What is the probability of rolling two six sided dice and obtaining at least one 3 Enter your answer as a fraction?

Well, the question is more complex than it seems at first glance, but you'll soon see that the answer isn't that scary! It's all about maths and statistics.

First of all, we have to determine what kind of dice roll probability we want to find. We can distinguish a few which you can find in this dice probability calculator.

Before we make any calculations, let's define some variables which are used in the formulas. n - the number of dice, s - the number of an individual die faces, p - the probability of rolling any value from a die, and P - the overall probability for the problem. There is a simple relationship - p = 1/s, so the probability of getting 7 on a 10 sided die is twice that of on a 20 sided die.

The probability of rolling the same value on each die - while the chance of getting a particular value on a single die is p, we only need to multiply this probability by itself as many times as the number of dice. In other words, the probability P equals p to the power n, or P = pⁿ = (1/s)ⁿ. If we consider three 20 sided dice, the chance of rolling 15 on each of them is: P = (1/20)³ = 0.000125 (or P = 1.25·10⁻⁴ in scientific notation). And if you are interested in rolling the set of any identical values, simply multiply the result by the total die faces: P = 0.000125 * 20 = 0.0025.
The probability of rolling all the values equal to or higher than y - the problem is similar to the previous one, but this time p is 1/s multiplied by all the possibilities which satisfy the initial condition. For example, let's say we have a regular die and y = 3. We want to rolled value to be either 6, 5, 4, or 3. The variable p is then 4 * 1/6 = 2/3, and the final probability is P = (2/3)ⁿ.
The probability of rolling all the values equal to or lower than y - this option is almost the same as the previous one, but this time we are interested only in numbers which are equal to or lower than our target. If we take identical conditions (s=6, y=3) and apply them in this example, we can see that the values 1, 2, & 3 satisfy the rules, and the probability is: P = (3 * 1/6)ⁿ = (1/2)ⁿ.
The probability of rolling exactly X same values (equal to y) out of the set - imagine you have a set of seven 12 sided dice, and you want to know the chance of getting exactly two 9s. It's somehow different than previously because only a part of the whole set has to match the conditions. This is where the binomial probability comes in handy. The binomial probability formula is:

P(X=r) = nCr * pʳ * (1-p)ⁿ⁻ʳ,

where r is the number of successes, and nCr is the number of combinations (also known as "n choose r").

In our example we have n = 7, p = 1/12, r = 2, nCr = 21, so the final result is: P(X=2) = 21 * (1/12)² * (11/12)⁵ = 0.09439, or P(X=2) = 9.439% as a percentage.

The probability of rolling at least X same values (equal to y) out of the set - the problem is very similar to the prior one, but this time the outcome is the sum of the probabilities for X=2,3,4,5,6,7. Moving to the numbers, we have: P = P(X=2) + P(X=3) + P(X=4) + P(X=5) + P(X=6) + P(X=7) = 0.11006 = 11.006%. As you may expect, the result is a little higher. Sometimes the precise wording of the problem will increase your chances of success.
The probability of rolling an exact sum r out of the set of n s-sided dice - the general formula is pretty complex:

$What is the probability of rolling two six sided dice and obtaining at least one 3 Enter your answer as a fraction?$

However, we can also try to evaluate this problem by hand. One approach is to find the total number of possible sums. With a pair of regular dice, we can have 2,3,4,5,6,7,8,9,10,11,12, but these results are not equivalent!

Take a look, there is only one way you can obtain 2: 1+1, but for 4 there are three different possibilities: 1+3, 2+2, 3+1, and for 12 there is, once again, only one variant: 6+6. It turns out that 7 is the most likely result with six possibilities: 1+6, 2+5, 3+4, 4+3, 5+2, 6+1. The number of permutations with repetitions in this set is 36. We can estimate the probabilities as the ratio of favorable outcomes to all possible outcomes: P(2) = 1/36, P(4) = 3/36 = 1/12, P(12) = 1/36, P(7) = 6/36 = 1/6.

The higher the number of dice, the closer the distribution function of sums gets to the normal distribution. As you may expect, as the number of dice and faces increases, the more time is consumed evaluating the outcome on a sheet of paper. Luckily, this isn't the case for our dice probability calculator!

The probability of rolling a sum out of the set, not lower than X - like the previous problem, we have to find all results which match the initial condition, and divide them by the number of all possibilities. Taking into account a set of three 10 sided dice, we want to obtain a sum at least equal to 27. As we can see, we have to add all permutations for 27, 28, 29, and 30, which are 10, 6, 3, and 1 respectively. In total, there are 20 good outcomes in 1,000 possibilities, so the final probability is: P(X ≥ 27) = 20 / 1,000 = 0.02.
The probability of rolling a sum out of the set, not higher than X - the procedure is precisely the same as for the prior task, but we have to add only sums below or equal to the target. Having the same set of dice as above, what is the chance of rolling at most 26? If you were to do it step by step, it would take ages to obtain the result (to sum all 26 sums). But, if you think about it, we have just worked out the complementary event in the previous problem. The total probability of complementary events is exactly 1, so the probability here is: P(X ≤ 26) = 1 - 0.02 = 0.98.

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

$What is the probability of rolling two six sided dice and obtaining at least one 3 Enter your answer as a fraction?$

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):

$What is the probability of rolling two six sided dice and obtaining at least one 3 Enter your answer as a fraction?$

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)

$What is the probability of rolling two six sided dice and obtaining at least one 3 Enter your answer as a fraction?$

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) = $\frac{4}{36}$

= $\frac{1}{9}$.

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