How to calculate Dice Probabilities?
Probability is the chance or likelihood of an event happening. It is represented by a number between 0 and 1. The higher the probability, the greater the chances of the event occurring.
- A probability of 0 means the event is impossible and cannot happen.
- A probability of 1 means the event is certain to happen.
- The probability value can never exceed 1.
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Probability Formula
The Formula for Probability is:
P(A)=\dfrac{Number \ of \ favorable \ outcomes}{Total \ number \ of \ possible \ outcomes}
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Here, we will discuss how to calculate the probabilities for rolling one die and two dice.
DICE
Dice are small, throwable objects used in games of chance, usually with six faces numbered 1 to 6. In probability, dice are used to calculate the chances of getting specific outcomes, like a certain number or a particular sum when rolling one or more dice.
One Die Roll
The simplest case of dice probability involves calculating the chance of rolling a specific number on a fair die. A standard die has six faces, numbered from 1 to 6, with each outcome being equally likely.
The basic formula is:
P(A) = {Number of favourable outcomes / Total number of possible outcomes}
For example, the probability of rolling a specific number, say a 1, is:
- P(1) = 1/6
This probability is the same for each of the numbers 1, 2, 3, 4, 5, or 6, since each face of the die is equally likely to land face up. Therefore, the probability of rolling any particular number between 1 and 6 is always:
P(n) = 1/6
Note:-
To express a probability as a percentage, multiply it by 100. For example, the probability of rolling a 6 on a fair six-sided die is 1/6 or approximately 0.167. When multiplied by 100, this gives about 16.7%.
Two or More Dice
When rolling two dice, the probability calculation becomes slightly more complex. For example, the probability of rolling two sixes involves multiplying the individual probabilities.
When rolling two standard dice, the total number of possible outcomes becomes 36. The total possible outcome sample space is given through the table below:
| Dice B ⇨ Dice A ⇩ | 1 | 2 | 3 | 4 | 5 | 6 |
|---|---|---|---|---|---|---|
| 1 | (1,1) | (1,2) | (1,3) | (1,4) | (1,5) | (1,6) |
| 2 | (2,1) | (2,2) | (2,3) | (2,4) | (2,5) | (2,6) |
| 3 | (3,1) | (3,2) | (3,3) | (3,4) | (3,5) | (3,6) |
| 4 | (4,1) | (4,2) | (4,3) | (4,4) | (4,5) | (4,6) |
| 5 | (5,1) | (5,2) | (5,3) | (5,4) | (5,5) | (5,6) |
| 6 | (6,1) | (6,2) | (6,3) | (6,4) | (6,5) | (6,6) |
Unconventional probabilities have the rule that one must multiply the individual probabilities jointly to attain the outcome. Therefore, the formula for this is,
Probability of both = Probability of result one × Probability of result two
For Example P(Two 6s) = P(6 on first die) x P(on second die) = 1/6 x 1/6 = 1/36
- Total Score from Two or More Dice
If you want to find the probability of getting a specific total(e.g, sum of 4) when rolling two dice, you need to count all possible combinations that give the total:
For example: to get sum as 4, the following combination are possible:
- (1,3)
- (2,2)
- (3,1)
So, there are 3 favourable outcomes out of 36 total possible outcomes(since each die has 6 faces:6 x 6 = 36).
(Sum of 4) = 3/36 = 1/12
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Similar Problems on Dice Probabilities
Question 1: Find the probability of retrieving a sum of 8 on throwing two dice.
Solution:
There are 36 total likely results on throwing two dice i.e., 6² = 6 × 6 = 36.
There are 5 total possibility of retrieving a sum of 8 on throwing two dice i.e., (2, 6), (3, 5), (4, 4), (5, 3), (6, 2).
Hence, the probability of retrieve a sum of 8 on throwing two dice is 5/36.
Question 2: Shawn tosses a die 400 times, and he documents the score of getting 6 30 times. What could be the probability of
a) retrieving a score of 6?
b) retrieving a score under 6?
Solution:
a) P (getting a score of 6)
= Number of times getting 6/total times
= 30/400
= 3/40b) P (getting a score under 6)
= number of times getting under 6/total times
= 370/400
= 37/40
Question 3: What is the probability of retrieving a sum of 6 if two dice are thrown?
Solution:
When two dice are rolled, n(S) = 36. Let, A be the event of getting a sum of 6. Then,
A = {(3, 3), (2, 4), (4, 2), (1, 5), (5, 1)}
n(A) = 5Hence, the required probability will be, P(A) = n(A)/n(S) = 5/36.
Question 4: Find the probability of throwing two dice and retrieving a sum of 4.
Solution:
The set of possible outcomes when we roll a die are {1, 2, 3, 4, 5, 6}
So, when two dice are rolled, there are 6 × 6 = 36 chances.When we roll two dice, the probability of retrieving number 4 is (1, 3), (2, 2), and (3, 1).
So, the number of favorable outcomes = 3Total number of possibilities = 36
Probability = {Number of likely affair } ⁄ {Total number of affair} = 3 / 36 = 1/12.Thus, 1/12 is the probability of rolling two dice and retrieving a sum of 4.
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