Three dice are rolled together the probability that different numbers will appear on each of them is


Three dice are rolled together the probability that different numbers will appear on each of them is

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Probability: Dice Rolling Examples

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Dice roll probability: 6 Sided Dice Example

It’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 Spaces

A 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: Steps

Step 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],
[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].

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!
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Two (6-sided) dice roll probability table

The 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:
3/36 + 6/36 = 9/36.

Roll a… Probability
2 1/36 (2.778%)
3 2/36 (5.556%)
4 3/36 (8.333%)
5 4/36 (11.111%)
6 5/36 (13.889%)
7 6/36 (16.667%)
8 5/36 (13.889%)
9 4/36 (11.111%)
10 3/36 (8.333%)
11 2/36 (5.556%)
12 1/36 (2.778%)

Probability of rolling a certain number or less for two 6-sided dice.

Roll a… Probability
2 1/36 (2.778%)
3 3/36 (8.333%)
4 6/36 (16.667%)
5 10/36 (27.778%)
6 15/36 (41.667%)
7 21/36 (58.333%)
8 26/36 (72.222%)
9 30/36 (83.333%)
10 33/36 (91.667%)
11 35/36 (97.222%)
12 36/36 (100%)

Dice Roll Probability Tables

Contents:
1. Probability of a certain number (e.g. roll a 5).
2. Probability of rolling a certain number or less (e.g. roll a 5 or less).
3. Probability of rolling less than a certain number (e.g. roll less than a 5).
4. Probability of rolling a certain number or more (e.g. roll a 5 or more).
5. Probability of rolling more than a certain number (e.g. roll more than a 5).

Probability of a certain number with a Single Die.

Roll a… Probability
1 1/6 (16.667%)
2 1/6 (16.667%)
3 1/6 (16.667%)
4 1/6 (16.667%)
5 1/6 (16.667%)
6 1/6 (16.667%)

Probability of rolling a certain number or less with one die

.

Roll a…or less Probability
1 1/6 (16.667%)
2 2/6 (33.333%)
3 3/6 (50.000%)
4 4/6 (66.667%)
5 5/6 (83.333%)
6 6/6 (100%)

Probability of rolling less than certain number with one die

.

Roll less than a… Probability
1 0/6 (0%)
2 1/6 (16.667%)
3 2/6 (33.33%)
4 3/6 (50%)
5 4/6 (66.667%)
6 5/6 (83.33%)

Probability of rolling a certain number or more.

Roll a…or more Probability
1 6/6(100%)
2 5/6 (83.333%)
3 4/6 (66.667%)
4 3/6 (50%)
5 2/6 (33.333%)
6 1/6 (16.667%)

Probability of rolling more than a certain number (e.g. roll more than a 5).

Roll more than a… Probability
1 5/6(83.33%)
2 4/6 (66.67%)
3 3/6 (50%)
4 4/6 (66.667%)
5 1/6 (66.67%)
6 0/6 (0%)

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Like the explanation? Check out our Practically Cheating Statistics Handbook for hundreds more solved problems.

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References

Dodge, Y. (2008). The Concise Encyclopedia of Statistics. Springer.
Gonick, L. (1993). The Cartoon Guide to Statistics. HarperPerennial.
Salkind, N. (2016). Statistics for People Who (Think They) Hate Statistics: Using Microsoft Excel 4th Edition.

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Three dice are rolled together the probability that different numbers will appear on each of them is
Three dice are rolled together the probability that different numbers will appear on each of them is

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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 outcomes
P(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 outcomes
P(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 outcomes
P(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 outcomes
P(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 outcomes
P(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 outcomes
P(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.

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