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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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Probability for rolling three dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each (three) dies. When three dice are thrown simultaneously/randomly, thus number of event can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its faces.Worked-out problems involving probability for rolling three dice: 1. Three dice are thrown together. Find the probability of: (i) getting a total of 5 (ii) getting a total of atmost 5 (iii) getting a total of at least 5. (iv) getting a total of 6. (v) getting a total of atmost 6. (vi) getting a total of at least 6. Solution: Three different dice are thrown at the same time. Therefore, total number of possible outcomes will be 63 = (6 × 6 × 6) = 216.(i) getting a total of 5: Number of events of getting a total of 5 = 6 i.e. (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (2, 1, 2) and (1, 2, 2) Therefore, probability of getting a total of 5 Number of favorable outcomesP(E1) = Total number of possible outcome = 6/216 = 1/36 (ii) getting a total of atmost 5: Number of events of getting a total of atmost 5 = 10 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1) and (1, 2, 2). Therefore, probability of getting a total of atmost 5 Number of favorable outcomesP(E2) = Total number of possible outcome = 10/216 = 5/108 (iii) getting a total of at least 5: Number of events of getting a total of less than 5 = 4 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1). Therefore, probability of getting a total of less than 5 Number of favorable outcomesP(E3) = Total number of possible outcome = 4/216 = 1/54 Therefore, probability of getting a total of at least 5 = 1 - P(getting a total of less than 5) = 1 - 1/54 = (54 - 1)/54 = 53/54 (iv)
getting a total of 6: Number of events of getting a total of 6 = 10 i.e. (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2). Therefore, probability of getting a total of 6 Number of favorable outcomesP(E4) = Total number of possible outcome = 10/216 = 5/108 (v) getting a total of atmost 6: Number of events of getting a total of atmost 6 = 20 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1), (1, 1, 3), (1, 3, 1), (3, 1, 1), (2, 2, 1), (1, 2, 2), (1, 1, 4), (1, 4, 1), (4, 1, 1), (1, 2, 3), (1, 3, 2), (2, 1, 3), (2, 3, 1), (3, 1, 2), (3, 2, 1) and (2, 2, 2). Therefore, probability of getting a total of atmost 6 Number of favorable outcomesP(E5) = Total number of possible outcome = 20/216 = 5/54 (vi) getting a total of at least 6: Number of events of getting a total of less than 6 (event of getting a total of 3, 4 or 5) = 10 i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1), (2, 1, 1) (1, 1, 3), (1, 3, 1), (3, 1, 1), (1, 2, 2), (2, 1, 2), (2, 2, 1). Therefore, probability of getting a total of less than 6 Number of favorable outcomesP(E6) = Total number of possible outcome = 10/216 = 5/108 Therefore, probability of getting a total of at least 6 = 1 - P(getting a total of less than 6) = 1 - 5/108 = (108 - 5)/108 = 103/108 These examples will help us to solve different types of problems based on probability for rolling three dice. Probability Probability Random Experiments Experimental Probability Events in Probability Empirical Probability Coin Toss Probability Probability of Tossing Two Coins Probability of Tossing Three Coins Complimentary Events Mutually Exclusive Events Mutually Non-Exclusive Events Conditional Probability Theoretical Probability Odds and Probability Playing Cards Probability Probability and Playing Cards Probability for Rolling Two Dice Solved Probability Problems Probability for Rolling Three Dice 9th Grade Math From Probability for Rolling Three Dice to HOME PAGE
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