What is the probability that at least one 6 appears when 6 fair dice are rolled round your answer to the second decimal point )?

Inhaltsverzeichnis Show

Dice roll probability: 6 Sided Dice Example
Dice Roll Probability for 6 Sided Dice: Sample Spaces
Dice Rolling Probability: Steps
Two (6-sided) dice roll probability table
Dice Roll Probability Tables
Probability of a certain number with a Single Die.
Probability of rolling a certain number or less with one die
Probability of rolling less than certain number with one die
Probability of rolling a certain number or more.
Probability of rolling more than a certain number (e.g. roll more than a 5).

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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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 an 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). And if you are interested in rolling the set of any identical values, simply multiply the result by the total die faces: P = 0.000125 * 20 = 0.0025.
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:

P(X=r) = nCr * pʳ * (1-p)ⁿ⁻ʳ,

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 r out of the set of n s-sided dice - the general formula is pretty complex:

However, we can also try to evaluate this problem by hand. 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.