Expected value is a useful tool for analysis in business and in game theory. However, it is often presented in the context of dice rolling, where probabilities are uniform and calculations are easier to perform.
So, what is the expected value of a dice roll? The expected value of a dice roll is 3.5 for a standard 6-sided die. This assumes a fair die – that is, there is a 1/6 probability of each outcome 1, 2, 3, 4, 5, and 6. The expected value of the sum of two 6-sided dice rolls is 7. Dice with a different number of sides will have other expected values.
Of course, we can also look at products, minimums, and maximums of dice rolls and get different expected values for these as well.
In this article, we’ll talk about the expected value for various dice rolls: 4, 6, 8, 12, 20, and N-sided dice.
Let’s get started.
(You can also watch a video summary of this article on YouTube!)
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What Is The Expected Value Of A Dice Roll?
The expected value of a dice roll is 3.5 for a standard 6-sided die (a die with each of the numbers 1 through 6 appearing on exactly one face of the die).
Remember that the expected value of an action is a weighted average of each possible outcome. The weight for each outcome is the probability of that outcome.
In this case, for a fair die with 6 sides, the probability of each outcome is the same: 1/6. The possible outcomes are the numbers 1 through 6: 1, 2, 3, 4, 5, and 6.
We take the product of each outcome and its probability and then add up those products to get the expected value (EV) of rolling a 6 sided die:
- EV(roll 6-sided die) = 1(1/6) + 2(1/6) + 3(1/6) + 4(1/6) + 5(1/6) + 6(1/6) [sum of products: probability times outcome]
- = 1/6 + 2/6 + 3/6 + 4/6 + 5/6 + 6/6
- = (1 + 2 + 3 + 4 + 5 + 6) / 6
- = 21/6
- = 3.5
The table below illustrates the concept of expected value of a 6-sided die roll in another way: with outcomes, probabilities, and their products.
| Die Roll | Proba bility | Product |
|---|---|---|
| 1 | 1/6 | 1/6 |
| 2 | 1/6 | 2/6 |
| 3 | 1/6 | 3/6 |
| 4 | 1/6 | 4/6 |
| 5 | 1/6 | 5/6 |
| 6 | 1/6 | 6/6 |
| Total | 1 | 21/6 =3.5 (EV) |
expected value of a 6-sided die roll
in another way: with outcomes,
probabilities, and their products.
Note that an outcome of 3.5 is not possible with a single die roll. However, if we roll a 6-sided die many times, the average outcome will be 3.5.
As an example, if we roll a 3 and then a 4, the average of the two die rolls is (3 + 4) / 2 = 7 / 2 = 3.5.
Of course, multiple dice rolls (or dice rolls for a die with more or less than 6 sides) will give us different expected values. Let’s start with 2 rolls of a 6-sided die (or rolling 2 6-sided dice) and taking the sum of the outcomes.
What Is The Expected Value Of 2 Dice Rolls?
The expected value of 2 dice rolls is 7 (assuming we take the sum of the two dice rolls).
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 D1 and the second die roll D2, then the expected value of their sum D1 + D2 is:
- E(D1 + D2) [take the expected value of the sum]
- = 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]
- = 7
The table below shows the outcome for both die rolls and the sum of the two outcomes. The probability for each outcome in the table is the same: 1/36 (which comes from 1/6 times 1/6).
Notice that the main diagonal (from bottom left to top right) has the value of 7, which is repeated 6 times. This value (7) is the most common value in the table (mode) and also the expected value (mean).
| 1st/2nd die roll | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 2 | 3 | 4 | 5 | 6 | 7 |
| 2 | 3 | 4 | 5 | 6 | 7 | 8 |
| 3 | 4 | 5 | 6 | 7 | 8 | 9 |
| 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | 9 | 10 | 11 |
| 6 | 7 | 8 | 9 | 10 | 11 | 12 |
rolls and the sum of the two outcomes. The
probability for each outcome in the table is the
same: 1/36 (which comes from 1/6 times 1/6).
What Is The Expected Value Of 3 Dice Rolls?
The expected value of 3 dice rolls is 10.5 (assuming we take the sum of the three dice rolls).
When we roll 3 6-sided dice and sum the results (or roll one die three times and sum the results), we get a larger expected value than if we just rolled one or two dice.
If we call the first die roll D1, the second die roll D2, and the third die roll D3, then the expected value of their sum D1 + D2 + D3 is:
- E(D1 + D2 + D3) [take the expected value of the sum]
- = E(D1) + E(D2) + E(D3) [expected value of a sum is the sum of the expected values, since expected value is a linear operator]
- = 3.5 + 3.5 + 3.5 [since the expected value of rolling a 6-sided die is 3.5]
- = 10.5
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. The most common sum is 10.5 (the expected value).
The probability of rolling a 3 is 1/216 (which comes from 1/6 times 1/6 times 1/6). The same goes for rolling an 18.
What Is The Expected Value Of N Dice Rolls?
The expected value of N dice rolls is 3.5N (assuming we take the sum of the N dice rolls).
When we roll N 6-sided dice and sum the results (or roll one die N times and sum the results), we get a larger expected value than if we just rolled fewer dice.
If we call the first die roll D1, the second die roll D2, the third die roll D3, etc., then the expected value of their sum D1 + D2 + D3 + … + DN is:
- E(D1 + D2 + D3+ … + DN) [take the expected value of the sum]
- = E(D1) + E(D2) + E(D3) + … + E(DN) [expected value of a sum is the sum of the expected values, since expected value is a linear operator]
- = 3.5 + 3.5 + 3.5 + … + 3.5 [since the expected value of rolling a 6-sided die is 3.5]
- = 3.5N [since we added 3.5 N times]
The “table” for rolling N dice would need to be N-dimensional, since there are N die rolls to account for. There are sums ranging from N (rolling a 1 on all N dice) to 6N (rolling a 6 on all N dice).
The probabilities of these events vary. The most common sum is 3.5N (the expected value).
The probability of rolling a total of N is 1/6N (which comes from 1/6 times 1/6 times 1/6 …). The same goes for rolling a total of 6N.
What Is The Expected Value Of The Product Of Two Dice Rolls?
The expected value of the product of two dice rolls is 12.25 for standard 6-sided dice.
When we take the product of two dice rolls, we get different outcomes than if we took the sum of the two dice rolls. The table below illustrates the possible products of two dice rolls:
| 1st/2nd die roll | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 2 | 3 | 4 | 5 | 6 |
| 2 | 2 | 4 | 6 | 8 | 10 | 12 |
| 3 | 3 | 6 | 9 | 12 | 15 | 18 |
| 4 | 4 | 8 | 12 | 16 | 20 | 24 |
| 5 | 5 | 10 | 15 | 20 | 25 | 30 |
| 6 | 6 | 12 | 18 | 24 | 30 | 36 |
of two dice rolls for 6-sided dice.
As before, the probability of each product in the table is 1/36 (which comes from 1/6 times 1/6). Taking the product of each table entry with 1/36 (or adding up the entries and dividing by 36) gives us the result of 12.25.
What Is The Expected Value Of The Minimum Of Two Dice Rolls?
The expected value of the minimum of two dice rolls is 91/36 (about 2.53) for standard 6-sided dice.
When we take the minimum of two dice rolls, we get different outcomes than if we took the sum or product of the two dice rolls. The table below illustrates the possible minimums of two dice rolls:
| 1st/2nd die roll | 1 | 2 | 3 | 4 | 5 | 6 |
| 1 | 1 | 1 | 1 | 1 | 1 | 1 |
| 2 | 1 | 2 | 2 | 2 | 2 | 2 |
| 3 | 1 | 2 | 3 | 3 | 3 | 3 |
| 4 | 1 | 2 | 3 | 4 | 4 | 4 |
| 5 | 1 | 2 | 3 | 4 | 5 | 5 |
| 6 | 1 | 2 | 3 | 4 | 5 | 6 |
minimums of two dice rolls for 6-sided dice.
As before, the probability of each product in the table is 1/36 (which comes from 1/6 times 1/6). Taking the product of each table entry with 1/36 (or adding up the entries and dividing by 36) gives us the result of 91/36 or about 2.53.
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What Is The Expected Value Of A 4 Sided Die?
The expected value of a dice roll is 2.5 for a standard 4-sided die (a die with each of the numbers 1 through 4 appearing on exactly one face of the die).
In this case, for a fair die with 4 sides, the probability of each outcome is the same: 1/4. The possible outcomes are the numbers 1 through 4: 1, 2, 3, and 4.
We take the product of each outcome and its probability and then add up those products to get the expected value (EV) of rolling a 4 sided die:
- EV(roll 4-sided die) = 1(1/4) + 2(1/4) + 3(1/4) + 4(1/4) [sum of products: probability times outcome]
- = 1/4 + 2/4 + 3/4 + 4/4
- = (1 + 2 + 3 + 4) / 4
- = 10 / 4
- = 2.5
The table below illustrates the concept of expected value of a 4-sided die roll in another way: with outcomes, probabilities, and their products.
| Die Roll | Proba bility | Product |
|---|---|---|
| 1 | 1/4 | 1/4 |
| 2 | 1/4 | 2/4 |
| 3 | 1/4 | 3/4 |
| 4 | 1/4 | 4/4 |
| Total | 1 | 10/4 =2.5 (EV) |
expected value of a 4-sided die roll
in another way: with outcomes,
probabilities, and their products.
Note that an outcome of 2.5 is not possible with a single die roll. However, if we roll a 4-sided die many times, the average outcome will be 2.5.
As an example, if we roll a 2 and then a 3, the average of the two die rolls is (2 + 3) / 2 = 5 / 2 = 2.5.
What Is The Expected Value Of An 8 Sided Die?
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.
We take the product of each outcome and its probability and then add up those products to get the expected value (EV) of rolling an 8 sided die:
- EV(roll 8-sided die) = 1(1/8) + 2(1/8) + 3(1/8) + 4(1/8) + 5(1/8) + 6(1/8) + 7(1/8) + 8(1/8) [sum of products: probability times outcome]
- = 1/8 + 2/8 + 3/8 + 4/8 + 5/8 + 6/8 + 7/8 + 8/8
- = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8) / 8
- = 36 / 8
- = 4.5
The table below illustrates the concept of expected value of a 8-sided die roll in another way: with outcomes, probabilities, and their products.
| Die Roll | Proba bility | Product |
|---|---|---|
| 1 | 1/8 | 1/8 |
| 2 | 1/8 | 2/8 |
| 3 | 1/8 | 3/8 |
| 4 | 1/8 | 4/8 |
| 5 | 1/8 | 5/8 |
| 6 | 1/8 | 6/8 |
| 7 | 1/8 | 7/8 |
| 8 | 1/8 | 8/8 |
| Total | 1 | 36/8 =4.5 (EV) |
expected value of a 8-sided die roll
in another way: with outcomes,
probabilities, and their products.
Note that an outcome of 4.5 is not possible with a single die roll. However, if we roll an 8-sided die many times, the average outcome will be 4.5.
As an example, if we roll a 4 and then a 5, the average of the two die rolls is (4 + 5) / 2 = 9 / 2 = 4.5.
What Is The Expected Value Of A 12 Sided Die?
The expected value of a dice roll is 6.5 for a standard 12-sided die (a die with each of the numbers 1 through 12 appearing on exactly one face of the die).
In this case, for a fair die with 12 sides, the probability of each outcome is the same: 1/12. The possible outcomes are the numbers 1 through 12: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, and 12.
We take the product of each outcome and its probability and then add up those products to get the expected value (EV) of rolling a 12 sided die:
- EV(roll 12-sided die) = 1(1/12) + 2(1/12) + 3(1/12) + 4(1/12) + 5(1/12) + 6(1/12) + 7(1/12) + 8(1/12) + 9(1/12) + 10(1/12) + 11(1/12) + 12(1/12) [sum of products: probability times outcome]
- = 1/12 + 2/12 + 3/12 + 4/12 + 5/12 + 6/12 + 7/12 + 8/12 + 9/12 + 10/12 + 11/12 + 12/12
- = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12) / 12
- = 78 / 12
- = 6.5
The table below illustrates the concept of expected value of a 12-sided die roll in another way: with outcomes, probabilities, and their products.
| Die Roll | Proba bility | Product |
|---|---|---|
| 1 | 1/12 | 1/12 |
| 2 | 1/12 | 2/12 |
| 3 | 1/12 | 3/12 |
| 4 | 1/12 | 4/12 |
| 5 | 1/12 | 5/12 |
| 6 | 1/12 | 6/12 |
| 7 | 1/12 | 7/12 |
| 8 | 1/12 | 8/12 |
| 9 | 1/12 | 9/12 |
| 10 | 1/12 | 10/12 |
| 11 | 1/12 | 11/12 |
| 12 | 1/12 | 12/12 |
| Total | 1 | 78/12 =6.5 (EV) |
expected value of a 12-sided die roll
in another way: with outcomes,
probabilities, and their products.
Note that an outcome of 6.5 is not possible with a single die roll. However, if we roll a 12-sided die many times, the average outcome will be 4.5.
As an example, if we roll a 6 and then a 7, the average of the two die rolls is (6 + 7) / 2 = 13 / 2 = 6.5.
What Is The Expected Value Of A 20 Sided Die?
The expected value of a dice roll is 10.5 for a standard 20-sided die (a die with each of the numbers 1 through 20 appearing on exactly one face of the die).
In this case, for a fair die with 20 sides, the probability of each outcome is the same: 1/20. The possible outcomes are the numbers 1 through 20: 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18, 19, and 20.
We take the product of each outcome and its probability and then add up those products to get the expected value (EV) of rolling a 20 sided die:
- EV(roll 20-sided die) = 1(1/20) + 2(1/20) + 3(1/20) + 4(1/20) + 5(1/20) + 6(1/20) + 7(1/20) + 8(1/20) + 9(1/20) + 10(1/20) + 11(1/20) + 12(1/20) + 13(1/20) + 14(1/20) + 15(1/20) + 16(1/20) + 17(1/20) + 18(1/20) + 19(1/20) + 20(1/20) [sum of products: probability times outcome]
- = 1/20 + 2/20 + 3/20 + 4/20 + 5/20 + 6/20 + 7/20 + 8/20 + 9/20 + 10/20 + 11/20 + 12/20 + 13/20 + 14/20 + 15/20 + 16/20 + 17/20 + 18/20 + 19/20 + 20/20
- = (1 + 2 + 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + 20) / 20
- = 210 / 20
- = 10.5
The table below illustrates the concept of expected value of a 20-sided die roll in another way: with outcomes, probabilities, and their products.
| Die Roll | Proba bility | Product |
|---|---|---|
| 1 | 1/20 | 1/20 |
| 2 | 1/20 | 2/20 |
| 3 | 1/20 | 3/20 |
| 4 | 1/20 | 4/20 |
| 5 | 1/20 | 5/20 |
| 6 | 1/20 | 6/20 |
| 7 | 1/20 | 7/20 |
| 8 | 1/20 | 8/20 |
| 9 | 1/20 | 9/20 |
| 10 | 1/20 | 10/20 |
| 11 | 1/20 | 11/20 |
| 12 | 1/20 | 12/20 |
| 13 | 1/20 | 13/20 |
| 14 | 1/20 | 14/20 |
| 15 | 1/20 | 15/20 |
| 16 | 1/20 | 16/20 |
| 17 | 1/20 | 17/20 |
| 18 | 1/20 | 18/20 |
| 19 | 1/20 | 19/20 |
| 20 | 1/20 | 20/20 |
| Total | 1 | 210/20 =10.5 (EV) |
expected value of a 20-sided die roll
in another way: with outcomes,
probabilities, and their products.
Note that an outcome of 10.5 is not possible with a single die roll. However, if we roll a 20-sided die many times, the average outcome will be 10.5.
As an example, if we roll a 10 and then an 11, the average of the two die rolls is (10 + 11) / 2 = 21 / 2 = 10.5.
What Is The Expected Value Of An N Sided Die?
The expected value of a dice roll is (N + 1) / 2 for a standard N-sided die (a die with each of the numbers 1 through N appearing on exactly one face of the die).
In this case, for a fair die with N sides, the probability of each outcome is the same: 1/N. The possible outcomes are the numbers 1 through N: 1, 2, … , N.
We take the product of each outcome and its probability and then add up those products to get the expected value (EV) of rolling an N sided die:
- EV(roll N-sided die) = 1(1/N) + 2(1/N) + … + N(1/N) [sum of products: probability times outcome]
- = 1/N + 2/N + … + N/N
- = (1 + 2 + … + N / N
- (N(N+1)/2) / N [the sum of the numbers 1 to N is given by N(N+1) / 2]
- N(N+1) / 2N
- (N+1) / 2 [N cancels in numerator and denominator]
You can test this formula for the expected values of each of the die rolls we looked at earlier.
For example, the expected value of a 6-sided die (N = 6) is 3.5, since (N+1)/2 = (6+1)/2 = 7/2 = 3.5.
The table below shows the expected value of an N sided die for various values of N.
| N = number of sides on die | Expected Value Of A Dice Roll |
|---|---|
| 4 | 2.5 |
| 6 | 3.5 |
| 8 | 4.5 |
| 10 | 5.5 |
| 12 | 6.5 |
| 14 | 7.5 |
| 16 | 8.5 |
| 18 | 9.5 |
| 20 | 10.5 |
| N | (N+1)/2 |
an N sided die for various values of N.
Conclusion
Now you know more about the expected value of a dice roll for various N-sided dice. You also know how to do the calculations to verify the expected values.
You might also be interested to learn more about variance in my article here.
You can learn more about the uses of expected value in my article here.
I hope you found this article helpful. If so, please share it with someone who can use the information.
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~Jonathon
