The expected value of the sum of the numbers shown in rolling a dice twice is 7. This is because each roll has an equal probability of landing on any number from 1 to 6, and the sum of those numbers has an equal chance of being any value from 2 to 12. Since the probabilities of each outcome are evenly distributed, the average or expected value is the midpoint, which is 7.
The expected value of the sum of the numbers shown in rolling a dice twice can be calculated by considering the probabilities of each possible outcome and their corresponding values.
To understand this concept better, let’s delve into the details. Each roll of a fair six-sided dice has an equal probability of landing on any number from 1 to 6. So, when rolling the dice twice, there are 36 possible outcomes (6 numbers on the first roll multiplied by 6 numbers on the second roll). The sum of those numbers can range from a minimum of 2 (when rolling two ones) to a maximum of 12 (two sixes).
By evaluating the probabilities and their corresponding values, we can derive the expected value. For instance, the probability of rolling a 2 is 1/36 because it can only occur when two ones are rolled. Similarly, the probability of rolling a 3 is 2/36 since it can happen when one of the dice show a 1 and the other shows a 2, or vice versa. Following this pattern, we can determine the probabilities for each possible sum from 2 to 12.
Using this information, we can calculate the expected value by multiplying each possible sum by its corresponding probability and summing them all up. However, to keep things concise and visually appealing, let’s represent this information in a table:
| Sum | Probability |
|---|---|
| 2 | 1/36 |
| 3 | 2/36 |
| 4 | 3/36 |
| 5 | 4/36 |
| 6 | 5/36 |
| 7 | 6/36 |
| 8 | 5/36 |
| 9 | 4/36 |
| 10 | 3/36 |
| 11 | 2/36 |
| 12 | 1/36 |
Now, let’s calculate the expected value by taking the sum of each product of the sum and its corresponding probability:
Expected Value = (2 * 1/36) + (3 * 2/36) + (4 * 3/36) + (5 * 4/36) + (6 * 5/36) + (7 * 6/36) + (8 * 5/36) + (9 * 4/36) + (10 * 3/36) + (11 * 2/36) + (12 * 1/36)
After performing the calculations, we find that the expected value of the sum of the numbers shown in rolling a dice twice is indeed 7.
To add an intriguing quote on the topic, let’s go with this one by the renowned physicist Albert Einstein:
“In the middle of difficulty lies opportunity.”
Interesting Facts:
- The concept of expected value is widely used in probability theory and statistics to calculate the long-term average of a random variable.
- The expected value is sometimes referred to as the mathematical expectation or mean.
- The expected value is not always an achievable outcome; it represents the average value over many trials.
- The expected value can be calculated for various random variables, not just sums of dice rolls.
- The expected value is a useful tool in decision-making, as it helps in assessing potential outcomes and risks.
By understanding the probabilities and expected values, we can gain valuable insights into various scenarios and make informed decisions. Remember, in the world of uncertainty, probabilities provide us with the key to analyzing and predicting outcomes.
See related video
This YouTube video discusses the calculation of the expected value, variance, and standard deviation of rolling a fair six-sided die. The expected value is determined by multiplying each possible value of the die roll by its corresponding probability and summing the results, resulting in an expected value of 7/2. The variance is found by subtracting the square of the expected value from the expected value of the squared value of the die roll. The variance is calculated to be 91/6. Finally, the standard deviation is estimated as the square root of the variance, approximately 1.7.
There are alternative points of view
For example, if a fair 6-sided die is rolled, the expected value of the number rolled is 3.5. The expectation of the sum of two (independent) dice is the sum of expectations of each die, which is 3.5 + 3.5 = 7. Similarly, for N dice throws, the expectation of the sum should be N * 3.5.
Surely you will be interested
What is the expected value of a dice roll?
The answer is: When you roll a fair die you have an equal chance of getting each of the six numbers 1 to 6. The expected value of your die roll, however, is 3.5.
Also question is, How do you find the expected value of a dice game?
In reply to that: To calculate the expected value, we multiply the value times it probability and sum the results. So the expected value of this game is: (100 * 1/216) + (-1 * 215/216) = -115/216 = -53 cents, approximately. So you can expect to lose about 53 cents on average for every roll of the dice!
Regarding this, What is the expected value of rolling 3 dice? The expected value of one die is 3.5, which is the average of the dots on the six faces. Rolling three dice would give you an expected outcome of 10.5. (It may be interesting to note that the expected value is an impossible occurrence, since dice have a whole number of dots.
When two dice are rolled what is the probability that two numbers will have a sum of 8? As a response to this: 5/36
The chance of rolling an 8 with 2 dice is 5/36. In the case of 3 dice: if the first die rolls a 1, 2, 3, 4, 5, or 6, then the next two dice must sum to 7, 6, 5, 4, 3, or 2, respectively, and there are 6 + 5 + 4 + 3 + 2 + 1 ways for these scenarios to occur.
What is the expected value of a dice roll?
Response will be: The expected value of a dice roll is 4.5 for a standard 8-sided die (a die with each of the numbers 1 through 8 appearing on exactly one face of the die). In this case, for a fair die with 8 sides, the probability of each outcome is the same: 1/8. The possible outcomes are the numbers 1 through 8: 1, 2, 3, 4, 5, 6, 7, and 8.
Herein, How do you find the expected value of a 6 sided dice?
In reply to that: When we roll 2 6-sided dice and sum the results (or roll one die twice and sum the results), we get a larger expected value than if we just rolled one die once. If we call the first die roll D 1 and the second die roll D 2, then the expected value of their sum D 1 + D 2 is:
How do you find the expected value of a die roll? If we call the first die roll D 1 and the second die roll D 2, then the expected value of their sum D 1 + D 2 is: = E (D1) + E (D2) [expected value of a sum is the sum of the expected values, since expected value is a linear operator] = 3.5 + 3.5 [since the expected value of rolling a 6-sided die is 3.5]
How many dice can you roll on a 6 sided die?
As an answer to this: 3.5 + 3.5 + 3.5 [since the expected value of rolling a 6-sided die is 3.5] The “table” for rolling 3 dice would need to be three-dimensional, since there are 3 die rolls to account for. There are sums ranging from 3 (rolling a 1 on all three dice) to 18 (rolling a 6 on all three dice). The probabilities of these events vary.
How do you find the expected value of a 6 sided dice?
When we roll 2 6-sided dice and sum the results (or roll one die twice and sum the results), we get a larger expected value than if we just rolled one die once. If we call the first die roll D 1 and the second die roll D 2, then the expected value of their sum D 1 + D 2 is:
What is the expected value of a dice roll?
The expected value of a dice roll is 4.5 for a standard 8-sided die (a die with each of the numbers 1 through 8 appearing on exactly one face of the die). In this case, for a fair die with 8 sides, the probability of each outcome is the same: 1/8. The possible outcomes are the numbers 1 through 8: 1, 2, 3, 4, 5, 6, 7, and 8.
What is the expectation of the sum of two (independent) dice?
Response to this: The expectation of the sum of two (independent) dice is the sum of expectations of each die, which is 3.5 + 3.5 = 7. Similarly, for N dice throws, the expectation of the sum should be N * 3.5. If you’re taking only the maximum value of the two dice throws, then your answer 4.47 is correct. This has been proven here in multiple ways.
What is the expected value for a n dice throw?
The answer is: Why are you taking the maximum value. Should it not be the sum. If it is the sum then the expected value for a n dice throw is N ∗ 3.5. Your solution looks correct. Did you overlook anything? ∑ k = 0 n − 1 ( n k) ( h − 1) k = ( 1 + ( h − 1)) n ( n n) ⏞ sum for k = 0 … n − ( n n) ( h − 1) n ⏞ k = n (1) = h n − ( h − 1) n