Probability For Dummies
Every so often someone asks a question like, "What's the probability of getting at least one 6 rolling two dice?" A common answer is 1/3, but this is not correct. The thinking here is that there's a 1/6 probability of rolling a 6 on each die, and 1/6 + 1/6 = 1/3. So I thought I'd talk about it.
Let's start with the basics.
The probability of any event occurring ranges from 0 (will not occur) to 1 (will occur). Probabilities can be expressed as a fraction, a decimal value, or a percent (which really just a decimal times 100).
The most common kinds of randomness in games are rolling dice and drawing cards from a shuffled deck. In both of these cases, the results are independent. This is a very important concept.
An independent event occurs without relation to any other event. For example, the results of any die roll are not affected by any past die rolls. This means that if you roll ten 6's in a row, the probability of rolling another 6 is still 1/6.
The other basic assumption is that the randomizing mechanism is fair. In the case of a 6-sided die, it is assumed to be balanced in such a way that the probability of each result is equal. This does not mean that if you roll 1,2,3,4,5 that the next roll will be a 6. It means that over a very large number of rolls, the number of each will be very close. The assumption also holds for cards. A well-shuffled deck implies that each card has an equal chance to be on the top.
To calculate the probability of a single fair event, divide the number of desired outcomes by the number of possible outcomes.
For example, the probability of rolling exactly a 3 on a d6 is 1 (one outcome in question) divided by 6 (six possible outcomes) making 1/6. This is probably intuitive for most people.
The probability of rolling an odd number is 3 (1, 3, 5) divided by 6, which is 3/6 or 1/2. This true of any 3 possible outcomes (rolling an even number, rolling 3 or less, rolling 2 3 or 5, etc.)
Now for more complex stuff.
The probability of multiple independent events ALL occurring is the product of their individual probabilities. However, you must be careful with the exact meaning of what you are looking for.
What is the probability of rolling two 6's on 2d6? The probabilities for each are 1/6. 1/6 * 1/6 = 1/36 (or 2.77%). Easy!
What is the probability of rolling 3 or more (both times) on 2d6? The probabilities for each are 4/6 (or 2/3). 2/3 * 2/3 = 4/9 (or 44.44%).
What is the probability of rolling a 6, then rolling an even number? 1/6 * 1/2 = 1/12 (or 8.33%).
What is the probability of rolling a 1 and a 5? This is the kind of question where you can easily go wrong. Do you mean rolling a 1 first, and rolling a 5 second? If so, it's 1/6 * 1/6 = 1/36 (or 2.77%). If you mean rolling two dice and having a 1-5 appear then what you really mean is rolling a 1 then a 5, or rolling a 5 then a 1. This leads us to:
The probability of one of several independent events occurring is the sum of their individual probabilities.
In the previous example, the probabilities of both 1-5 and 5-1 rolls are 1/36, so the probability of either of them occurring is 1/36 + 1/36 = 2/36 or 1/18 (or 5.55%).
What is the probability of rolling an odd/even pair on 2d6? This is a similar question. The probability of rolling odd then even is 1/2 * 1/2 = 1/4. The same for even-odd. So the total is 1/4 + 1/4 = 1/2 (or 50%).
Now let's kick it up a notch.
What is the probability of rolling 9 or more on 2d6? Rather than iterating all the probabilities for 9's, 10's 11's and 12's, you can do this by making a table:
You can see that 10 of the 36 results are 9 or more. And, as you guessed, the probability is 10/36 or 5/18 (or 27.77%).
What is the probability of rolling doubles (not triples) on 3d6? This one is tricky. Let's break it down by first examining case one: the first two rolls match and third does not. The first roll can be anything (probability 1). The second roll must match the first (probability 1/6). The third roll must not match (probability 5/6, do you see this?). So to total probability for case one is 1 * 1/6 * 5/6 = 5/36.
The second case is the same as the first one, but rearranged: any roll, no match, match first roll. Again we have 1 * 5/6 * 1/6 = 5/36. Likewise with the third case: anything, no match, match second roll = 1 * 5/6 * 1/6 = 5/36.
The final answer is the sum of the three separate cases: 5/36 + 5/36 + 5/36 = 15/36 (41.66%).
What is the probability of rolling three completely different numbers on 3d6? This one should be easy by now. The first number can be anything (1), the second must be anything but the first (5/6), and the third must be anything but the first two (4/6 or 2/3), resulting in 1 * 5/6 * 2/3 = 10/18 or 5/9 (or 55.55%).
Interesting note. Look at the previous two examples. One is for none of the three rolls to match, and other is for two of the rolls to match. The only other possibility is 3-of-a-kind. 1 (36/36) - 5/9 (20/36) - 15/36 = 1/36, which is the answer! The sum of the probabilities of all possible outcomes is 1.
What is the probability of drawing 2 aces from the top of a shuffled standard deck of cards? In this case, the second drawn card is going to be affected by the first drawn card, because the number of cards remaining in the draw pile has been changed. There are 4 aces in the deck and 52 cards. The probability of the first card drawn being an ace is therefore 4/52 or 1/13. Now what's left? There are 3 possible aces out of 51 cards: 3/51 or 1/17. So the answer is 1/13 * 1/17 = 1/221 (or 0.45%).
Now back to the original question. What's the probability of getting at least one 6 rolling two dice? Can you do it? There are two cases: 6-anything, and anything-6. However, notice that these two cases overlap. To make them distinct, we need to change the wording a bit: 6-anything, and not6-6. Case one is 1/6 * 1 = 1/6. Case two is 5/6 * 1/6 = 5/36. The final answer is 1/6 (6/36) + 5/36 = 11/36 (or 30.55%)!