Transcript My Presentation in Excel Format

Talk about the ethics of calculating probability.
Probability – How likely something will occur
Probability is usually expressed as a decimal number. A value of 1 means that an event will
always happen, while a probability of zero means an event will never happen.
Converting Percent to decimal. Percent = Decimal * 100
The Mathy Definition:
Given a set of possible events, the probability of an event occurring is equal to:
The Number of times the event occurs in the set of possible events / total number of events in the
set.
Here is a simple example: What is the Probability of a six being rolled on a single d6?
Set of possible events {1,2,3,4,5,6}
Number of times 6 occurs in the set: 1
Total Number of Events: 6
Probability = 1 / 6 = .166666666 or about 16%
Probability of a set of events occurring.
These probabilities are common in Warhammer 40K. Lets say something has an armor save of
3+. What is a probability of the unit passing its armor save?
Lets define two groups of events, each group will contain a subset of the possible results.
By Definition of a 3+ Armor Save, a roll of 3,4,5, or 6 means the wound is absorbed and ignored.
A roll of 1 or 2 means the save fails and the wound is taken.
Total Set = {1,2,3,4,5,6}
A success = {3,4,5,6} = 4 occurances
A Failure = {1,2} = 2 occurances
Probability of a success = 4 / 6 = .666666667
Probability of a failure = 2 / 6 = .3333333333
If you know the probability of when an event will occur, then you also know the probability of the
event not occurring. The probability of an event not occurring = 1 – probability of the event
occurring.
This can be flipped around to produce:
The probability of an event occurring = 1 – probability of the event not occurring.
General Extension of Calculating the Probability of sets
Other Dice Sizes:
The math I just showed applies to dice of all size. The difference would be the number of
possible events. With a d10,
There are ten items in the set of possible events – {0,1,2,3,4,5,6,7,8,9}
To calculate the probability of a single number it would be 1/10 = .1
To calculate the probability of a probability of a 5+ roll it would be
Success = {5,6,7,8,9} = 5 Items
Fail = {0,1,2,3,4} = 5 Items
Probability = 5 / 10 = .5 or 50%
What about “custom dice”?
Heroscape uses a die where 3 sides are skulls, 2 sides are shields, and one side is blank.
This works the same except that our possible set of events is
{skull,skull,skull,shield,shield,blank}
In order to score a hit in Heroscape, a skull must be rolled.
The success set is {skull,skull,skull} or 3 items
The Fail set is {shield, shield, blank} or 3 items
Probability of hit = 3 /6 or .5
Probability and multiple dice.
Many times a game will require a player to roll multiple dice and total the result. The success or
failure of an event depends on the total of the dice. Wizkids utilizes this mechanic in many of its
games.
Fundamentally, how probability is calculated for multiple dice is the same as a single die except
that the possible event set is more difficult and time consuming to generate. Also, multiple dice
probabilities will follow normal distribution instead of uniform distribution.
A single die will follow uniform distribution. This means that each number has an equal probability
of being rolled. Normal distributions follow a bell shaped curve instead of a flat line. Therefore,
some numbers will come up much more frequently then others.
Calculating the Possible Event Set of multiple dice
This is the harder and more time consuming process when calculating the probability of multiple
dice. Essentially, you have two independent events. The value which appears on one die has no
effect on what appears on the other die. Therefore, for each value that appears on one die, the
other die can display any of its values.
That probably seemed quite confusing, so lets take a look at an example with 2d6. If one die rolls
a one, the other die can be a 1,2,3,4,5, or 6. The following rolls are part of the possible event set:
{1,1} , {1,2} , {1,3} , {1,4} , {1,5} , {1,6}
Since the first die is not affected by the second die, if the first die rolls two, then the second die
can be either a 1,2,3,4,5, or 6. Therefore, the following rolls are also part of the event set:
{2,1} , {2,2} , {2,3} , {2,4} , {2,5} , {2,6}
Extending this confusing idea out, all the possible outcomes of two dice become
{3,1} , {3,2} , {3,3} , {3,4} , {3,5} , {3,6}
{4,1} , {4,2} , {4,3} , {4,4} , {4,5} , {4,6}
{5,1} , {5,2} , {5,3} , {5,4} , {5,5} , {5,6}
{6,1} , {6,2} , {6,3} , {6,4} , {6,5} , {6,6}
Keep in mind that since the rolls are two independent events, the sets {1,2} and {2,1} are the not
the same result and need to be counted separately. Also, you will probably discover that the
total number of possible outcomes is equal to 36 or 6 x 6. As a general rule, the total number of
Now keep in mind, we were interested in calculating the probability of a certain total roll coming
up, so the next step is to determine the total of each of the rows.
After doing the Math, you are left with the following the table.
2 3 4 5 6 7
3 4 5 6 7 8
4 5 6 7 8 9
5 6 7 8 9 10
6 7 8 9 10 11
7 8 9 10 11 12
From this point, it is as simple as counting up the number times a particular sum occurs and
divide it by the total number of possible sums. For example, what is the probability that a roll of 7
will occur on two dice?
7 shows up six times on the chart above, so the # of items in the success set is six. The total
number of possible sums is 36, so the probability of seven occurring is 6 / 36 or .166666667
Solving the probability of rolling a target is identical to the process used for a single die. Lets find
out what the probability of rolling a 5+ is:
Success = {5,6,7,8,9,10,11,12). Find out how many times each number occurrs:
# of Times = {4,5,6,5,4,3,2,1} = 30 success results
Probability of Success = 30 / 36 or .83333
Finding the probability of larger groups of dice is done the same way, but it becomes much more
All of that is great, but all of that is for a single event. What about for a chain of events?
In Warhammer 40K several d6's are rolled to resolve attacks.
The attack must first hit, then wound, then the target must fail its armor save.
In a situation where several independent events must occur in order for a single larger event to
occur, probability is determined by:
Total Probability = Probability of event 1 * probability of event 2 * ... * probability of event n
For an example, lets say a figure will hit on a 3+, wound on a 2+, and the target has an armor
save of 4+. What is the probability of a wound?
There are three events to consider:
Event one success set = {3,4,5,6} = 4 items Probability = 4/6 or .6666666666
Event two success set = {2,3,4,5,6} = 5 items Probability = 5/6 or .8333333333
Event three success set = {1,2,3} = 3 items
Probability = 3/6 or .5
The Total Probability = E1 * E2 * E3 = .66666666 * .83333333 * .5 = .27777777
Venn Diagrams
In order to understand our next topic, it is helpful to understand the concept of a Venn Diagram. A
Venn Diagram is a way to graphically represent probability.
A simple chart will consist of a circle inside of a square. This simple chart is used to represent the
probability of a single event. The square represents all of the possible outcomes of an event.
The circle represents all of the outcomes which would be labeled a success. In theory, probability
can be calculated by taking the the area of the circle and dividing it by to the area of the square.
Lets take Venn Diagrams to the next step and add in a second set of successful outcomes. This
would produce one of four types of diagrams. The first is two separate circles within the square.
While this diagram shows that there are two sets of successful outcomes, the two sets of
successful outcomes share no common values. In this case it is impossible for both events to be
successful at the same time.
The second diagram shows the second circle completely contained within the first circle. In this
case, the second set of successful values is completely contained within the first set of successful
values. Therefore, Event 2 cannot be successful without Event 1 also being successful.
The third diagram is just a single circle. This indicates that the successful outcomes sets of Event
1 and Event 2 are identical. Either both events will be successful or neither event will be
successful.
The fourth diagram shows two circles with an overlapping section. This indicates that either event
1 can succeed, event 2 can succeed, or both can succeed.
The reason I took that last diversion is to help explain our next concept which is figuring out the
probability that one or more events can succeed.
Imagine you have two separate events and you need to figure out what is the probability that at
least one of them will succeed. The process of calculating this probability is not as simple as
adding the two probabilities together. It is calculated by The probability of event 1 success + the
probability of event 2 success – (The probability of event 1 success * the probability of event 2
success)
The reason for this crazy equation is best demonstrated by looking at the Venn Diagram.
remember how I said that the probability can be calculated by taking the area of the circles and
dividing it by the area of the circle? Well if you just add the probability of each event together,
then the area where the two circles intersect is counted twice. As a result, an incorrect area is
counted for the circles and the probability will be wrong.
However, if we subtract the area of the intersection after adding the area of both circles, then are
of the intersection is only counted once and the proper area of the circles is counted and the
probability will be correct.
Lets take a look at an example:
A twin linked las cannon hits on a 3+ on a single d6. For all non-40K players out there, twin linked
weapons allow the player to reroll their chance to hit if the first attack misses.
Essentially this equates to a player rolling 2d6 and if either one has a 3,4,5, or 6 on it, the attack
hits.
The success set of event 1 is {3,4,5,6} or 4 items
The success set of event 2 is also {3,4,5,6) or 4 items
Since the total number of items in each set is 6, the probability of each event is
4 / 6 = .66666666
Plugging that back into our equation we get
Total Probability = E1 + E2 – (E1 * E2)
= .6666666 + .66666666 - (.66666666 * .66666666) = .888888 or 88%
What about two events which have different probabilities?
The same equation still works. In a recent article for the Army Showcase I compared how well
the Tau Hammerhead works in 5th edition vs how it works in 4th Edition.
In fifth edition, the Tau Hammerhead will hit exactly where it wants if either a blast marked is
rolled on a scatter or 2d6 roll a four or less.
There are two events there:
1) A blast marker is rolled on a scatter die
2) A Four or less is rolled on 2d6
success Set for event 1 is {Blast,Blast} or 2 items out of a possible six
success Set for event 2 contains 6 items out of a possible six. (Jump back to earlier in the
podcast to see a refresh on how I calculated that number).
Probability of event 1 = 2/6 =.3333333
Probability of event 2 = 6/36 = .166666666
Total Probability = EV1 + EV2 – (EV1 * EV2) = .3333333 + .16666666 - (.3333333 * .16666666) =
.44444
What about more then two events?
Lets say I have three marines which hit on 3+ roll of a d6. What is the probability that at least one
will hit?
This is where things start to get messy. Its quite solvable if you take the time and draw out a
Venn Diagram. What you are going to be using is something call the inclusion/exclusion principle
and Wikipedia has an article on the topic. The previous two examples have been using the
inclusion/exclusion principle for two items. Beyond two items that math gets really long I just don't
have enough time to explain it right now. Maybe in the future I will go into an even deeper look at
probability.
The last note on probability is that probability has no memory. What this means is that one
independent event and no effect on another independent event.
While the probability of rolling 2 sixes on 2d6 is really small, each die has a 1 in 6 chance of
rolling a six. Therefore if the first die rolls a six, it does not decrease the probability of rolling a six
on the second die.
Also, here are some important numbers:
The average roll of a 1d6 is 3.5
The average roll of 2d6 is 7
The average roll of 3d6 is 10.5
Keep those numbers in mind if you ever have to check someones dice for being loaded.
Also, I have heard that some people claim that rolling lots of dice produce higher rolls, but that
idea is simply disproven by math. Perhaps that myth persists to keep people from rolling their 24
Tau Firewarrior attacks one die at a time......