How many shots can I expect from my FT? Zero, but that is just me. But others should expect an almost equally low number. See what I mean.

How Many Shots With A Weapon?

By Robert Delwood
(c) 2005, Wayward Publications

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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 programmer-skeptics 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.