introduction to probability

The characters included the main character, Lin Yu; the main character, the system, Xiao Shu; the probability control system; the supporting character, Xiao Ai, a silicon-based life form; the supporting character, Pyramid, a silicon-based life form; the male supporting character, Lin Xuan, who was 'Infinite Everything' since birth; the female supporting character, Tushan Xue, a ten-tailed celestial fox; the female supporting character, Ye Xin, an electric power user; and the female supporting character, Lin Yu, Lin Xuan's sister. "Probability-based Control System" Author: Tao in My Heart. It's an urban/supernatural novel. [User recommendation: The protagonist of the biggest cheating novel I've read so far can easily create an infinite multiverse. It's very awesome.] I hope you will like this book.

For two dice, there were several possibilities: 1. ** Single number probability **: - Each die had six sides, and the numbers were 1, 2, 3, 4, 5, and 6. When a die is rolled, the probability of each number appearing is 1/6. When two dice were thrown, for example, the probability of both dice rolling a 1 was 1/6. Therefore, the probability of both dice rolling a 1 (the combination of 1 and 1) was (1/6)×(1/6)=1/36. Similarly, any particular combination of numbers (such as 3 and 5) has a probability of 1/36. 2. ** Point sum probability **: - The sum of the points was 2 (1 + 1), and there was only one possibility. The probability was (1/6)×(1/6)=1/36. - The sum of the points is 3 (1+2 or 2 + 1). There are two possibilities, and the probability is 2/36. - The sum of the points was 4 (1+3, 2+2, 3+1). There were 3 possibilities, and the probability was 3/36. - The sum of the points is 5 (1+4, 2+3, 3+2, 4+1). There are 4 possibilities, and the probability is 4/36. 3. ** Dice size probability (1 - 6 is small, 7 - 12 is big)**: - There were a total of 6x6 = 36 combinations of the two dice. Among them, there were 15 situations where the sum of points was 1 - 6 (small), and 21 situations where the sum of points was 7 - 12 (large). However, since it was impossible for two dice to have a point, the probability of a small point appearing was 5/11 (about 45.45%), and the probability of a big point appearing was 6/11 (about 54.55%). 4. ** The probability that the sum of the two dice numbers is odd (assuming the first die numbers are 1, 2, 3, 3, 5, 6, and the second die numbers are 1, 2, 4, 4, 5, 6)**: - For the first die, the probability of an odd number appearing was 4/6 = 2/3, and the probability of an even number appearing was 1/3. For the second die, the probability of an odd number appearing was 2/6 = 1/3, and the probability of an even number appearing was 2/3. To get the odd sum, there are two situations: if the first die is odd and the second die is even, the probability is 2/3×2/3 = 4/9; if the first die is even and the second die is odd, the probability is 1/3×1/3 = 1/9. Therefore, the total probability was 4/9+1/9 = 5/9. Hurry up and click on the link below to return to the super classic "Lord of the Mysteries"!

The probability of a normal non-safe buff was between 10% and 30%, depending on the equipment's quality, level, enchantment, and strengthening level. The probability of a safe buff was between 3% and 10%, but there was no separate probability of a 10 to 11 buff.

The condition probability refers to the probability of event A happening under the condition that event B has already happened. It is recorded as P(A| B)。The following is an example analysis of the probability of dice rolling. Suppose we have two dice. Event A is that the sum of the two dice is 8, and Event B is that the first die rolls a 3. First, calculate the probability of event A without any conditions. There were a total of 6×6 = 36 possibilities for the outcome of the two dice. There are five cases where the sum of points is 8:(2,6),(3,5),(4,4),(5,3), and (6,2), so P(A)=5/36. Then, the probability of event B is calculated. The probability of the first die rolling a 3 is 1/6, which is P(B)=1/6. The only situation where both events A and B occur at the same time is (3,5), so P(A Empyrean B)=1/36. According to the formula P(A| B)=P(A Empyrean B)/P(B). Under the condition that the first die rolls a 3, the probability of the sum of the two dice points being 8 is P(A| B)=(1/36)/(1/6)=1/6。 For example, Event C was the difference between the two dice being 1, and Event D was the second die rolling a 4. Similarly, P(C) was first found. There were 10 cases where the difference between the points was 1:(1,2),(2,3),(3,4),(4,5),(5,6),(2,1),(3,2),(4,3),(5,4), and (6,5), so P(C)=10/36. The probability of event D is P(D)=1/6. There are two cases where events C and D occur at the same time: (3,4) and (5,4), so P(C Empyrean D)=2/36. Under the condition that the second die rolls a 4, the probability of the difference between the two dice being 1 is P(C| D)=(2/36)/(1/6)=1/3。 In short, when calculating the conditioned probability of rolling the dice, the key was to clarify the relationship between the events, separately calculate the probability of the event occurring alone and the probability of them occurring at the same time, and then use the conditioned probability formula to calculate. Hurry up and click on the link below to return to the super classic "Lord of the Mysteries"!

Rolling dice was a classic example of probability. For a single standard die, because it had six sides, each marked with the six numbers 1, 2, 3, 4, 5, and 6, the chances of each number appearing were equal. Therefore, when throwing a single die, the probability of any number appearing in the result was 1/6. When there were multiple dice involved, the probability calculation would be more complicated. Take two dice as an example, if you consider the size of the dice,(1 - 6 is small, 7 - 12 is big), because there are a total of 6×6 = 36 combinations of the number of points on the two dice. Among them, there are 15 cases where the sum is 1 - 6 (small), and there are 21 cases where the sum is 7 - 12 (big). However, since it is impossible for the two dice to appear, the actual probability of small appearing is 5/11 (about 45.45%), and the probability of big appearing is 6/11 (about 54.55%). If there were three dice, each die would have six situations, and there would be a total of 6x6x6 = 216 combinations. For example, if one calculated the probability of rolling a big die, there would be three kinds of big situations (namely, 4, 5, and 6 points) for each die. There were a total of 3×3×3 = 27 kinds. Therefore, the probability of a big situation appearing was 27 × 216 = 1/8. Similarly, the probability of a small situation appearing was also 1/8. The probability of a leopard appearing (three dice with the same number of points, such as 111, 222, etc.) was 6 × 216 = 1/36. In addition, in theory, if the structure of the die was uneven, for example, the dots were hollowed out, and the six small round pits on the six dots were lighter than the one on the opposite side, then the probability of the six appearing would be greater than the one. However, in the ideal standard die situation, the probability of each number was 1/6. At the same time, there were many other situations in the probability calculation of dice rolling, such as the probability calculation of the sum of two dice numbers being odd, which could be solved by different probability methods (such as the step-by-step probability calculation method in junior high school or the combination knowledge method in high school). Hurry up and click on the link below to return to the super classic "Lord of the Mysteries"!