Monday, October 5, 2020

Isn't it better to use a dice deck in Catan?

First of all, better is a value judgement. Let's examine the properties and you can evaluate whether you like them better or not.

The idea of a dice deck in Catan is that you have a deck of 36 cards, where instead of rolling 2d6 you flip over a card and it gives you the number for resource generation. (See here, although I'll ignore the events and the New Year card.) The values on the cards match the distribution of 2d6, such that the first draw has the same probability as dice. However, if you draw the deck to exhaustion, then you're guaranteed to get exactly five results of "6".

Let's start with a question about using dice: if the probability of rolling a six is $5/36$, then do we expect to roll 5 sixes out of 36 rolls? First, let me say that in probability, expect is a word with particular meaning. It refers to the average, or mean value, or something. We write it as an operator using a special capital E. It being an operator means that we write it to the left and it operates on whatever is to the right of it. So when we write $\mathbb{E} X$, we're taking the expectation of the random variable $X$. I'm writing with special typesetting tools (Well, just $\LaTeX$.), but if you're writing with pencil and paper, on a blackboard, or on an internet forum, you may not have access to such special symbols. Then just a capital E will do.

So our question, reformulated is whether the expected value of the number of occurrences of the number six is $5/36$—i.e. $P(\text{rolling a six on 2d6})$—or is it some other value? To find out, let's calculate the expected value of the number of occurrences of the number six on 2d6. Wow, that's long winded. Let's assign it a variable name, $Y$.

\begin{align} Y = \text{number of results equal to six across 36 rolls of 2d6}. \end{align}

How do we count this? Let's assign names to these 36 rolls. We'll call the results of the 36 2d6 rolls $X_1$—$X_{36}$.

As we've been saying, we know from our previous analysis of 2d6 that

\begin{align} P(X_i = 6) = \frac{5}{36} \quad i \in \{1,2,\cdots,36\}. \end{align}

Recall that the probability of something not happening is equal to 1 minus the probability of it happening, or

\begin{align} P(\text{not some thing}) = 1 - P(\text{some thing}). \end{align}


\begin{align} P(X_i \neq 6) = 1 - P(X_i = 6) = 1 - \frac{5}{36} = \frac{31}{36}. \end{align}

Let's define the random variables $Y_1$—$Y_{36}$ to be 1 if $X_i=6$ and 0 if $X_i \neq 6$. We write this as,

\begin{align} Y_i = \begin{cases} 1 & X_i = 6 \\ 0 & X_i \neq 6 \end{cases} \quad i \in \{1,2,\cdots,36\}. \end{align}

From this definition, we can see that the sum of these $Y_i$ is the count $Y$ we are looking for.

\begin{align} Y = \sum^{36}_{i=1} Y_i \end{align}

What then, is the expected number of sixes $\mathbb{E}Y$? Expectation is a linear operator. That means a few things, that we'll go into elsewhere, but what's important here is that it means that since $Y_i$ are independent, we can distribute the expectation across the different terms in the summation $Y_i$. Basically, we can move it from the left to the right of the summation, like this

\begin{align} \mathbb{E}Y = \mathbb{E} \sum^{36}_{i=1} Y_i = \sum^{36}_{i=1} \mathbb{E}Y_i. \end{align}

As mentioned above, the expected value of a random variable is the average so,

\begin{align} \mathbb{E}Y_i &= \sum^1_{y=0} y \cdot P(Y_i = y). \end{align}

Note that the summation above is from 0 to 1 because those are the only two values possible for $Y_i$. Using the probabilities of $X_i = 6$ and $X_i \neq 6$ above, we can find

\begin{align} \mathbb{E}Y_i &= 0 \cdot P(Y_i = 0) + 1 \cdot P(Y_i = 1)\\ \mathbb{E}Y_i &= 0 \cdot P(X_i \neq 6) + 1 \cdot P(X_i = 6)\\ \mathbb{E}Y_i &= 0 \cdot \frac{31}{36} + 1 \cdot \frac{5}{36}\\ \mathbb{E}Y_i &= \frac{5}{36}. \end{align}

Plugging this back into our equation for $\mathbb{E}Y$ above yields,

\begin{align} \mathbb{E}Y &= \sum^{36}_{i=1}\frac{5}{36} \\ \mathbb{E}Y &= 36 \cdot\frac{5}{36} \\ \mathbb{E}Y &= 5 . \end{align}

So we do expect to get 5 sixes when rolling 2d6 36 times. Note that above, we changed from summing $5/36$ 36 times to multiplies $5/36$ by 36. That's because those are equivalent. That's what multiplication is! We can do this whenever we are summing a constant value.

The same thing also holds for all other result here, not just six.

We can also ask, what's the distribution of $Y$? Or, what's the probability of getting 5 sixes, or 4, or 3, or however many? We could try to follow the approaches we've used to find the distribution of the sums of dice. But we have 36 rolls here, that doesn't sound like a fun calculation. On the other hand, each $Y_i$ has a simple distribution, so maybe it's not so bad.

Fortunately, this is a problem with an existing solution that has been well studied: the binomial distribution. We've used it before, but this time we'll derive what the distribution of $Y$ is. It's probably not that bad. Let's change our notation a little bit, though. The binomial distribution is a parameterized distribution, which means it takes parameters. Basically, there's a bunch of different versions of the distribution. A normal (I mean, regular) single die has a uniform distribution, but the particulars of the uniform distribution depend on the number of faces of the die as a parameter. Similarly a binomial distribution, written as $Y ~ B(n, p)$ takes two parameters: the number of trials, $n$, and the probability of the success of each one, $p$. The number of trials is the number of times that we roll the die. Here we'll use 36 trials to match the 36 cards in the dice deck. The probability, $p$, depends on which face we're focused on. For a result of six, the probability of getting that value in each trial, which we call a success, is $5/36$.

So then in each trial we either get a success, with probability $p$, or failure with probability $1-p$. We're interested in the count of the successes. The extremes are relatively easy to calculate. The probability that we get $n$ successes in $n$ trials is the probability that we succeed on every roll.

\begin{align} f_Y(n) = p^n \end{align}

Similarly, the probability that we get no successes is the probability that we fail on every roll.

\begin{align} f_Y(0) = (1-p)^n \end{align}

Now, let's consider some result in between, where we get $k$ successes. This could happen by the first $k$ rolls succeeding and the last $n-k$ rolls failing.

\begin{align} p^k \cdot (1-p)^{n-k} \end{align}

However, we could also fail the first $n-k$ rolls, and succeed in the last $k$.

\begin{align} (1-p)^{n-k} \cdot p^k \end{align}

Note that these occur with the same probability. We can construct several different distinct orders of success and failure that result in $k$ successes. All of them share this same $p^k \cdot (1-p)^{n-k}$ term, but we need to multiply this by the number of ways that we can construct a sequence with $k$ successes in $n$ trials. We need the number of ways to select $k$ elements out of a set of size $n$. We know that this is $\binom{n}{k} = \frac{n!}{k! (n-k)!}$. Thus, we can write the pmf of $Y$ as follows.

\begin{align} f_Y(k) &= \binom{n}{k} \cdot p^k \cdot (1-p)^{n-k} \\ f_Y(k) &= \binom{36}{k} \cdot \left(\frac{5}{36}\right)^k \cdot \left(1-\frac{5}{36}\right )^{n-k} \end{align}

By replacing $p$ with the probabilities for each valid result on 2d6, we can get the distributions for the number of times that each of those results appear across a sequence of 36 rolls. These distributions are plotted in Fig. 1. Note that these random variables are not independent. We analyzed each separately, but it's clear that in the case when there are an excess of one value, it decreases the probability of getting a large number of another value.

Figure 1: Probabilities of rolling 2d6 

We can contrast that with the certainty of a dice deck expressed in terms of the probability mass functions shown in Fig. 2. Here we know the number of times that each value appears, and so each value has a probability of 1 at the number of times that it occurs in the deck. While these values correspond to both the average and peak values of the distribution of 2d6, it does not capture the sizable variation. This is particularly evidence for the results that come up more rarely. It's almost as likely to get no twos across 36 rolls as to get 1 two. At the other extreme, if we consider the number of sevens, we see that the probability of getting exactly the expected number of sevens is significantly lower than for twos. Getting 4 or 7 of them, instead of 6,may seem close to the average and within desired variation, but neither is possible with the dice deck.

Figure 2: Probabilities using a dice deck

There are a couple of extreme cases where what we get compared to what we want may break down with the dice deck. First, if near the end of the game enough of the deck remains that it's pretty clear that it will not get reshuffled, but certain values have already been exhausted, the benefits of building certain settlements and cities is quite different than with dice. With dice you can build on a twelve and hope that it pays out, but with the dice deck you may know that it never will. The other extreme case is when the deck does get reshuffled very near the end of the game. Through most of the game you've been playing with the knowledge that the deck ensures that the rolls across several turns produce a set distribution. However, if you end shortly after a reshuffle, you now have very similar probability properties to that of dice, where there's the possibility that something more unexpected may come up.

By modifying the simple dice deck that we considered you can adjust the properties between these two extremes. For example, could reshuffle before reaching the end of the deck. By adjusting the time of reshuffling, you can slide on a spectrum between a our pure example and using dice. Of course, if you shuffle after every draw you get the same probabilities as dice. In that case, dice are better, because they're so much faster to use.

So far, we've covered differences between dice and a deck of cards which basically boil down to this: a deck has memory. What we draw on one term influences the next, because the state of the deck is changed. If you draw a seven, it's less likely you'll draw another until the deck is reshuffled. On the other hand, each roll of the dice is independent. There's no influence of one roll on the next.

There's another fundamental difference between the two: the deck is shuffled at the beginning. This means that the result of each draw is predetermined (unless you believe the entire deck can be in a state of superposition). If you make a poor choice, you can wallow in regret, as you knew that if you had made a different choice you could have had a better outcome. Not so with dice. Following the butterfly effect, any decisions you make before a roll influence that roll. This is not in a controllable way, but just that minute differences in the timing or action of a die roll may yield a different result. Thus, no second guessing is necessary, as if you had decided to do something different, the dice may still have been against you.

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