Chance of rolling a 7

what is the probablity of rolling two dice and it coming out as seven?

please give me the correct awnser and explain please.

One way to approach a probability problem is to make a list or description of all the possible outcomes of the experiment, and then calculate the proportion of these possible outcomes satisfy the condition you are looking for, here that the sum is seven. To make this work you need to find a way to list the possible outcomes so that every item on the list has the same probability of occurring.

In the experiment of rolling two dice think of one as red and the other as green and list the possible result of the roll in a table. Here, for example, the (3,5) in third row and fifth column means a 3 was rolled on the red die and a 5 on the green die. As the table shows there are 36 possible outcomes.

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) |

For each of the possible outcomes add the numbers on the two dice and count how many times this sum is 7. If you do so you will find that the sum is 7 for 6 of the possible outcomes. Thus the sum is a 7 in 6 of the 36 outcomes and hence the probability of rolling a 7 is 6/36 = 1/6.

I hope this helps,

Harley Go to Math Central

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Chance of rolling a 7 what is the probablity of rolling two dice and it coming out as seven? please give me the correct awnser and explain please. One way to approach a probability problem## Dice Probability Calculator

The dice probability calculator is a great tool if you want to estimate the dice roll probability over numerous variants. There are may different polyhedral die included, so you can explore the probability of a 20 sided die as well as that of a regular cubic die. So, just evaluate the odds, and play a game! In the text, you’ll also find a short descriptions of each of the options.

### Polyhedral dice

Everybody knows what a regular 6 sided die is, and, most likely, many of you have already played thousands of games where the one (or more) was used. But, did you know that there are **different types of die**? Out of the countless possibilities, the most popular dice are included in the **Dungeons & Dragons dice set**, which contains seven different polyhedral dice:

**4 sided dice**, also known as a**tetrahedron**– each face is an equilateral triangle**6 sided dice**, a classic**cube**– each face is a square**8 sided dice**, also known as an**octahedron**– each face is an equilateral triangle**10 sided dice**, also known as a**pentagonal trapezohedron**– each face is a kite**12 sided dice**, also known as a**dodecahedron**– each face is a regular pentagon**20 sided dice**, also known as an**icosahedron**– each face is an equilateral triangle

Don’t worry, we take each of these dice into account in our dice probability calculator. You can choose whichever you like, and e.g. pretend to roll five 20 sided dice at once!

### How to calculate dice roll probability?

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 a 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).

**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:

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 out of the set** – unfortunately, there isn’t a single formula for this problem. 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 .