What is the probability of getting the same number on both dices Round your answer to two decimal places?

When rolling two dice, distinguish between them in some way: a first one and second one, a left and a right, a red and a green, etc. Let (a,b) denote a possible outcome of rolling the two die, with a the number on the top of the first die and b the number on the top of the second die. Note that each of a and b can be any of the integers from 1 through 6. Here is a listing of all the joint possibilities for (a,b):

(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)
Note that there are 36 possibilities for (a,b). This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36. The set of all possible outcomes for (a,b) is called the sample space of this probability experiment.

With the sample space now identified, formal probability theory requires that we identify the possible events. These are always subsets of the sample space, and must form a sigma-algebra. In an example such as this, where the sample space is finite because it has only 36 different outcomes, it is perhaps easiest to simply declare ALL subsets of the sample space to be possible events. That will be a sigma-algebra and avoids what might otherwise be an annoying technical difficulty. We make that declaration with this example of two dice.

With the above declaration, the outcomes where the sum of the two dice is equal to 5 form an event. If we call this event E, we have

E={(1,4),(2,3),(3,2),(4,1)}.
Note that we have listed all the ways a first die and second die add up to 5 when we look at their top faces.

Consider next the probability of E, P(E). Here we need more information. If the two dice are fair and independent , each possibility (a,b) is equally likely. Because there are 36 possibilities in all, and the sum of their probabilities must equal 1, each singleton event {(a,b)} is assigned probability equal to 1/36. Because E is composed of 4 such distinct singleton events, P(E)=4/36= 1/9.

In general, when the two dice are fair and independent, the probability of any event is the number of elements in the event divided by 36.

What if the dice aren't fair, or aren't independent of each other? Then each outcome {(a,b)} is assigned a probability (a number in [0,1]) whose sum over all 36 outcomes is equal to 1. These probabilities aren't all equal, and must be estimated by experiment or inferred from other hypotheses about how the dice are related and and how likely each number is on each of the dice. Then the probability of an event such as E is the sum of the probabilities of the singleton events {(a,b)} that make up E.

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.

  1. 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.

  2. 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.

  3. 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.

  4. 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.

  1. 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.

  2. The probability of rolling an exact sum r out of the set of n s-sided dice - the general formula is pretty complex:

What is the probability of getting the same number on both dices Round your answer to two decimal places?

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.

🔎 Our ratio calculator can be quite helpful in finding the missing term of such ratios!

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!

  1. 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.

  2. 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.

When 2 dice are rolled find the probability of getting?

Probability of rolling a certain number or less for two 6-sided dice. ... Two (6-sided) dice roll probability table..

What is the probability that rolling 2 dice will result in 2 even numbers?

P(E)=636=16. Show activity on this post. Probability of two different even numbers = Probability of both even * Probability of both different given they are both even. Probability of both even is 12×12=14.

What are the number of events in getting a two equal number in rolling two dice?

This total number of possibilities can be obtained from the multiplication principle: there are 6 possibilities for a, and for each outcome for a, there are 6 possibilities for b. So, the total number of joint outcomes (a,b) is 6 times 6 which is 36.

When two dice are thrown find the probability of getting same numbers on both dice * 1 point 2 3 1 6 1 12 1?

` P(getting the same number on both dice) =`6/36 = 1/6`.