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 Contents: Watch the video for three examples: Probability: Dice Rolling Examples Watch this video on YouTube. Can’t see the video? Click here. Need help with a homework question? Check out our tutoring page! Dice roll probability: 6 Sided Dice ExampleIt’s very common to find questions about dice rolling in probability and statistics. You might be asked the probability of rolling a variety of results for a 6 Sided Dice: five and a seven, a double twelve, or a double-six. While you *could* technically use a formula or two (like a combinations formula), you really have to understand each number that goes into the formula; and that’s not always simple. By far the easiest (visual) way to solve these types of problems (ones that involve finding the probability of rolling a certain combination or set of numbers) is by writing out a sample space. Dice Roll Probability for 6 Sided Dice: Sample SpacesA sample space is just the set of all possible results. In simple terms, you have to figure out every possibility for what might happen. With dice rolling, your sample space is going to be every possible dice roll. Example question: What is the probability of rolling a 4 or 7 for two 6 sided dice? In order to know what the odds are of rolling a 4 or a 7 from a set of two dice, you first need to find out all the possible combinations. You could roll a double one [1][1], or a one and a two [1][2]. In fact, there are 36 possible combinations. Dice Rolling Probability: StepsStep 1: Write out your sample space (i.e. all of the possible results). For two dice, the 36 different possibilities are: [1][1], [1][2], [1][3], [1][4], [1][5], [1][6], [2][1], [2][2], [2][3], [2][4], [2][5], [2][6], [3][1], [3][2], [3][3], [3][4], [3][5], [3][6], [4][1], [4][2], [4][3], [4][4], [4][5], [4][6], [5][1], [5][2], [5][3], [5][4], [5][5], [5][6], [6][1], [6][2], [6][3], [6][4], [6][5], [6][6]. Step 2: Look at your sample space and find how many add up to 4 or 7 (because we’re looking for the probability of rolling one of those numbers). The rolls that add up to 4 or 7 are in bold: [1][1], [1][2], [1][3], [1][4], [1][5], [1][6], There are 9 possible combinations. Step 3: Take the answer from step 2, and divide it by the size of your total sample space from step 1. What I mean by the “size of your sample space” is just all of the possible combinations you listed. In this case, Step 1 had 36 possibilities, so: 9 / 36 = .25 You’re done! Two (6-sided) dice roll probability tableThe following table shows the probabilities for rolling a certain number with a two-dice roll. If you want the probabilities of rolling a set of numbers (e.g. a 4 and 7, or 5 and 6), add the probabilities from the table together. For example, if you wanted to know the probability of rolling a 4, or a 7:
Probability of rolling a certain number or less for two 6-sided dice.
Dice Roll Probability TablesContents: Probability of a certain number with a Single Die.
Probability of rolling a certain number or less with one die.
Probability of rolling less than certain number with one die.
Probability of rolling a certain number or more.
Probability of rolling more than a certain number (e.g. roll more than a 5).
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Visit out our statistics YouTube channel for hundreds of probability and statistics help videos! ReferencesDodge, Y. (2008). The Concise Encyclopedia of Statistics. Springer.
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Assuming the die is just marked even and odd rather than with numbers, there are eight orderings. They range from all odd to all even: OOO, OOE, OEO, EOO, OEE, EOE, EEO, EEE. In your formula: $$\frac{n!}{(n-r)!}$$ You are missing that you don't care about the orders of the duplicates. So you need $$\frac{n!}{(n-r)!r!}$$ The $n$ is the total number of rolls, the $r$ is the number of either evens or odds (it's symmetric). But to get the total number of orderings, you need to add these: $$\sum\limits_{r=0}^n\frac{n!}{(n-r)!r!}$$ Now substitute 3 for $n$. $$\sum\limits_{r=0}^3\frac{3!}{(3-r)!r!}$$ Unrolling that, we get $$\frac{3!}{3!0!} + \frac{3!}{2!1!} + \frac{3!}{1!2!} + \frac{3!}{0!3!}$$ or $$1 + 3 + 3 + 1$$ So we have one all odds, three with one even, three with two evens, and one all even. That's eight total. Another way of thinking of this is that there is only one ordering of all odd numbers or all even numbers while there are three places where the lone odd or even number can be. Of those eight, how many fit your parameters? Exactly one, OOE. So one in eight or $\frac{1}{8}$. As others have already noted, you could get that much more easily by simply figuring that you have a one in two chance of getting the result you need for each roll. There's three rolls, so $(\frac{1}{2})^3 = \frac{1}{8}$. If you want to treat 1, 3, and 5 as different values and 2, 4, and 6 as different values, you can. But it is much easier to think of them as just odd or even. Because you don't want to try write this out for $6^3 = 216$ orderings. And in the end, you will get the same basic result. You will have twenty-seven OOE orderings, which is again one eighth of the total. This is because there are three possible values for each, 1, 3, and 5 for the two odds and 2, 4, and 6 for the even. And $\frac{27}{216} = \frac{1}{8}$. Permutations leads you down a harder path. It's easier to think just in terms of probability or even ordering. |
If a die is rolled, what is the probability of getting an even number or a number greater than 3
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