Two dice are thrown together 4 times. the probability that both dice will show same numbers twice is


Two dice are thrown together 4 times. the probability that both dice will show same numbers twice is

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%)

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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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In other words, probability is known as a possibility. It is a math of prospect, that deals with the happening of a random event. The value is shown from zero to one. In mathematics, Probability has been described to guess how likely happening events are to occur. The meaning of possibility is primarily the scope to which something is to be expected to occur.

Probability

To understand probability more accurately, 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 equally likely events is 1/6. As the probability of happening any of a possible event is the same so there are equal chances of getting any likely number in this case it is either 1/6 or 50/3.

Formula of Probability

Probability of an equally likely event = Number of favorable outcome/Total number of possible outcome 

P(A) = {Number of ways A occurs} ⁄ {Total number of outcomes}

There are unlikely types of events based on an unlike basis. One type is likely event and complimentary event. Then there are impossible and sure events. One type is a simple and compound event. There are independent and dependent events, mutually exclusive, exhaustive events, etc. Let’s understand these events in detail.

  • Equally Likely Events: After rolling a dice, the possibility of getting any of the equally likely events is 1/6. As the happening of an event is an equally likely event so there is equal or the same possibility of obtaining any number, in this case, it is either 1/6 in fair dice rolling.
  • Complementary Events: There is a chance or possibility of only two results which is an event will occur or not. Like a person will study or not study, cleaning a car or not cleaning a car, etc. are examples of complementary events.
  • Impossible and Sure Events: If the possibility of happening any of an equally likely event is 0, such an event is called an impossible event and if the possibility of happening any of an equally likely event is 1, it is called a sure event. In other words, the empty set ϕ is an impossible event and the sample space S is a sure event.
  • Simple Events: Any happening carrying a single point of the sample space is known as a simple event in probability. For example, if S = {46 , 75 , 86 , 64 , 99} and E = {75} then E is a simple event.
  • Compound Events: Opposite to the simple event, if any event contains more than one single point of the sample space then such an event is called a compound event. Considering the  example, if S = {56, 78, 96, 54, 89}, E1 = {56, 54 }, E2 = {78, 56, 89} then, E1 and E2 represent two compound events.
  • Independent Events and Dependent Events: If the possibility of any happening is completely unaffected by the possibility of any other happening, such events are known as independent events in probability the happening which is affected by other happening is known as dependent events.
  • Mutually Exclusive Events: If the occurrence of one event keeps out the occurrence of another event, such events are mutually exclusive events i.e. two events don’t have any common number. For example, if S = {4,5,6,7,8,9} and E1, E2 are two happening such that E1 consists of numbers less than 7 and E2 consists of numbers greater than 8. So, E1 = {4,5,6,7} and E2 = {8,9} . Then, E1 and E2 are mutually exclusive.
  • Exhaustive Events: A set of events is called exhaustive, which describes that one of them must happen.
  • Events Associated with “OR”: If two happening E1 and E2 are related with OR then it means that either E1 or E2 or both. The merging symbol (∪) is used to show OR in probability. Thus, the event E1 U E2 shows E1 OR E2. If we have mutually exhaustive events E1, E2, E3 …En related with sample space S then, E1 U E2 U E3 U … En = S
  • Events Associated with “AND”: If two happening E1 and E2 are related with AND then it describes that the joining of elements that is common to both the events. The intersection symbol (∩) is used to show AND in probability. Thus, the event E1 ∩ E2 shows E1 and E2.
  • Event E1 but not E2: It shows the difference between two of the happening. Event E1 but not E2 represent all the end results that are present in E1 but not in E2. Thus, the event E1 but not E2 is shown as E1, E2 = E1 – E2

Solution:

Dice Roll Probability

Total of two dice

Score

2

2.78%

3

5.56%

4

8.33%

5

11.11%

6

13.89%

7

16.67%

8

13.89%

9

11.11%

10

8.33%

11

5.56%

12

2.78%

There’s only one amalgamation that give a total of 2—when each die shows a 1. Likewise, there is only one amalgamation that give a total of 12—when each die shows a 6. They are the least possible likely amalgamation to take place.

As you can see, 7 is the most common roll with two six-sided dice. There are six times more likely chances to roll a 7 than a 2 or a 12, which is a huge difference. There are twice as likely chances to roll a 7 than a 4 or a 10. However, it’s only 1.2 times more likely chances to roll a 7 than a 6 or an 8.

Therefore, 7 is the most likely score from throwing two dice.

Similar Questions

Question 1: Two dice are thrown simultaneously. Find the probability of, getting six as a product?

Answer:

Two different dice are thrown simultaneously the possible happening events are 1, 2, 3, 4, 5 and 6. The total number of favorable outcomes is (6 × 6) = 36.

Let E1 = possibility of occurring four as a product. The number whose product is four will be 

E1 = [(3, 2), (2, 3), (1, 6), (6,1)] = 4

Therefore, probability of getting ‘four as a product’

P(E1) =  Number of favorable outcome/Total number of possible outcome 

         = 4/36

         = 1/9

Question 2: Two dice are thrown simultaneously. Find the probability of, getting a sum ≤ 3?

Answer:-

Two different dice are thrown simultaneously the possible happening events are 1, 2, 3, 4, 5 and 6. The total number of favorable outcomes is (6 × 6) = 36.

Let A = event of getting sum ≤ 3. 

The number whose sum ≤ 3 will be A = [(1, 1), (1, 2), (2, 1)] = 3

Therefore, probability of getting ‘sum ≤ 3’

P(A) = Number of favorable outcome/Total number of possible outcome

        = 3/36

        = 1/12

Question 3: Two dice were thrown simultaneously. Then find the probability of getting the product of outcomes multiple of 4.

Solution:

Two different dice are thrown simultaneously the possible happening events are 1, 2, 3, 4, 5 and 6. The total number of favorable outcomes is (6 × 6) = 36.

They are :  

(1,1),(2,1),(3,1),(4,1),(5,1),(6,1),

(1,2),(2,2),(3,2),(4,2),(5,2),(6,2),

(1,3),(2,3),(3,3),(4,3),(5,3),(6,3),

(1,4),(2,4),(3,4),(4,4),(5,4),(6,4),

(1,5),(2,5),(3,5),(4,5),(5,5),(6,5),

(1,6),(2,6),(3,6),(4,6),(5,6),(6,6)

Number of getting multiple of 4 possible outcomes =15

[i.e.(1,4)(2,2)(2,4)(2,6)(3,4)(4,1)(4,2)(4,3)(4,4)(4,5)(4,6)(6,2)(6,4)(6,6)]

P(A)= Number of favorable outcome/Total number of possible outcome

 P(A)= 15/36

       = 5/12

Question 4: Two dice are thrown simultaneously What is the probability of getting two numbers whose product is even?

Solution:

Two different dice are thrown simultaneously the possible happening events are 1, 2, 3, 4, 5 and 6. The total number of favorable outcomes is (6 × 6) = 36.

Let A = possibility of happening two numbers whose product is even

Then A={(1,2),(1,4),(1,6),(2,1),(2,2),(2,3),(2,4),(2,5),(2,6),(3,2),(3,4),(3,6),(4,1),(4,2),(4,3),

(4,4)(4,5),(4,6),(5,2),(5,4),(5,6),(6,1),(6,2),(6,3),(6,4),(6,5),(6,6)}

Number of getting product is even = 27

P(A)=Number of favorable outcome/Total number of possible outcome

​      =  27/36

      =  3/4