What is the probability that you roll a 5 on at least one of the dice?

Dice and the Laws of Probability

Edward D. Collins

For more questions and problems regarding dice (and coin) probabilities, please see this page.

Let's imagine you are playing a game which uses dice. You are about to roll three of them. You NEED to roll at least one 6. A 6 appearing on any one (or more) of the three dice will win the game for you! What are your chances?

33.3 %
42.1 %
50 %
66.6 %

Quite some time ago, I was over at a friend's house watching him and another friend play a board game called Axis & Allies. At one point this exact scenario came up - Kent was planning on rolling three dice and really wanted at least one 6 to appear. He made a comment that with three dice, his chances were 3/6 or 50%.

Kent's reasoning was, with one die, the chances of rolling a 6 were 1/6 which is correct. He also believed if he were to roll two dice, his chances were double this or 2/6. This is INCORRECT and this is where his faulty reasoning begins.

Knowing a little bit about the laws of probability, I quickly knew the fraction "2/6" for two dice and "3/6" for three dice was incorrect and spent a brief moment computing and then explaining the true percentages.

Unfortunately, I do not believe I was successful in explaining to Kent why my figures were correct. Maybe I can do so here.

Obviously, with Kent's logic above, if the chances of rolling a 6 with two dice is 2/6 and the chances of rolling a 6 with three dice is 3/6, then the chances of rolling a 6 with six dice would be 6/6 !! 100%?? Of course, this is obviously incorrect. I don't care how many dice you roll, the chances of rolling a 6 will never be 100%.

When you roll just one die, there are six different ways the die can land, as shown by the following graphic:

When two dice are rolled, there are now 36 different and unique ways the dice can come up. This figure is arrived at by multiplying the number of ways the first die can come up (six) by the number of ways the second die can come up (six). 6 x 6 = 36.

This graphic shows this very nicely. I've used two different colored dies to help show a roll of 2-1 is different from a roll of 1-2.

If you use the above graphic and count the number of times is 6 appears when two dice are rolled, you will see the answer is eleven. Eleven times out of 36 or 30.5 %, slightly less than the 33.3% (2/6) Kent thought. When you roll two dice, you have a 30.5 % chance at least one 6 will appear.

This figure can also be figured out mathematically, without the use of the graphic. One way to do so is to take the number of ways a single die will NOT show a 6 when rolled (five) and multiply this by the number of ways the second die will NOT show a 6 when rolled. (Also five.) 5 x 5 = 25. Subtract this from the total number of ways two dice can appear (36) and we have our answer...eleven.

So, let's use this same method to answer our question and determine the chances of at least one 6 appearing when three dice are rolled.

Take the chances of a 6 NOT appearing on the first die...

5 / 6

and multiply this by the chances of a 6 NOT appearing on the second die...

5 / 6 x 5 / 6 = 25 / 36

and multiply this by the chances of a 6 NOT appearing on the third die...

25 / 36 x 5 / 6 = 125 / 216

So, there are 125 out of 216 chances of a 6 NOT appearing when three dice are rolled. Simply subtract 125 from 216 which will give us the chances a 6 WILL appear when three dice are rolled, which is 91. 91 out of 216 or 42.1 %, not quite the 50% Kent originally thought.

Here is a table showing the fractions and percentages of a 6 appearing (or any other single digit for that matter) and not appearing with several different numbers of dice:


...with just one die

Notice you have just over a 50 % chance (51.77) of rolling at least one 6 when rolling four dice. With six dice it's just over 66 %. Also notice even when you roll nine dice, rolling a 6 is far from a certainty - you have just a little better than 80 % chance of rolling this 6. This means, on the average, even when you roll nine dice, two times out of ten you will not roll a single 6!

Hope this helped clear it up, Kent!

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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