In this series, we’ve already looked at the probabilities of rolling straights versus multiple dice of a kind, and the probabilities of rolling different sums with multiple dice. In this installment of Probability in Games, we’ll combine aspects from each of these techniques to analyze a mechanic from classic games like Can’t Stop and Kingsburg (both available on iOS). In each of these games, you roll some dice, choose a subset of those dice to sum, and that sum then influences your turn in some way. In the example of Can’t Stop, players roll four dice. They then divide those dice into two groups of two dice, and calculate the sum of each pair. These two sums determine which, if any, scoring track(s) they can progress along on their turn. What do you think this probability distribution should look like? In other words, how often will your roll allow you to choose two dice with a specific sum, for each possible sum?

Modern gamers tend to be familiar with the probability distribution of sums rolled on two six-sided dice. An abbreviated histogram of this distribution curve is depicted in the small pips below each numbered token in the Settlers of Catan.

Here’s the full probability distribution for rolling 2d6:

And at first glance, this graph also looks similar to the board in the game Can’t Stop. Are the probability distributions in these two games related? And if so, how?

In this article, we’ll analyze the player’s chances of being able to add two of four randomly rolled dice to a particular sum. In fact we’ll do this calculations for every possible sum to explore the distribution of these probabilities. To do this, I’d like to start with a somewhat naïve approach that will lead us to a rough approximation of the answer. From there, we can discuss the short-fallings of this naïve approach and use those insights to develop a more complete solution.

Let’s think about the number of ways that a player might choose two of the four dice that they have rolled. You might recall our discussion from a previous blog post about calculating combinations, and recognize that this is 4-choose-2 = 4!/(2!*2!) = 6 ways to choose two of the four dice. If not, you can just as quickly list out the six ways of choosing two of four dice. From this, we might naïvely expect that choosing any two of four randomly rolled dice should be equivalent to choosing any one of six randomly rolled pairs of dice. In both cases we have six choices, and each choice consists of two randomly rolled dice.

So let’s find the likelihood that one of six randomly rolled pairs of dice hit each possible sum… starting with the most common sum of seven. We’ll start by working out the odds that a single pair of dice will add to seven. No matter what you roll on the first die in any pair, there is always one value for that second die that can be added to the first to reach seven (in fact, it is always the value on the bottom of the first die). So there is a 1/6 chance that any pair of randomly rolled dice will add up to seven.

But what if we roll six pairs of dice? This calculation might remind you of the previous blog post where we wanted to know how often we’d roll multiple dice with a particular value. We have a 1/6 chance that the first pair adds to seven, and the other 5/6 of the time we need to check another pair. The same goes for each subsequent pair. Expanding this out, we get:

1/6 + 5/6*(1/6 + 5/6*(1/6 + 5/6*(1/6 + 5/6*(1/6 + 5/6*(1/6))))) = 66.5%

So we have a 66% chance of rolling a pair of dice that add up to seven, when rolling six pairs of dice. We can substitute into this equation, the probabilities of rolling other sums (as listed on the table above), to find our chances of rolling them across six different attempts to build the following table.

With this distribution in mind, let’s now return to our initial assumption about how choosing two of four randomly rolled dice should be equivalent to choosing one of six randomly rolled pairs of dice. Can you think of any ways or reasons why these two mechanisms should differ?

In my mind, there are two reasons that these distributions should diverge. The first has to do with rolling multiple dice of a kind. For instance when you roll four of a kind, you have no choice about the sum that you choose. And although this could happen when rolling six pairs of dice, it will happen much less often with unlikely sums like two and twelve. Also notice that the sum of two equal dice is always even, so it seems reasonable to expect some more variation between odd and even sums. The second issue in my mind has to do with rolling dice with different values. For instance imagine rolling dice with four different values. Whenever this happens, you are guaranteed to have rolled two dice that add up to seven. There is just no way to pick four different numbers between one and six without getting at least one pair that adds up to seven. Give it a try, if you don’t believe me. In contrast to this, it is quite possible to roll six pairs of dice with different sums without any of them being equal to seven.

Alright, so if these techniques result in different probability distributions, how do we formulate a better solution? To get at the issue of our options being limited by rolling multiple equal dice, let’s start by breaking down the different ways in which we might roll duplicate values. For each type of roll, we’ll make a note of our chances of rolling it.

**Four Singles (ABCD): 6/6 * 5/6 * 4/6 * 3/6 = 27.7%
**In order to roll four singles, the first die can have any value (6/6), the second can be anything besides the first (5/6), the third dice can be anything other than the previous two (4/6), etc.

**One Pair (AABC): 6/6 * 1/6 * 5/6 * 4/6 * 4C2 = 55.5%
**The first die can be anything (6/6), another must match that (1/6), and we need two more that are not equal to the first nor to each other (5/6 and 4/6). However the pair can really fall between any two of the four dice (4C2).

**Two Pairs (AABB): 6/6 * 1/6 * 5/6 * 1/6 * 4C2/2 = 6.94%
**The first die can be anything (6/6), and the second must match (1/6). The third die must be different from the first two (5/6), and the fourth must match the third (1/6). However each pair could really fall between any two of the four dice (4C2). Since choosing one pair automatically fixes the other, this is actually counting every possible order twice so we divide by two (/2).

**Three of a Kind (AAAB): 6/6 * 1/6 * 1/6 * 5/6 * 4C1 = 9.25%
**The first three dice must all be the same (6/6, 1/6, and 1/6), and the fourth die must be different (5/6). However the single unique value rolled could appear on any of the four dice (4C1).

**Four of a Kind (AAAA): 6/6 * 1/6 * 1/6 * 1/6 = 0.462%
**The first die can be anything (6/6), and every other dice must be the same (1/6, 1/6, and 1/6).

With these different types of rolls and their likelihoods in mind, we’ll now turn to the distribution of sums available within each type of roll. These distributions can be broken down into four basic distributions: the sum of two equal values (Equal), the sum of two unique values (Two Unique), the sum of two from three unique values (Three Unique), and the sum of two from four unique values (Four Unique). Here’s how each of our roll types can be decomposed into these distributions.

Four Singles (ABCD): Four Unique (AB, AC, AD, BC, BD, CD) One Pair (AABC): Equal (AA) + Three Unique (AB, AC, BC) – overlap (AA = BC) Two Pairs (AABB): 2 * Equal (AA, BB) + Two Unique (AB) Three of a Kind (AAAB): Equal (AA) + Two Unique (AB) Four of a Kind (AAAA): Equal (AA)

Hopefully these letters designating unique dice values help you see how these rolls can be decomposed into these different distributions. Once we work out these four distributions, we’ll be able to calculate the full probability distribution for every possible sum in Can’t Stop with them. As an example of how this is done, let’s consider computing the probability of being able to choose two dice that sum to seven. In terms of expressing this, I’ll write Four Unique(7) below to mean the chances of four unique values including two that add up to 7, Equal(7) to mean the chances of two equal values summing to seven, etc. Also, don’t worry about that overlap term above for now. It’s zero in this case, and we’ll discuss it in more detail near the end of this discussion.

27.7% * Four Unique(7) + 55.5% * (Equal(7) + Three Unique(7)) + 6.94% * (2*Equal(7) + Two Unique(7)) + 9.25% * Equal(7)

Fortunately these distributions are relatively easy to write out. Equal is the easiest, since any die value added to itself is an even value. So each even value have a likelihood of 1/6, and every other value has a chance of 0/6. For the sums of unique values, there fortunately aren’t too many of them. There are 6C2=15, 6C3=20, and 6C4=15 ways to choose two, three, and four unique values from the six possible dice values. So I wrote each of these combinations out, and counted the number of times each sum could be made from a combination. I’d certainly love to hear, if anyone knows of a formula to generate these distributions. In the mean time, here are the values that I calculated:

By plugging these values into our formula above, we can find the chances of rolling four dice so that a pair of them add to seven.

27.7% * 100% + 55.5% * (0% + 60%) + 6.94% * (2*0% + 20%) + 9.25% * 0% = 62.3% (or without the rounding errors is 64.35%)

We can perform a similar calculation for the ten other possible sums, but first lets discuss that overlap that we’ve been ignoring. When rolling one pair, it’s possible that the two unique values have the same sum as the the pair of matching dice values. The formula above is counting each of these occurrences twice: once in the Equal distribution, and once in the Three Unique distribution. However this can only happens with even sums. Specifically, this happens each time the two non-matching dice add up to the even sum in question. And you might recall that we just counted the number of ways that two unique values can add up to each possible sum while generating the distribution for Two Unique. So for even sums, the overlap that we must subtract is Two Unique*15/60. Multiplying by fifteen gets us back to the number of ways that two unique values can add up to the sum, and dividing by sixty ensures that we only count each way once out of the sixty possible ways to roll exactly one pair among four dice: 6C3 * 3 = 60. With all of this work behind us, we can now finally compute the probability distributions for rolling every possible sums in Can’t Stop:

This post has now gotten much longer than I originally anticipated, so I will stop here. If you’ve enjoyed this exploration, I encourage you to take it further. What really matters when playing Can’t Stop is that your roll produces one of the three values that you are currently scoring. Try extending these techniques to come up with a distribution that covers every possible combination of three sums, instead of for each sum individually. As always, please post any questions or request that you have for this or future installments of the blog in the comments below. Thanks for reading!

[…] En Can’t Stop las probabilidades son ligeramente diferentes, la mayoría de las veces que tiramos los dados, es posible conseguir alguna combinación que coincida con los tres números que jugamos. La simple práctica en el juego nos hace dar cuenta que la posibilidades de conseguir combinaciones de los números es más alta en en Catan. Esto se debe a que utilizamos cuatro dados para emparejar y esto aumenta considerablemente las probabilidades. Aquí tenéis dos gráficas donde podéis ver la diferencia entre el Can’t Stop y Catan. I consultar el estudio que han hecho en esta sugarpillstudios. […]

Thanks for the comment. This might be getting lost in translations, but I’m not sure how you are calculating the probabilities in your graph for Can’t Stop. I’m certainly interested in learning what we are doing differently.