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There are $7776$ possible outcomes of rolling $5$ six-sided dice, of this total there are $651$ possible outcomes where the $5$ dice equal exactly $15$.
How could you calculate the number of possible outcomes where the top three dice equal $15$?
$\endgroup$ 1Probability is also known as a possibility. This means math of chance, that trade in the happening of a likely event. The value is designated from zero to one. In math, Probability has been shown to guess how likely events are to occur. Basically, the probability is the scope to which something is to be anticipated to happen.
Probability
To understand probability more correctly, take an example as rolling a dice, the possible outcomes are 1, 2, 3, 4, 5, and 6. The possibility of occurring any of the favorable outcomes is 1/6. As the probability of occurring any of the events is the same so there are similar chances of getting any likely number, in this case, it is either 1/6 or 50/3.
Formula of Probability
Probability of an event = {Number of favourable events } ⁄ {number of total events}
P(A) = {Number of ways A occurs} ⁄ {Total number of events}
DICE
Dice is a small block that has between one and six marks or dots on its edge and is used in games to give a random figure. Dice are small, tossable blocks with marked sides that can pause in several figures. They are used to give rise to random figures, often as part of sideboard games, as well as dice games, board games, role-playing games, and games of chance.
A customary die is a cube with each of its six faces noticeable with a different number of figures from one to six. When tossable or rolled, the die comes to pause appear a random number from one to six on its higher side, with the occurrence of each event being equally likely. Dice may also have concave or uneven shapes and may have faces marked with figures or characters instead of pips. Loaded dice are drawn to service some results over others for escape or entertainment.
Solution:
Possibility for throwing six sided three dice will be 1, 2, 3, 4, 5 and 6 dots in each (three) dies.
Three dice are roll at the same time, number of attainable outcome can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.
Now, let’s consider the possible sums from rolling three dice. The smallest attainable sum occur when all of the dice are the smallest, or one each. This gives a sum of three when we are throwing three dice i.e., (1, 1, 1). The greatest number on a die is six, which means that the greatest possible sum happens when all three dice are sixes is 18 i.e., (6, 6, 6). When n dice are rolled, the minimal attainable sum is n and the greatest attainable sum is 6n. There is only one way when three dice can total 3,
- 3 ways for 4
- 6 for 5
- 10 for 6
- 15 for 7
- 21 for 8
- 25 for 9
- 27 for 10
- 27 for 11
- 25 for 12
- 21 for 13
- 15 for 14
- 10 for 15
- 6 for 16
- 3 for 17
- 1 for 18
Forming Sums
As shown above, for three dice the possible sums include every number from three to 18. Calculate probability by using add up plan and acknowledging that we are considering ways to separate a integer into absolutely three whole numbers. For example, to obtain a sum of three there is only one way i.e., 3 = 1 + 1 + 1. Since each die is individualistic from the others, a sum such as four can be acquired in three different ways:
- 1 + 1 + 2
- 1 + 2 + 1
- 2 + 1 + 1
Forward add up arguments can be used to find the number of ways of creating the other sums. The partitions for each sum follow:
- 3 = 1 + 1 + 1
- 4 = 1 + 1 + 2
- 5 = 1 + 1 + 3 = 2 + 2 + 1
- 6 = 1 + 1 + 4 = 1 + 2 + 3 = 2 + 2 + 2
- 7 = 1 + 1 + 5 = 2 + 2 + 3 = 3 + 3 + 1 = 1 + 2 + 4
- 8 = 1 + 1 + 6 = 2 + 3 + 3 = 4 + 3 + 1 = 1 + 2 + 5 = 2 + 2 + 4
- 9 = 6 + 2 + 1 = 4 + 3 + 2 = 3 + 3 + 3 = 2 + 2 + 5 = 1 + 3 + 5 = 1 + 4 + 4
- 10 = 6 + 3 + 1 = 6 + 2 + 2 = 5 + 3 + 2 = 4 + 4 + 2 = 4 + 3 + 3 = 1 + 4 + 5
- 11 = 6 + 4 + 1 = 1 + 5 + 5 = 5 + 4 + 2 = 3 + 3 + 5 = 4 + 3 + 4 = 6 + 3 + 2
- 12 = 6 + 5 + 1 = 4 + 3 + 5 = 4 + 4 + 4 = 5 + 2 + 5 = 6 + 4 + 2 = 6 + 3 + 3
- 13 = 6 + 6 + 1 = 5 + 4 + 4 = 3 + 4 + 6 = 6 + 5 + 2 = 5 + 5 + 3
- 14 = 6 + 6 + 2 = 5 + 5 + 4 = 4 + 4 + 6 = 6 + 5 + 3
- 15 = 6 + 6 + 3 = 6 + 5 + 4 = 5 + 5 + 5
- 16 = 6 + 6 + 4 = 5 + 5 + 6
- 17 = 6 + 6 + 5
- 18 = 6 + 6 + 6
Specific Probabilities
Probability of an event = {Number of favourable events } ⁄ {number of total events},or 216. The results are:
- Possibility of getting a sum of 3: 1/216 = 0.0046 × 100= 0.5%
- Possibility of getting a sum of 4: 3/216 = 0.0138 × 100 = 1.4%
- Possibility of getting a sum of 5: 6/216 = 0.0277 × 100 = 2.8%
- Possibility of getting a sum of 6: 10/216 = 0.0462 × 100 = 4.6%
- Possibility of getting a sum of 7: 15/216 = 0.069 × 100 = 7.0%
- Possibility of getting a sum of 8: 21/216 = 0.097 × 100 = 9.7%
- Possibility of getting a sum of 9: 25/216 = 0.115 × 100 = 11.6%
- Possibility of getting a sum of 10: 27/216 = 0.125 × 100 = 12.5%
- Possibility of getting a sum of 11: 27/216 = 0.125 × 100 = 12.5%
- Possibility of getting a sum of 12: 25/216 = 0.115 × 100 = 11.6%
- Possibility of getting a sum of 13: 21/216 = 0.097 × 100 = 9.7%
- Possibility of getting a sum of 14: 15/216 =0.069 × 100 = 7.0%
- Possibility of getting a sum of 15: 10/216 = 0.0462 × 100 = 4.6%
- Possibility of getting a sum of 16: 6/216 = 0.0277 × 100 = 2.8%
- Possibility of getting a sum of 17: 3/216 = 0.013 × 100 = 1.4%
- Possibility of getting a sum of 18: 1/216 = 0.0046 × 100 = 0.5%
As can be seen, the extreme values of 3 and 18 are least probable. The sums which are in the middle are most probable.
Sample Problems
Question 1: Three dice are thrown together. Find the probability of getting a total of 5.
Solution:
Three dice are roll at the same time, number of attainable outcome can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.
Number of favourable 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)
So, probability of occurring a total of 5
P(A) = {Number of favourable events } ⁄ {number of total events}
= 6/216
= 1/36
Question 2: Three dice are thrown together. Find the probability of getting a total of almost 5.
Solution:
Three dice are roll at the same time, number of attainable outcome can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.
Number of affair of occurring 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, possibility of happening a total of atmost 5
P(E) = {Number of favourable events } ⁄ {number of total events}
= 10/216
= 5/108
Question 3: Three dice are thrown together. Find the probability of getting a total of at least 5.
Solution:
Three dice are roll at the same time, number of attainable outcome can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.
Number of affair of happening a total of less than 5 = 4
i.e. (1, 1, 1), (1, 1, 2), (1, 2, 1) and (2, 1, 1).
Therefore, possibility of happening a total of less than 5
P(E) = {Number of favourable events } ⁄ {number of total events}
= 4/216
= 1/54
Therefore, possibility of occurring a total of at least 5 = 1 – P (occurring a total of less than 5)
= 1 – 1/54
= (54 – 1)/54
= 53/54
Question 4: Three dice are thrown together. Find the probability of getting a total of 6.
Solution:
Three dice are roll at the same time, number of attainable outcome can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.
Number of affair of happening 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, possibility of happening a total of 6
P(E) = {Number of favourable events } ⁄ {number of total events}
= 10/216
= 5/108
Question 5: Three dice are thrown together. Find the probability of getting a total of 8.
Solution:
Three dice are roll at the same time, number of attainable outcome can be 63 = (6 × 6 × 6) = 216 because each die has 1 to 6 number on its dots.
Number of affair of happening a total of 8 = 21
i.e. (2, 2, 4), (4, 2, 2), (2, 4, 2), (1, 5, 2), (2, 5, 1), (5, 1, 2), (2, 1, 5), (1, 2, 5), (3, 3, 2), (3, 2, 3), (2, 3, 3) and so on…
Therefore, possibility of happening a total of 8
P(E) = {Number of favourable events } ⁄ {number of total events}
= 21/216
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.
| 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.
| 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.
| 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
.
| 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
.
| 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.
| 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).
| 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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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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