You roll two die what is the probability you will roll a 4, then roll a number less than 3

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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.

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:

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!

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.

One popular way to study probability is to roll dice. A standard die has six sides printed with little dots numbering 1, 2, 3, 4, 5, and 6. If the die is fair (and we will assume that all of them are), then each of these outcomes is equally likely. Since there are six possible outcomes, the probability of obtaining any side of the die is 1/6. The probability of rolling a 1 is 1/6, the probability of rolling a 2 is 1/6, and so on. But what happens if we add another die? What are the probabilities for rolling two dice?

To correctly determine the probability of a dice roll, we need to know two things:

• The size of the sample space or the set of total possible outcomes
• How often an event occurs

In probability, an event is a certain subset of the sample space. For example, when only one die is rolled, as in the example above, the sample space is equal to all of the values on the die, or the set (1, 2, 3, 4, 5, 6). Since the die is fair, each number in the set occurs only once. In other words, the frequency of each number is 1. To determine the probability of rolling any one of the numbers on the die, we divide the event frequency (1) by the size of the sample space (6), resulting in a probability of 1/6.

Rolling two fair dice more than doubles the difficulty of calculating probabilities. This is because rolling one die is independent of rolling a second one. One roll has no effect on the other. When dealing with independent events we use the multiplication rule. The use of a tree diagram demonstrates that there are 6 x 6 = 36 possible outcomes from rolling two dice.

Suppose that the first die we roll comes up as a 1. The other die roll could be a 1, 2, 3, 4, 5, or 6. Now suppose that the first die is a 2. The other die roll again could be a 1, 2, 3, 4, 5, or 6. We have already found 12 potential outcomes, and have yet to exhaust all of the possibilities of the first die.

The possible outcomes of rolling two dice are represented in the table below. Note that the number of total possible outcomes is equal to the sample space of the first die (6) multiplied by the sample space of the second die (6), which is 36.

The same principle applies if we are working on problems involving three dice. We multiply and see that there are 6 x 6 x 6 = 216 possible outcomes. As it gets cumbersome to write the repeated multiplication, we can use exponents to simplify work. For two dice, there are 62 possible outcomes. For three dice, there are 63 possible outcomes. In general, if we roll n dice, then there are a total of 6n possible outcomes.

With this knowledge, we can solve all sorts of probability problems:

1. Two six-sided dice are rolled. What is the probability that the sum of the two dice is seven?

The easiest way to solve this problem is to consult the table above. You will notice that in each row there is one dice roll where the sum of the two dice is equal to seven. Since there are six rows, there are six possible outcomes where the sum of the two dice is equal to seven. The number of total possible outcomes remains 36. Again, we find the probability by dividing the event frequency (6) by the size of the sample space (36), resulting in a probability of 1/6.

2. Two six-sided dice are rolled. What is the probability that the sum of the two dice is three?

In the previous problem, you may have noticed that the cells where the sum of the two dice is equal to seven form a diagonal. The same is true here, except in this case there are only two cells where the sum of the dice is three. That is because there are only two ways to get this outcome. You must roll a 1 and a 2 or you must roll a 2 and a 1. The combinations for rolling a sum of seven are much greater (1 and 6, 2 and 5, 3 and 4, and so on). To find the probability that the sum of the two dice is three, we can divide the event frequency (2) by the size of the sample space (36), resulting in a probability of 1/18.

3. Two six-sided dice are rolled. What is the probability that the numbers on the dice are different?

Again, we can easily solve this problem by consulting the table above. You will notice that the cells where the numbers on the dice are the same form a diagonal. There are only six of them, and once we cross them out we have the remaining cells in which the numbers on the dice are different. We can take the number of combinations (30) and divide it by the size of the sample space (36), resulting in a probability of 5/6.