What is the probability that the total of two dice will be greater than 9 Given that the first die is a 6?

Índice Show

Dice roll probability: 6 Sided Dice Example
Dice Roll Probability for 6 Sided Dice: Sample Spaces
Dice Rolling Probability: Steps
Two (6-sided) dice roll probability table
Dice Roll Probability Tables
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).

Contents:

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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%)

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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Probability for rolling two dice with the six sided dots such as 1, 2, 3, 4, 5 and 6 dots in each die.

When two dice are thrown simultaneously, thus number of event can be 62 = 36 because each die has 1 to 6 number on its faces. Then the possible outcomes are shown in the below table.

Probability – Sample space for two dice (outcomes):

Note:

(i) The outcomes (1, 1), (2, 2), (3, 3), (4, 4), (5, 5) and (6, 6) are called doublets.

(ii) The pair (1, 2) and (2, 1) are different outcomes.

Worked-out problems involving probability for rolling two dice:

1. Two dice are rolled. Let A, B, C be the events of getting a sum of 2, a sum of 3 and a sum of 4 respectively. Then, show that

(i) A is a simple event

(ii) B and C are compound events

(iii) A and B are mutually exclusive

Solution:

Clearly, we haveA = {(1, 1)}, B = {(1, 2), (2, 1)} and C = {(1, 3), (3, 1), (2, 2)}.

(i) Since A consists of a single sample point, it is a simple event.

(ii) Since both B and C contain more than one sample point, each one of them is a compound event.

(iii) Since A ∩ B = ∅, A and B are mutually exclusive.

2. Two dice are rolled. A is the event that the sum of the numbers shown on the two dice is 5, and B is the event that at least one of the dice shows up a 3. Are the two events (i) mutually exclusive, (ii) exhaustive? Give arguments in support of your answer.

Solution:

When two dice are rolled, we have n(S) = (6 × 6) = 36.

Now, A = {(1, 4), (2, 3), (4, 1), (3, 2)}, and

B = {(3, 1), (3, 2), (3, 3), (3, 4), (3, 5), (3, 6), (1,3), (2, 3), (4, 3), (5, 3), (6, 3)}

(i) A ∩ B = {(2, 3), (3, 2)} ≠ ∅.

Hence, A and B are not mutually exclusive.

(ii) Also, A ∪ B ≠ S.

Therefore, A and B are not exhaustive events.

More examples related to the questions on the probabilities for throwing two dice.

3. Two dice are thrown simultaneously. Find the probability of:

(i) getting six as a product

(ii) getting sum ≤ 3

(iii) getting sum ≤ 10

(iv) getting a doublet

(v) getting a sum of 8

(vi) getting sum divisible by 5

(vii) getting sum of atleast 11

(viii) getting a multiple of 3 as the sum

(ix) getting a total of atleast 10

(x) getting an even number as the sum

(xi) getting a prime number as the sum

(xii) getting a doublet of even numbers

(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die

Solution:

Two different dice are thrown simultaneously being number 1, 2, 3, 4, 5 and 6 on their faces. We know that in a single thrown of two different dice, the total number of possible outcomes is (6 × 6) = 36.

(i) getting six as a product:

Let E1 = event of getting six as a product. The number whose product is six will be E1 = [(1, 6), (2, 3), (3, 2), (6, 1)] = 4

Therefore, probability of getting ‘six as a product’

Number of favorable outcomes
P(E1) = Total number of possible outcome = 4/36 = 1/9

(ii) getting sum ≤ 3:

Let E2 = event of getting sum ≤ 3. The number whose sum ≤ 3 will be E2 = [(1, 1), (1, 2), (2, 1)] = 3

Therefore, probability of getting ‘sum ≤ 3’

Number of favorable outcomes
P(E2) = Total number of possible outcome = 3/36 = 1/12

(iii) getting sum ≤ 10:

Let E3 = event of getting sum ≤ 10. The number whose sum ≤ 10 will be E3 =

[(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),

(6, 1), (6, 2), (6, 3), (6, 4)] = 33

Therefore, probability of getting ‘sum ≤ 10’

Number of favorable outcomes
P(E3) = Total number of possible outcome = 33/36 = 11/12

(iv) getting a doublet: Let E4 = event of getting a doublet. The number which doublet will be E4 = [(1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)] = 6

Therefore, probability of getting ‘a doublet’

Number of favorable outcomes
P(E4) = Total number of possible outcome = 6/36 = 1/6

(v) getting a sum of 8:

Let E5 = event of getting a sum of 8. The number which is a sum of 8 will be E5 = [(2, 6), (3, 5), (4, 4), (5, 3), (6, 2)] = 5

Therefore, probability of getting ‘a sum of 8’

Number of favorable outcomes
P(E5) = Total number of possible outcome = 5/36

(vi) getting sum divisible by 5:

Let E6 = event of getting sum divisible by 5. The number whose sum divisible by 5 will be E6 = [(1, 4), (2, 3), (3, 2), (4, 1), (4, 6), (5, 5), (6, 4)] = 7

Therefore, probability of getting ‘sum divisible by 5’

Number of favorable outcomes
P(E6) = Total number of possible outcome = 7/36

(vii) getting sum of atleast 11:

Let E7 = event of getting sum of atleast 11. The events of the sum of atleast 11 will be E7 = [(5, 6), (6, 5), (6, 6)] = 3

Therefore, probability of getting ‘sum of atleast 11’

Number of favorable outcomes
P(E7) = Total number of possible outcome = 3/36 = 1/12

(viii) getting a multiple of 3 as the sum:

Let E8 = event of getting a multiple of 3 as the sum. The events of a multiple of 3 as the sum will be E8 = [(1, 2), (1, 5), (2, 1), (2, 4), (3, 3), (3, 6), (4, 2), (4, 5), (5, 1), (5, 4), (6, 3) (6, 6)] = 12

Therefore, probability of getting ‘a multiple of 3 as the sum’

Number of favorable outcomes
P(E8) = Total number of possible outcome = 12/36 = 1/3

(ix) getting a total of atleast 10:

Let E9 = event of getting a total of atleast 10. The events of a total of atleast 10 will be E9 = [(4, 6), (5, 5), (5, 6), (6, 4), (6, 5), (6, 6)] = 6

Therefore, probability of getting ‘a total of atleast 10’

Number of favorable outcomes
P(E9) = Total number of possible outcome = 6/36 = 1/6

(x) getting an even number as the sum:

Let E10 = event of getting an even number as the sum. The events of an even number as the sum will be E10 = [(1, 1), (1, 3), (1, 5), (2, 2), (2, 4), (2, 6), (3, 3), (3, 1), (3, 5), (4, 4), (4, 2), (4, 6), (5, 1), (5, 3), (5, 5), (6, 2), (6, 4), (6, 6)] = 18

Therefore, probability of getting ‘an even number as the sum

Number of favorable outcomes
P(E10) = Total number of possible outcome = 18/36 = 1/2

(xi) getting a prime number as the sum:

Let E11 = event of getting a prime number as the sum. The events of a prime number as the sum will be E11 = [(1, 1), (1, 2), (1, 4), (1, 6), (2, 1), (2, 3), (2, 5), (3, 2), (3, 4), (4, 1), (4, 3), (5, 2), (5, 6), (6, 1), (6, 5)] = 15

Therefore, probability of getting ‘a prime number as the sum’

Number of favorable outcomes
P(E11) = Total number of possible outcome = 15/36 = 5/12

(xii) getting a doublet of even numbers:

Let E12 = event of getting a doublet of even numbers. The events of a doublet of even numbers will be E12 = [(2, 2), (4, 4), (6, 6)] = 3

Therefore, probability of getting ‘a doublet of even numbers’

Number of favorable outcomes
P(E12) = Total number of possible outcome = 3/36 = 1/12

(xiii) getting a multiple of 2 on one die and a multiple of 3 on the other die:

Let E13 = event of getting a multiple of 2 on one die and a multiple of 3 on the other die. The events of a multiple of 2 on one die and a multiple of 3 on the other die will be E13 = [(2, 3), (2, 6), (3, 2), (3, 4), (3, 6), (4, 3), (4, 6), (6, 2), (6, 3), (6, 4), (6, 6)] = 11

Therefore, probability of getting ‘a multiple of 2 on one die and a multiple of 3 on the other die’

Number of favorable outcomes
P(E13) = Total number of possible outcome = 11/36

4. Two dice are thrown. Find (i) the odds in favour of getting the sum 5, and (ii) the odds against getting the sum 6.

Solution:

We know that in a single thrown of two die, the total number of possible outcomes is (6 × 6) = 36.

Let S be the sample space. Then, n(S) = 36.

(i) the odds in favour of getting the sum 5:

Let E1 be the event of getting the sum 5. Then,
E1 = {(1, 4), (2, 3), (3, 2), (4, 1)}
⇒ P(E1) = 4
Therefore, P(E1) = n(E1)/n(S) = 4/36 = 1/9
⇒ odds in favour of E1 = P(E1)/[1 – P(E1)] = (1/9)/(1 – 1/9) = 1/8.

(ii) the odds against getting the sum 6:

Let E2 be the event of getting the sum 6. Then,
E2 = {(1, 5), (2, 4), (3, 3), (4, 2), (5, 1)}
⇒ P(E2) = 5
Therefore, P(E2) = n(E2)/n(S) = 5/36
⇒ odds against E2 = [1 – P(E2)]/P(E2) = (1 – 5/36)/(5/36) = 31/5.

5. Two dice, one blue and one orange, are rolled simultaneously. Find the probability of getting

(i) equal numbers on both

(ii) two numbers appearing on them whose sum is 9.

Solution:

The possible outcomes 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)

Therefore, total number of possible outcomes = 36.

(i) Number of favourable outcomes for the event E

= number of outcomes having equal numbers on both dice

= 6 [namely, (1, 1), (2, 2), (3, 3), (4, 4), (5, 5), (6, 6)].

So, by definition, P(E) = \(\frac{6}{36}\)

= \(\frac{1}{6}\)

(ii) Number of favourable outcomes for the event F

= Number of outcomes in which two numbers appearing on them have the sum 9

= 4 [namely, (3, 6), (4, 5), (5, 4), (3, 6)].

Thus, by definition, P(F) = \(\frac{4}{36}\)

= \(\frac{1}{9}\).

These examples will help us to solve different types of problems based on probability for rolling two dice.

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