Monday, April 4, 2022

How often does each number come up in Space Base?

 In Space Base, every turn the active player rolls two six-sided dice. Each player then chooses to treat these as separate numbers or first sum them together and get the corresponding rewards or reward, respectively. Thus, in some sense, the question is not answerable, because it depends on each player's choice. The rulebook does a decent job on laying this out and speaking somewhat about the probabilities.

We previously found the probability of rolling each result on 2d6 for Catan, so we won't revisit that. Is there a good way to layer on top of that the idea that we can take either the sum of the dice or the results individually? The rulebook makes one choice. A quick glance at the table there may be misleading, as it shows the distribution of a regular 2d6 roll with die faces, but then show the numbers for a combination of sum and individual results.

If we want to get a single probability for how often each number comes up, we have to make a choice of how to handle the case where we roll doubles. Should we count that as two occurrences of the single die value or just one? It depends on what we care about. If we're trying to figure out the average return then I think it makes sense to choose, as the rulebook did, to count that as two results. You can see that in the number of times that they list a 1 showing up: 12 out of the 36 possible results of rolling two 6-sided dice, as shown in Table 1. Note that one of these times is when both dice are rolled together. Thus, there are only 11 rolls that produce these 12 ones. While the double roll makes up for there being only 11 outcomes that include a 1 being rolled in the average reward over many rolls, there may be cases where we are more interested in how many rolls include a given number and don't benefit from rolling doubles. In Space Base there are some cards that gain a charge cube when rolled, but can only hold one charge cube. Thus, rolling doubles of that number can be worse overall, as we don't have access to rewards from a second number.

\begin{align*} \begin{array}{c|cccccc} & 1 & 2 & 3 & 4 & 5 & 6 \\ \hline 1& {1},{1} & {1},2 & {1},3 & {1},4 & {1},5 & {1},6\\ 2& 2,{1} & 2,2 & 2,3 & 2,4 & 2,5 & 2,6\\ 3& 3,{1} & 3,2 & 3,3 & 3,4 & 3,5 & 3,6\\ 4& 4,{1} & 4,2 & 4,3 & 4,4 & 4,5 & 4,6\\ 5& 5,{1} & 5,2 & 5,3 & 5,4 & 5,5 & 5,6\\ 6& 6,{1} & 6,2 & 6,3 & 6,4 & 6,5 & 6,6\\ \end{array} \end{align*}

Table 1: Rolling ones on two 6-sided dice

We also need to choose how to handle the option to take the sum of the dice or the individual values. The rulebook includes both at full weight. That is, out of the 36 possible results of rolling two dice, it includes all the times that a 6 is rolled on individual dice as well as all the times that the sum of the two dice is 6. If this were the only number for which we received rewards, or if it were significantly better than our other rewards, then this makes sense, as we'd always choose 6 when it's an option. However, in many cases it could be that getting both a 2 and a 4, for example, is better than just getting the sum of 6. That may be an obvious choice or a difficult one. A way to handle this is to say that we don't know which way players are going to choose (though this may vary with strategy) and treat it as a random event with each option having equal probability. We can think of this as reducing the number of times each number comes up by a factor of 2.

Fortunately, we are spared some complexity by the fact that the sum and individual numbers are always different, since each die has a minimum value of 1. Otherwise, we'd have to consider what to do when the sum matches a single die. Actually, I think we'd always choose the single values, unless it was to use an ability to adjust the number that depends on using the sum. However, we'd still have to account for it in the math to avoid double counting.

There's also a choice of how to combine these two factors. I see two natural pairings. First, if looking at the average amount of return, I think it makes sense to assume that the sum and individual values are each used half the time. This gives something like a weighted probability that you get the rewards from a given number. (I say weighted, because it may sometimes involve double rewards, as discussed above.) Second, if looking at how often something is available, it makes sense to count double rolls as single, but include both individual values and the sum at full weight. This is essentially answering how many turns a given number is an option.

Those both give the middling option for the probabilities. Interestingly, the rulebook went with neither and instead prints the more optimistic combination of counting a double value as two result and including both individual numbers and sums. This does make sense if you're looking at the average reward of a number that is strictly better than everything else, but it ignores the counterfactual or opportunity cost of not choosing the other option. That is, if you're choosing between the sum 5 and the individual numbers 2 and 3, if you're looking at 5, the benefit of what's there has to be compared to the alternative of the rewards for both 2 and 3.

I suppose I should say something about the last of the four options: what does pairing no doubles with random sum or individual? It seems appropriate to use when you only get benefit from a single reward per turn, but it's a tossup as to whether the number would be chosen given the other choices.

\begin{align*} \begin{array}{c|cccc} \text{Method} & 1 & 2 & 3 & 4 \\ \hline P(\text{sum}) = P(\text{individual}) & 0.5 & 0.5 & 1 & 1 \\ \text{# single} & 11 & 12 & 11 & 12 \\ \hline 1 & \frac{5.5}{36} & \frac{6}{36} & \frac{11}{36} & \frac{12}{36}\\ 2 & \frac{6}{36} & \frac{6.5}{36} & \frac{12}{36} & \frac{13}{36}\\ 3 & \frac{6.5}{36} & \frac{7}{36} & \frac{13}{36} & \frac{14}{36}\\ 4 & \frac{7}{36} & \frac{7.5}{36} & \frac{14}{36} & \frac{15}{36}\\ 5 & \frac{7.5}{36} & \frac{8}{36} & \frac{15}{36} & \frac{16}{36}\\ 6 & \frac{8}{36} & \frac{8.5}{36} & \frac{16}{36} & \frac{17}{36}\\ 7 & \frac{3}{36} & \frac{3}{36} & \frac{6}{36} & \frac{6}{36}\\ 8 & \frac{2.5}{36} & \frac{2.5}{36} & \frac{5}{36} & \frac{5}{36}\\ 9 & \frac{2}{36} & \frac{2}{36} & \frac{4}{36} & \frac{4}{36}\\ 10 & \frac{1.5}{36} & \frac{1.5}{36} & \frac{3}{36} & \frac{3}{36}\\ 11 & \frac{1}{36} & \frac{1}{36} & \frac{2}{36} & \frac{2}{36}\\ 12 & \frac{0.5}{36} & \frac{0.5}{36} & \frac{1}{36} & \frac{1}{36}\\ \end{array} \end{align*}

Table 2: Probability of reward in Space Base

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