Introduction to Combinatorics and Probability for Gaming (Part I)

The modern study of probability was sponsored by a French noble, the Chevalier de Mere, who employed Blaise Pascal to determine why he was consistently losing at a particular dice game. Together, they analyzed two games that the noble played:

  1. Roll a six sided die four times, and bet on whether any of the results will be a 6.
  2. Roll a pair of dice twenty-four times, betting on whether a pair of any sixes show up on any roll.

The noble had tried to compute the probabilities naively, with the following logic:

  1. There is a $ \frac{1}{6} $ chance of any given die roll turning up 6. With four rolls, there must be a $ 4 \cdot \frac{1}{6} = \frac{4}{6} = \frac{2}{3} $ chance of a 6 appearing.
  2. There's a $ \frac{1}{6} $ chance of a six appearing on either die, so a $ \frac{1}{6} \cdot \frac{1}{6} = \frac{1}{36} $ chance of double sixes appearing. (So far so good!) Over 24 rolls, we then have $ 24 \cdot \frac{1}{36} = \frac{24}{36} = \frac{2}{3} $ chance of a double six appearing.

Given these apparent probabilities, the noble would frequently bet in both cases on the six or double six appearing, and while he'd usually make money off the first game, he frequently lost in the second case.

The error in the noble's thinking can be revealed if we consider what happens in the first game if we were to roll six or eight times. Would the resultant probabilities then be 6/6 = 1, or $ \frac{8}{6} = \frac{4}{3} $? Of course not.

There are two approaches to computing the probability of rolling at least one six in four rolls. The first is to consider every case. First, some notation. [6] indicates a roll of 6. [1-5] indicates any roll that landed on any value between one and five. [6][1-5] would indicate that the first roll was a six, and the second was not. Here are the different rolls that would have won the noble his bet, and their associated probabilities:
[6][1-5][1-5][1-5] 1/6 * 5/6 * 5/6 * 5/6 = 125 / 1296
[1-5][6][1-5][1-5] 5/6 * 1/6 * 5/6 * 5/6 = 125 / 1296
[1-5][1-5][6][1-5] 5/6 * 5/6 * 1/6 * 5/6 = 125 / 1296
[1-5][1-5][1-5][6] 5/6 * 5/6 * 5/6 * 1/6 = 125 / 1296
[6][6][1-5][1-5] 1/6 * 1/6 * 5/6 * 5/6 = 25 / 1296
[6][1-5][6][1-5] 1/6 * 5/6 * 1/6 * 5/6 = 25 / 1296
[6][1-5][1-5][6] 1/6 * 5/6 * 5/6 * 1/6 = 25 / 1296
[1-5][6][6][1-5] 5/6 * 1/6 * 1/6 * 5/6 = 25 / 1296
[1-5][6][1-5][6] 5/6 * 1/6 * 5/6 * 1/6 = 25 / 1296
[1-5][1-5][6][6] 5/6 * 5/6 * 1/6 * 1/6 = 25 / 1296
[6][6][6][1-5] 1/6 * 1/6 * 1/6 * 5/6 = 5 / 1296
[6][6][1-5][6] 1/6 * 1/6 * 5/6 * 1/6 = 5 / 1296
[6][1-5][6][6] 1/6 * 5/6 * 1/6 * 1/6 = 5 / 1296
[1-5][6][6][6] 5/6 * 1/6 * 1/6 * 1/6 = 5 / 1296
[6][6][6][6] 1/6 * 1/6 * 1/6 * 1/6 = 1 / 1296
Summing these up, we get $ \frac{671}{1296} $ ~= 51.7%. This is quite a tedious calculation, even though with a bit of knowledge of combinatorics we could have skipped most of the enumeration. Fortunately, there's an easier way, which is a blessing when considering the work that would otherwise be necessary for the 24 die case.

There is one set of outcomes absent from the list above, and it's a very simple one: [1-5][1-5][1-5][1-5], with a probability of $ (\frac{5}{6})^{4}) = \frac{625}{1296} $. This is the probability that we get no 6s. Since we only care that we get at least one 6, but don't object to getting more, this case represents the complement to what we're looking for. Therefore, if instead of calculating probability of making a roll with at least one 6, we instead calculate the probability that we won't roll all values between 1 and 5. These two situations are the same, but the latter makes our approach clearer.

If the probability of an event X happening is x, then the probability of all events that aren't X is 1 - x. If we consider our event to be "Only roll 1-5 on all dice", with a probability of $ \frac{625}{1296} $, then our probability of getting at least one 6 is $ 1 - \frac{625}{1296} = \frac{671}{1296} $, the value that we calculated above.


Armed with this trick, we can help our Chevalier friend calculate the probability of winning the second game. As he observed, the probability of a double-six is $ \frac{1}{36} $, so the probability that the roll will not be a double-six is \frac{35}{36}. We multiply this value with itself 24 times (i.e. $ (\frac{35}{36})^{24} $) to calculate the probability of not getting a double-six, and get the rather ugly value of $ \frac{11419131242070580387175083160400390625}{22452257707354557240087211123792674816} $, which we can approximate with the much more approachable percentage chance of 50.86%. Therefore the probability of a double-six appearing is 1 - 50.86% = 49.14%. Assuming even odds being given on the game, for every 100 francs he bet, he could expect to get only about 98 of them back.

Many times, this trick is enough to calculate the desired value, but sometimes the number of 6s (or success in Shadow Run 4 or new World of Darkness) matter. In my next post, I'll describe ways to calculate these values more easily.