Now that I’m not working 60+ hours a week, the SigFig is no longer relegated to single-player gaming. A few nights ago we teamed up for an adventure through Castle Ravenloft: a Dungeons & Dragons board game with pre-constructed characters/scenarios and randomly generated maps and monsters instead of a dungeon master.
You can’t play DnD without dice, and you can’t play with dice without considering probability. Dice, coins, and decks of cards are the canonical examples for basic probability exercises. However, I’ve observed that more recent K-12 texts shy away from cards and refer to “number cubes” instead of dice, probably to avoid any links to gambling. (Side note: when I was an associate editor working on a middle school science program for a major publisher, I was shot down when I tried to use an ice cream scoop as an example of a lever. The reasoning behind that decision? We weren’t supposed to promote junk food. I’m of the opinion that anything that links science to ice cream is a good thing.) The gambling connotation could be avoided and more interesting calculations could be performed if polyhedral role-playing game (RPG) dice were used as examples, but I’m willing to bet that many of those who are convinced that using the term “dice” will drive children to casinos also think that DnD and its ilk are instruments of Beelzebub.
A standard set of RPG dice includes one four-sided die (“d4” in standard notation), a d6, a d8, one or two d10s, a d12, and a d20. Since the probability of any event (or set of events) occurring depends on the number of possible outcomes, RPG dice clearly allow for far more interesting situations than standard six-sided dice.
The standard definition for the probability of any given outcome is the number of ways that particular outcome can occur divided by the total number of possible outcomes. Since there is only one way for a coin to land heads up and there are only two possible ways that a coin can land, the probability of a coin landing heads up is 1/2. When calculating the probability of an outcome that consists of a specific set of different outcomes, we have to break out our possibility formula, as we will see shortly.
During our Castle Ravenloft adventure, the SigFig and I had to disarm a “trap” that could deal damage to our characters if they ended up in a certain region of the board. In order to disarm the trap, we had to roll a 10 or higher on a d20. Since there are 11 numbers on a d20 that are greater than or equal to 10, the probability of success was 11/20, or 55%. (It was probably sometime in junior high, where tests/quizzes were often out of twenty points, that I picked up on the quick trick of multiplying by five to convert a point total into a percentage.) However, the SigFig had a card that allowed any player to re-roll a die once. Since the card had to be discarded after it was used, we had to figure out if an extra roll increased the probability of success enough to justify using it on a trap-disarming attempt.
There are two possible ways to figure out the probability of success when an extra roll is available:
Method 1) Calculate the probability of failing on the first roll and succeeding on the second roll. Add this to the probability of succeeding on the first roll (which would negate the need for the second roll).
Method 2) Take advantage of the fact that all of the probabilities for a situation must add up to one. The probability of success is then simply one minus the probability that both rolls fail.
I opted for Method 2.
Here’s where our possibility formula comes in handy. There are nine ways for the first roll to fail and nine ways for the second roll to fail, resulting in 81 different double-failure possibilities. There are 20 possible outcomes for the first roll and 20 possible outcomes for the second roll, resulting in 400 different possible outcomes. The probability of both rolls failing is thus 81/400, or just over 20%; this means that the probability of success is 319/400, or just under 80%. Given how high this probability was, we chose to disarm the trap and ended up succeeding.
There are two important things to note here.
Note 1) The possibility formula we used is the version in which the different options are simply multiplied together. The more elaborate version discussed in these posts is not necessary because we don’t have to get rid of duplicate possibilities: the order of your fro-yo toppings or Dominion cards doesn’t matter, but getting a 1 on the first roll and a 2 on the second roll is NOT the same outcome as getting a 2 on the first roll and a 1 on the second roll.
Note 2) The same result would have been achieved had you multiplied the probability of the first roll failing (9/20) by the probability of the second roll failing (9/20). This isn’t a coincidence, and multiplying the individual probabilities together is indeed the standard way to calculate the probability of two independent events both occurring. However, this method works because it calculates the number of possible ways our desired events can occur and the total number of possible outcomes, and I wanted to make that connection clear.
Verifying that Method 1 produces the same results as Method 2 is left as an exercise for the reader.
I have ALWAYS wanted to say that.