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

The reason it's confusing is that you haven't stated the problem
clearly. You have taken the re-roll portion out of context and
re-applied it. For instance, your example, for some reason, adds more
dice rolls past 4, they should be compared directly to 4 dice rolls- it
doesn't matter if the 5th or subsequent rolls are a 1 or a 6, only the that
the first 4 were 6's (by the assumed definition of "EXACTLY 4 6's" and not 4
6's and a couple of other numbers...). You've added points that are irrelevant
to generate your numbers so they don't compare directly with each other. Along
those lines, your example has you rolling many more times in your 4 dice
example than the single die example which produces an artificially high
probability. To make it comparable you have to compare rolling a set of 4 dice
once (4 rolls total) against rolling 4 single dice (4 rolls total). Your
example is
rolling a single die and re-rolling only sixes, vs. rolling a set of 4
dice and re-rolling 6's. It's a clear case where you are comparing 1
against 4.

The chance to roll 4 sixes with a die rolled 4 times is 1 in 1296. It doesn't
matter if you take one die, roll it 4 times or take 4 dice, roll them once
each or 2 dice and roll one 3 times and one once, you still roll a total of 4
dice that are independent of each other and produce
the exact same number of permutations - 1296. Of those permutations only
one has 4 sixes in it. Therefore the odds of rolling 4 sixes and EXACTLY 4 6's
is 1 in 1296 no matter which order or grouping of dice you use.

The dice are only linked because the FT rule says that to get a re-roll
you have to roll a 6, this narrows the search to only dice rolls that contain
a 6, except for the last one, which is looking for a 4 or 5. This means to
produce 19 points of damage you need 9 rolls of 6 followed by a roll of 4 or
5. The problem is carefully delineated and already
has taken the re-rolls into account.  There are exactly 2 possibilities
that follow this definition 9,9,9,9,9,9,9,9,9,5 and 9,9,9,9,9,9,9,9,9,4 out of
60,466,176 permutations.

By defining the problem as 9 6's we have already taken the re-roll in
account, in addition, the exact number of re-rolls was already
determined (9). Your example seems to imply that if you get a six you are
allowed an additional roll even past the stated number of rolls (4).

You don't have to consider multiple batteries for the following reasons-

Part 1)The total is an odd number of points (19). You can only get a
re-roll on a 6 and that produces 2 points (an even number). So to
generate an odd number of points you need an odd point roll (4 or 5). Two odd
rolls make an even and three would not be possible since you
wouldn't get the reqired number of re-rolls. Therefore only 1 die can
roll a 4 or 5 to produce a single or odd number of points.

Part 2) Batteries are batteries - a class 1 beam battery gets one die,
no special modifiers or rules. Same batteries are interchangeable (i.e. a
class 1 is the exact same as any other class 1) each produces 1 die of fire
that has the same to hit chance as any other Class 1.

Part 3) If batteries are the same then it doesn't matter which die roll
is associated with which battery as they give the same result - a 6 is a
6, and a 6 by a class 1 battery is the same as a 6 by a class 1 battery.

Part 4) if the 6's are equivalent then it doesn't matter if they come from 1
battery or 2. And it doesn't matter which battery rolls the 4 or
5.

Final - All the rolls can be treated as having come from a single Class
1 battery.

If there were three or more batteries involved, then the calculation would
then have to include partial hits from multiple batteries and would be much
more complicated.

--Binhan

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