Anyone who has ever
played ASL has faced this situation at one time or another: A gun
malfunctions and inevitably he asks in disgusted resignation "what
were the chances?" For as common and perhaps simple as the question
is, the answer is less direct and requires an explanation. Ironically
it goes back to something as basic as fifth grade math.
Let us define the question more precisely. The
Breakdown of a weapon is the minimum number needed on two dice (DR)
for the weapon to malfunction. Naturally, the higher number, the more
times the weapon may be expected to fire. 12 is the default but other
weapons are often marked explicitly with different values. The FT of
most nationalities has an X10 and will be used in our examples. The
player keeps firing until the first time a 10 or higher is rolled and
we will count the shot in which it breaks down. 10 is also a
convenient number since there is a one in six of its occurrence so to
further simplify the problem we will assume rolling a single die
(dr)
The question at the most basic level:
How may die rolls can be expected before rolling the first
six?
For those thinking the answer is six, that is,
they are expecting to get to six die rolls before losing their
weapon, are mathematically correct but critically misguided. Let me
explain. We are all familiar with the standard bell curve of rolling
two dice. We know at such a fundamental level that the most common
roll (the mode) will be seven, the average roll (the mean) will be
seven and halfway point of the distribution curve (the median) will
also be seven. The median is point along the curve that exactly half
of the occurrences will be to the left and half will be to the right.
In our case, we know half the time we will roll less than seven and
half the time we will roll above a seven, ignoring the case of seven
itself for the moment. In fact, we are too familiar with this
standard bell curve. It will cloud our thinking as we will see later
on.
But our problem does not follow the standard bell
curve. The average no longer equals neither the median nor the mode. In our
case, the median is weighted near one end of the probability curve.
We are guaranteed one shot but to fire our weapon a second time, we
have the additional requirement of not rolling a six on the first
attempt. To fire a third time, we have to not roll a six on both the
first and second attempts. And so on. In fact, it becomes less likely
to fire the weapon at least a certain amount of times because we have
to not roll a six each time. Mathematically, it looks like
this:
First successful shot: (5/6) or 84%
Second successful shot: (5/6) x (5/6) or 68%
Third successful shot: (5/6) x (5/6) x (5/6) or 59%
Fourth successful shot: (5/6) x (5/6) x (5/6) x (5/6) or 49%
Fifth successful shot: (5/6) x (5/6) x (5/6) x (5/6) x (5/6) or 40%
Sixth successful shot: (5/6) x (5/6) x (5/6) x (5/6) x (5/6) x (5/6)or 33%
While it does not get harder to fire each
subsequent time (it still breaks down on a six regardless of how many
times it has fired previously), it obviously becomes harder to reach
a certain point. For instance, a player will get at least six shots
only 33% of the time. Keep that in mind, too, for a moment.
Notice that player will get at least four shots
49% of the time. This is close enough to 50% that we will call it the
halfway point. Stated another way, players will get at least four
shots half the time and more than four shots the other half. It is
this midway, or median, point that is of interest to us. Therefore
players should consider themselves lucky if they get at least five
attacks and unlucky if they get only three. So four is the magic
number of shots for which the weapon is good.
But we said the average is six. Why is there a
difference? The average assumes there is no upper limit to the number of times
the die can be rolled. If this game is repeated enough times,
occasionally a player will roll the die a million times before the
first six. While unlikely, it skews the average enough to push it
toward six. By a more common measure, games of 50, 25 or 15 die rolls
will also be recorded, with each game contributing toward the average
of six. Stated another way, if you roll a die six times, one of those rolls will
likely be a six. However, when firing a weapon, when you roll a six matters. If
you roll it on the first attempt, you won't get another try later.
Getting back to the case of rolling a die an unlimited number of time, the error of our assumption is becoming apparent. The
possibility of getting a million rolls is zero since scenarios do not last that
long. Assuming most games are eight turns, the most a flame thrower can fire is
16 times. This new limitation reduces the average from the theoretical six to
just over five. Now, add a more realistic limitation and assume in an eight turn
game, a flame thrower is only going to fire six times. One can reasonably expect
the unit has to move into position, survive opposing combat and have a target
available, for instance; all of which further restricts the FT unit from firing
for reasons other than breakdown. The average of six attempts comes very close
to our magic number of four. Additionally, if the game conditions change and we
do not count the occurrence of the breakdown number (for examples, radios do not
maintain contact if the breakdown number is rolled), the average drops another
entire point. We are not longer guaranteed getting even a single usage of the
weapon. In the case of a X10 radio, the expected number of fires is reduced to
three  half of what conventional wisdom maintains.
This becomes confusing since we inherently know
the average is six. But that is assuming an unlimited number of rolls
 a factor that has to be restricted in any meaningful discussion of
weapon breakdown. In the standard case, since there is no upper limit
to the number of rolls allowed, the curve extends infinitely to the
right. But we do have a restriction: the length of the scenario.
Think of it this way. Since half the occurrences happen in the
limited area of the numbers one through four, the other half take up
the rest of the positive numbers ad infinitum. If the inclusion of
the numbers five through infinity count toward the theoretical
average of six, certainly the exclusion of those same numbers has to
reduce the average from six.
So what is the answer? Four; three if you do not
include the last die roll. If you played the game enough times and
each scenario is very long, you will get an average of six rolls. But
as pointed out before, players really only have very few instances to
fire. More indicative is the statement that the weapon will get no
more than four shots half the time. Also, do not falsely assume if
any given scenario were simply longer, the weapon becomes more
effective. The weapon average is always the same, merely if more
weapons are given more opportunities to fire, a few of the weapons
will get to fire many times. Because of those reasons, four is the
number in which ASLers are interested.
Even so, many will do not believe this. They have
tried small computer programs testing this and are convinced the
answer is still six. I challenge those programmerskeptics to try
their application again but this time insert our one restriction. In
the line testing the die roll value, add another test if the number
of times the die have been rolled for the current game to be six or
more and stop if it is. For example,
if( dieRoll == 6  numberOfRollsSoFar == 6 )
//exit game loop
For the mathematically inclined, the following
expression may be evaluated to both six and infinity. I leave the
homework to you.
As always, I encourage discussion. If you agree or
disagree, feel free to write me.
