Lauren Perry, PhD

Using LaTeX Equations in Github Pages

2022-04-06T00:00:00+00:00

While there are a variety of workarounds for using LaTeX in Github pages, particularly related to mathematical equations, I find the following to be the easiest approach.

Step 1. Open RStudio and create an RMarkdown file. If you’re not familiar with RMarkdown, check out the RMarkdown Cookbook at https://bookdown.org/yihui/rmarkdown-cookbook/. This also contains good information about using LaTeX in RMarkdown.

Step 2. Write your document using standard markdown syntax and LaTeX math!. Unlike a Markdown file, RMarkdown can also compile quite a bit of LaTeX, most notably mathematical equations. I haven’t tried to do anything too bizarre, but it seems to work well with everything I’ve wanted to do in math mode.

Step 3. Compile to HTML. This will save your work as an HTML document.

Step 4. Upload your HTML document to Github.

Step 5. Open the HTML file in an editor (not in a browser or in RStudio) and add your Github pages header information to the top of the file. If you don’t have a way to edit HTML files on your local machine, upload your file to Github and edit it directly on the Github website. I add my header as markdown, which necessitates changing the file extension to .md (markdown can handle html, so this works ok). If you change the file type to markdown, make sure to delete from the top of the file. You may also want to remove the title and header information from the RMarkdown file.

Step 6. Save and push any necessary changes. Et voila!

Average Character Stats

2022-01-25T00:00:00+00:00

Character Stats

In Dungeons and Dragons, each character has five primary abilities: strength, dexterity, constitution, intelligence, wisdom, and charisma. These allow your character to interact with the world. To determine ability scores, you roll four six-sided dice and record the total of the highest three; if you rolled 2, 6, 1, 4, one ability score would be \(2 + 6 + 4 = 12\).

Calculating the average ability score for ``roll four, drop the lowest" is a bit complicated (see the next section for details), but we can do a quick simulation study to find the average score. We call these averages expected values, which tell us the average value we would expect to see if we rolled out character stats millions and millions of times.

To do this simulation in R, we might use the following code:

  nrep <- 10000000
  scores <- numeric(length=nrep)
  for(i in 1:nrep){
    rolls <- sample(1:6, size=4, replace=TRUE)
    scores[i] <- sum(rolls[-which.min(rolls)])
  }
  mean(scores)

There are other ways to write this simulation, but I think this one has the easiest logic to follow! We set the number of repetitions (nrep) to be 10 million. Then, we create a variable to store all 10 million ability scores (scores). The next line starts a loop that runs 10 million times - the function sample represents the four rolls of the 6-sided dice and the sum function adds over the highest four (-which.min removes the lowest roll). Finally, we exit the loop and take the average of the 10 million simulated ability scores.

Running this with 100 million repetitions resulted in an average ability score of 12.24 with a standard deviation of 2.85. Of the 100 million simulated ability scores, 95% were between 6 and 17. This is a pretty simple system - possible scores are integers (whole numbers) ranging from 3 (by rolling 1, 1, 1, 1) to 18 (by rolling 6, 6, 6, and anything else) - so we should expect 100 million repetitions to give us an extremely precise picture of the distribution of rolls.

How much better is this than just rolling three six-sided dice and taking their sum? Well, the average roll for any given d6 is \[\frac{1+2+3+4+5+6}{6}=3.5\] so the average sum is \[3.5 + 3.5 + 3.5 = 10.5.\] The fifth edition D&D Player’s Handbook also offers the following ability scores as an alternative to rolling: 15, 14, 13, 12, 10, and 8. These have an average of 12, which is only slightly worse than the average using rolls of the dice.

Finally, the fifth edition D&D Player’s Handbook has a variant for customizing ability scores, where you have 27 points to spend increasing each score from a baseline of 8. Each time you increase a score by 1, you spend a point. (Increasing to 9 costs 1, to 10 costs 2, etc.). This method sets a maximum score of 15. Notice that the point cost for each ability score is (score\(-8\)). With 27 points to spend across 6 scores, \[\begin{align*} & (x_1 - 8) + (x_2 - 8) + (x_3 - 8) + (x_4 - 8) + (x_5 - 8) + (x_6 - 8) = 27 \\ & \implies x_1 + x_2 + x_3 + x_4 + x_5 + x_6 - 48 = 27 \\ & \implies x_1 + x_2 + x_3 + x_4 + x_5 + x_6 = 75 \end{align*}\] where \(x_i\) is the \(i\)th ability score. An ability score total of 75 means an average score of 12.5, which is actually slightly better on average than the other methods.

Advanced Concept: Order Statistics

The calculation of the expected value relies on something called order statistics. First, the sample space is defined as the collection of all possible unique outcomes. Here, we have a sample space of \(S = \{1,2,3,4,5,6\}\) for each die - all possible unique outcomes consist of all possible results from one roll of a die. We let \(X\) be the random variable associated with a single roll of a die. \(X\) has what is called a discrete uniform distribution, where each possible outcome has the same probability of occurring.

The \(\mathbf{k}\)th order statistic, denoted \(x_{(k)}\) is is the \(k\)th smallest value in a sample. The corresponding random variable \(X_{(k)}\) will have some distribution which is different from the distribution of \(X\). That is, we will find that each possible outcome is no longer equally likely.

In the setting where we roll four; drop the lowest, we drop \(x_{(1)}\) and take the sum \(x_{(2)} + x_{(3)} + x_{(4)}\). The expected value of this sum will be \[E[X_{(2)} + X_{(3)} + X_{(4)}] = E[X_{(2)}] + E[X_{(3)}] + E[X_{(4)}].\]

Another way to think about “roll four; drop the lowest” is by adding the four rolls and then subtracting the minimum, \(x_{(1)}\). This turns out to be a little easier to manage mathematically. Let \(X_i\) represent the \(i\)th (unordered) roll, where \(i\) takes on the values \(1, 2, 3, 4\). Then we can rewrite that expected value as \[E[X_1] + E[X_2] + E[X_3] + E[X_4] - E[X_{(1)}]\] Why is this easier? Well, the expected value of any unordered roll is just the average roll for a six-sided die, \(E[X_i] = 3.5\). So \[E[X_1] + E[X_2] + E[X_3] + E[X_4] - E[X_{(1)}] = 14- E[X_{(1)}]\] and we now only need to find the value of \(E[X_{(1)}]\), the expected value for the minimum roll.

In order to find this expected value mathematically, we will derive the \(k\)th order statistic. To start, we know that \[\begin{align*} P(X \le x) &= \frac{x}{6} \\ P(X=x) &= \frac{1}{6} \\ P(X < x) &= P(X \le x-1) = \frac{x-1}{6} \\ P(X > x) &= 1 - \frac{x}{6} = \frac{6-x}{6} \\ P(X \ge x) &= P(X > x-1) = 1 - \frac{x-1}{6} = \frac{7-x}{6} \end{align*}\] for \(x = 1,2,3,4,5,\) or \(6\). We can work out these quantities based on the discrete uniform distribution on (1,6) or we can work these out conceptually. For example, \[\begin{align*} P(X \le 2) &= P(\text{roll a 1 or roll a 2}) \\ &= \frac{\text{2 ways to get a roll that is less than or equal to two}}{\text{6 possible outcomes}}\\ &= \frac{2}{6}. \end{align*}\]

Consider \(P(X_{(1)} \ge x)\). If the minimum value is greater than or equal to \(x\), necessarily all of the other values will also be greater than or equal to \(x\). This idea will simplify our calculations significantly. \[\begin{align*} P(X_{(1)} \ge x) &= P(\text{all rolls are greater than or equal to }x) \\ &= P(X_1 \ge x \text{ and } X_2 \ge x \text{ and } X_3 \ge x \text{ and } X_4 \ge x) \\ &= P(X_1 \ge x)\times P(X_2 \ge x)\times P(X_3 \ge x)\times P(X_4 \ge x) \\ &= \frac{7-x}{6}\times\frac{7-x}{6}\times\frac{7-x}{6}\times\frac{7-x}{6} \\ &= \left(\frac{7-x}{6}\right)^4 \end{align*}\]

Similarly, \[\begin{align*} P(X_{(1)} > x) &= P(\text{all rolls are greater than }x) \\ &= P(X_1 > x \text{ and } X_2 > x \text{ and } X_3 > x \text{ and } X_4 > x) \\ &= P(X_1 > x)\times P(X_2 > x)\times P(X_3 > x)\times P(X_4 > x) \\ &= \frac{6-x}{6}\times\frac{6-x}{6}\times\frac{6-x}{6}\times\frac{6-x}{6} \\ &= \left(\frac{6-x}{6}\right)^4 \end{align*}\]

Finally, \[\begin{align*} P(X_{(1)} = x) &= P(X_{(1)} \ge x) - P(X_{(1)} > x) \\ &= \left(\frac{7-x}{6}\right)^4 - \left(\frac{6-x}{6}\right)^4 \end{align*}\]

With the probability distribution, or probability mass function, for the minimum, we are now ready to calculate the expected value of our roll 4, drop the lowest setting.

The probability distribution can be summarized as

x	1	2	3	4	5	6
\(P(X_{(1)} = x)\)	0.518	0.285	0.135	0.050	0.012	0.001

this probability distribution has expected value \[ E(X_{(1)}) = \sum_{j=1}^6 x_j P(X_{(1)} = x_j) = 1.755\]

Returning to the average value for the roll four and drop the lowest, \[\begin{align*} E[X_1] + E[X_2] + E[X_3] + E[X_4] - E[X_{(1)}] &= 14- E[X_{(1)}] \\ &= 14 - 1.755 \\ &= 12.2446 \end{align*}\] and we find that, in fact, the simulation study was quite accurate!

Rolling with Advantage and Disadvantage

2022-01-13T00:00:00+00:00

Rolling with Advantage and Disadvantage

How does probability change when you roll with advantage (or disadvantage)?

Disadvantage

When you roll with disadvantage, you roll two dice, usually d20s, and take the lower number. Say you need to roll a \(20\). If you roll with disadvantage, the only way to achieve that \(20\) is for both dice to roll a \(20\). Let’s say our two dice consist of a red die and a blue die. We will let

\(A\): roll a \(20\) on the blue die.
\(B\): roll a \(20\) on the red die.

To get my \(20\) with disadvantage, I need \(A\) and \(B\) to occur at the same time.

Let’s take a step back for a moment and think about rolling two d4s. What is the probability that both dice roll a \(4\)? (For a visual of all possible rolls, see A Gentle Introduction to Probability Using Dice) From the table, we can see that there are (4 sides \(\times\) 4 sides) 16 possible combinations of rolls and 1 of them satisfy our condition: (4,4). Then \[P(\text{roll 4 on red AND roll 4 on blue}) = \frac{1 \text{ way}}{16 \text{ possible combinations of rolls}} = 1/16\] Something else interesting is happening here. Note that \[P(\text{roll a 4}) = \frac{1 \text{ way}}{4 \text{ possible rolls}} = 1/4\] and that \[P(\text{roll 4 on red AND roll 4 on blue}) = 1/16 = (1/4)\times(1/4)\] It turns out this is not a coincidence! Dice rolls are what we call independent, meaning that each time I roll a die, it has no impact on other rolls of that (or any other) die.

Multiplication rule for two independent events. For two independent events, let’s call them \(A\) and \(B\), \[P(A \text{ and } B) = P(A)\times P(B).\]

Now return your attention to rolling a 20 (on a d20) with disadvantage. Recall

\(A\): roll a \(20\) on the blue die.
\(B\): roll a \(20\) on the red die.

The two dice roll independently, so \[P(A \text{ and } B) = P(A)\times P(B) = \frac{1}{20}\times \frac{1}{20} = \frac{1}{400} =0.0025\] There is a 0.25% (one quarter of one percent!) chance of rolling two 20s.

In general, an average roll of a d20 is 10.5 (this is the mean and the median). We can use order statistics to find the average roll with disadvantage, similar to the approach used in this post)… or we can simulate it using R:

  nrep <- 10000000
  roll.dis <- numeric(length=nrep)
  for(i in 1:nrep){
    roll.dis[i] <- min(sample(1:20, size=2, replace=TRUE))
  }
  mean(roll.dis)

We set the number of repetitions (nrep) to be 10 million. Then, we create a variable to store all 10 million rolls with disadvantage (roll.dis). The next line starts a loop that runs 10 million times - the function sample represents the two rolls of the d20 and the min function takes the smaller of the two. Finally, we exit the loop and take the average of the 10 million simulated rolls with disadvantage.

Running the simulation, we find that the average roll with disadvantage is 7.18, significantly lower than our usual average of 10.5! Even worse, the median roll with disadvantage is a 6!

Advantage

When you roll with advantage, you roll your two dice - usually d20s - and take the higher number. (You may also roll the same die twice, but this is equivalent to rolling two different dice! We will talk about rolling two dice for the sake of consistency.) Suppose again you need to roll a \(20\). Now, you need the blue die to roll a \(20\) or the red die to roll a \(20\).

Note: In statistics, when we say “or” we are using what is called an inclusive or. This means A or B or both. If I asked “Do you want pizza or tacos?” and you said “Yes”, that would be a perfectly valid response. You’re allowed to want both! Treat yourself!

Consider again rolling two d4s, the first red and the second blue. If I need to get a \(4\) but I have advantage, I will be happy with any combination of rolls that includes a \(4\). (See A Gentle Introduction to Probability Using Dice). There are seven such combinations: \((1, 4)\), \((2, 4)\), \((3, 4)\), \((4, 4)\), \((4, 1)\), \((4, 2)\), and \((4, 1)\) so the probability of getting a 4 on red or a 4 on blue is \[P(\text{roll 4 on red OR roll 4 on blue}) = \frac{7 \text{ ways}}{16 \text{ possible combinations of rolls}} = 7/16.\]

Naturally, we get another trick for this one. The probability of getting a \(4\) on red is \(1/4\), which is the same as the probability of a \(4\) on blue. We might be tempted to try adding these probabilities, but if we do this, we end up double counting the roll \((4,4)\).

Here’s another way to think about this: Consider a standard 52 card deck. Let \(A\) be the event that a card drawn is a diamond and let \(B\) be the event it is a face card. Then \(P(A) = 13/52\) and \(P(B) = 12/52\). If we add these, we double count the Jack of Diamonds, Queen of Diamonds, and King of Diamonds. So we need to account for that: \[\frac{13}{52} + \frac{12}{52} - \frac{3}{52}.\]

For our d4 with advantage, this translates to \[\frac{1}{4} + \frac{1}{4} - \frac{1}{16} = \frac{7}{16}.\] What is this “double counting” we are taking into account? How do we figure this out quickly? It turns out, if we add two probabilities, of say \(A\) and \(B\), we always double count \((A \text{ and } B)\).

Addition rule for two events. For events \(A\) and \(B\), \[P(A \text{ or } B) = P(A) + P(B) - P(A \text{ and } B).\] So let’s return to rolling two d20s with advantage. Again let

\(A\): roll a \(20\) on the blue die.
\(B\): roll a \(20\) on the red die.

The probability of getting a 20 with advantage, \(P(\text{roll a } 20 \text{ on blue OR roll a } 20 \text{ on red})\) is \[P(A \text{ or } B) = \frac{1}{20} + \frac{1}{20} - \frac{1}{400} = \frac{39}{400} = 0.0975\] where we found \(P(A \text{ or } B)\) in the subsection on disadvantage. There is a 9.75% chance of rolling a 20 with advantage.

What about an average roll with advantage? This is going to be really similar to how we thought about average rolls with disadvantage:

  nrep <- 10000000
  roll.adv <- numeric(length=nrep)
  for(i in 1:nrep){
    roll.adv[i] <- max(sample(1:20, size=2, replace=TRUE))
  }
  mean(roll.adv)

We set the number of repetitions (nrep) to be 10 million. Then, we create a variable to store all 10 million rolls with advantage (roll.adv). The next line starts a loop that runs 10 million times - the function sample represents the two rolls of the d20 and the max function takes the larger of the two. Finally, we exit the loop and take the average of the 10 million simulated rolls with advantage.

Running the simulation, we find that the average roll with advantage is 13.83. Our median is is even higher, coming in at 15, significantly better than that 10.5 average we get rolling straight.