March 2018
Pandemic: The Cure is a neat version of the game Pandemic that is almost entirely dice-based.
Like Pandemic, there are four diseases of different colors, and the goal is to cure all four. Unlike Pandemic, in Pandemic: The Cure you do this by collecting dice of the disease's color, and at the end of every turn you can roll them. If you roll at least a 13, then the disease is cured. If not, you can try again next turn, but the dice are tied up until then, which has some negative consequences in the game.
The dice are six-sided, but they're not standard; each color has different numbers on it:
| black | 0, 3, 3, 3, 4, 5 |
|---|---|
| red | 0, 1, 1, 4, 6, 6 |
| yellow | 0, 2, 2, 4, 5, 5 |
| blue | 0, 1, 2, 3, 6, 6 |
In addition, one character (the Scientist) can cure if she rolls an 11 or above.
After playing a few times, it seemed to me like having four dice had very good odds for rolling a 13, so my thought was the optimal strategy would be to collect four dice and then stop trying to collect more. The Scientist's special power seemed weak compared to the others, so my guess was that it made using the same strategy with three dice practical.
I started by calculating the results for three dice by hand. Let's look at the black dice first.
For three dice, there are \(6*6*6=216\) possibilities. So we need to find how many of these possibilities are 13 or greater. We can do this by finding all combinations of dice that satisfy this. For example, rolling a 5, 5, and 3 is 13. There's only one 5, but there are three 3's, and there are three ways to arrange the dice, so this accounts for \(1*1*3*3=9\) possibilities.
So the only tricky part is making sure you don't miss any combinations (like I did a few times :-/ ). It helps to have an order to iterate over the combinations - for example, all three dice are 5, then two dice are 5 and one is not, then one dice is 5 and two are not, etc.
Here's the breakdown for three black dice >= 13:
| Dice | Number |
|---|---|
| 5, 5, 5 | \(1*1*1*1=1\) |
| 5, 5, 4 | \(1*1*1*3=3\) |
| 5, 5, 3 | \(1*1*3*3=9\) |
| 5, 4, 4 | \(1*1*1*3=3\) |
So the total number of possibilities is \(1+3+9+3=\textbf{16}\), and the chance of getting at least 13 with three dice is \(\frac{16}{216}=\textbf{7.4%}\)
Similarly, for three black dice >= 11:
| Dice | Number |
|---|---|
| 5, 5, 5 | \(1*1*1*1=1\) |
| 5, 5, 4 | \(1*1*1*3=3\) |
| 5, 5, 3 | \(1*1*3*3=9\) |
| 5, 4, 4 | \(1*1*1*3=3\) |
| 5, 4, 3 | \(1*1*3*6=18\) |
| 5, 3, 3 | \(1*3*3*3=27\) |
| 4, 4, 4 | \(1*1*1*1=1\) |
| 4, 4, 3 | \(1*1*3*3=9\) |
So the total number of possibilities is \(1+3+9+3+18+27+1+9=\textbf{71}\), and the chance of getting at least 11 with three dice is \(\frac{71}{216}=\textbf{32.9%}\)
I actually did the rest of the three dice ones by hand, but doing four dice seemed like it would be a lot more trouble. So: to the computer!
For a change, I wrote a C# script to do these calculations (instead of Python). It was fun: I used Vim on Linux with Mono, although man have I been spoiled by Visual Studio and Intellisense! Anyway, here's the script.
| >= 13 | >= 11 | ||
|---|---|---|---|
| Black | |||
| 3 dice | 7.4% | 32.9% | |
| 4 dice | 46.9% | 70.1% | |
| 5 dice | 77.2% | 89.5% | |
| Red | |||
| 3 dice | 23.1% | 40.3% | |
| 4 dice | 45.8% | 62.7% | |
| 5 dice | 67.4% | 78.3% | |
| Blue | |||
| 3 dice | 20.4% | 34.3% | |
| 4 dice | 45.1% | 60.9% | |
| 5 dice | 67.3% | 79.9% | |
| Yellow | |||
| 3 dice | 12.0% | 34.7% | |
| 4 dice | 45.8% | 67.7% | |
| 5 dice | 73.5% | 86.5% |
Fancy math equations by MathJax.