[quoted original message omitted]

_______________________________________________
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:

_______________________________________________
Gzg-l mailing list
Gzg-l@lists.csua.berkeley.edu
http://lists.csua.berkeley.edu/mailman/listinfo/gzg-lActually no. You
are extrapolating backwards from the calculated result in a different
direction than the original question.

The odds (1 in 30,233,088) are for getting to the final total of 19 points.
Each step has a smaller probability, but when combined, reach the final ratio.
For each die roll, the chance to get a six is 1 in 6. So in your case, whether
you roll a single die twice, or two dice at the same time, each individual die
has the same chance to get a 6.

An example is listed below:

If you are proceeding from the beginning with two class one batteries. This
means you get two dice to roll. To get to 19 points of damage you
need to get a series of re-rolls.

For this example I will assume that the rolls end up even between the two
batteries, but the case works just as well for any combination of hits (4,5 or
6) between the two batteries that add up to ten dice total.

First roll - Both dice have to get a six, otherwise you don't get a
re-roll - chance of getting 2 6's on 2 dice, 1 in 36

Second roll, Both dice have to get a six otherwise you don't get a
re-roll - chance of getting 2 6's on 2 dice, 1 in 36

Third roll, Both dice have to get a six otherwise you don't get a
re-roll - chance of getting 2 6's on 2 dice, 1 in 36

Fourth roll, Both dice have to get a six otherwise you don't get a
re-roll - chance of getting 2 6's on 2 dice, 1 in 36

Fifth roll, One die gets a 6, the other die gets a 4 or 5 with no
re-roll, (1 in 6 and 1 in 3), overall 1 in 18

Technically there is a sixth roll, but by definition of the problem you don't
count the rolled miss. If the roll wasn't a miss, then the point total would
be higher and that's a different problem.

This is no different than rolling a single battery 9 times for a six and then
once for a 4 or 5.

You are asking a different question: Does using more batteries produce an
increased chance of causing 19 points, and the answer is yes. The above
analysis was for a specific event and limited itself to fixed
end-point - 19 points in 10 dice.  You are asking if given an unlimited
pool of dice, would more dice have a greater chance of scoring 19 points than
fewer and the answer is yes. If you limit the dice pool (10 dice) the answer
is no.

--Binhan

________________________________

From: gzg-l-bounces@lists.csua.berkeley.edu
[mailto:gzg-l-bounces@lists.csua.berkeley.edu] On Behalf Of Roger Books
Sent: Tuesday, October 25, 2005 11:55 AM
To: gzg-l@lists.csua.berkeley.edu
Subject: Re: Statistics - was RE: [GZG] Re: laser classes

Then 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 1.6 million.
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

[quoted original message omitted]

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.

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

Cheers, Tony.

> On 25-Oct-05, at 10:39 AM, B Lin wrote:

> Again you are confusing the question. It doesn't matter how many dice

> a

> You are asking a different question: Does using more batteries produce

I think in this case, we are dealing with a pool of 11 dice instead of 10, as
Beam A could miss totally (1-3 on the d6) and then Beam B hits with a 6
and so on for a total of 10 dice from B alone. Together that is 11 dice, so
the probability we are looking for is the chance to get 19 points of damage
with a pool of 11 dice.

You know, I'll admit I gave up before, but let me try this.

With two dice, you have better chance of starting that string of sixes; not
HUGELY better, but better.

Does that work? It's what I was saying before, but I let it get clouded in the
smaller case of boxcars.

So, the chance for the first six is not one in six, but eleven in
thirty-six, not quite one in three. One of those eleven is the boxcar
case,
so that in the one in thirty-six total, and in eleven of at least one
six rolled, chance, the next roll of two dice is not quite one in three. I
think calculus is needed, as this seems to be approaching a limit of doubling
your chances.

Ok, it doesn't require calc, but I don't want to work it out. ;->=

The_Beast

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

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

Technically no. The original statement was 19 points of damage from 2 class 1
"lasers" (assume beams). If Beam A had totally missed, then the statement
would have been 19 points from a single Class 1 "laser".

If you want to be picky there are 8 6's followed by a re-roll, one 6
followed by a miss and then a final die that is 4 or 5 with no re-roll.

The die roll that misses is moot since it doesn't add a point or a
re-roll.  It is similar to all the X batteries that fired the same turn
and missed, you don't count them in the calculation.

Including the miss die is also extraneous for this reason - let's say
for argument's sake that the opponent instead of rolling a miss, chose
not to re-roll a six.  Are you saying the probability of scoring 19
points changes because the 11th die wasn't rolled? The answer is no. Which
means the 11th die is completely irrelevant to the odds of scoring 19 points
with 2 class 1 beams.

Example with 3 dice - chance to roll a total of 18 on three dice (1 in
216)
A - 6,6,6
B - 6,6,6,(1)

If you didn't know what the 4th die roll was, do the odds of the first three
rolls change?

--Binhan

[quoted original message omitted]