Expected Value of One Die Face Has Roll Again
Expected value is a useful tool for analysis in business and in game theory. However, it is oft presented in the context of dice rolling, where probabilities are uniform and calculations are easier to perform.
And then, what is the expected value of a dice scroll? The expected value of a dice roll is 3.v for a standard 6-sided die. This assumes a off-white dice – that is, there is a one/6 probability of each event 1, 2, 3, 4, 5, and vi. The expected value of the sum of two half-dozen-sided dice rolls is 7. Dice with a different number of sides will have other expected values.
Of course, we can also expect at products, minimums, and maximums of dice rolls and get different expected values for these also.
In this article, we'll talk about the expected value for various dice rolls: 4, 6, 8, 12, xx, and Due north-sided dice.
Allow'south become started.
What Is The Expected Value Of A Dice Roll?
The expected value of a dice roll is 3.v for a standard 6-sided die (a die with each of the numbers 1 through 6 appearing on exactly i face of the die).
Remember that the expected value of an action is a weighted boilerplate of each possible outcome. The weight for each upshot is the probability of that outcome.
In this case, for a fair die with half-dozen sides, the probability of each outcome is the same: 1/6. The possible outcomes are the numbers one through half dozen: one, 2, 3, 4, 5, and 6.
We accept the product of each upshot and its probability and and so add up those products to get the expected value (EV) of rolling a 6 sided die:
- EV(ringlet six-sided die) = 1(ane/6) + 2(i/half dozen) + 3(i/half dozen) + 4(1/6) + 5(1/6) + six(1/half-dozen) [sum of products: probability times outcome]
- = 1/half dozen + 2/half dozen + three/6 + 4/half-dozen + five/vi + half-dozen/6
- = (1 + 2 + iii + 4 + 5 + half dozen) / half-dozen
- = 21/half-dozen
- = 3.v
The tabular array below illustrates the concept of expected value of a half dozen-sided die roll in some other way: with outcomes, probabilities, and their products.
| Die Curlicue | Proba bility | Product |
|---|---|---|
| 1 | 1/6 | 1/half-dozen |
| 2 | 1/6 | 2/6 |
| 3 | ane/six | three/vi |
| 4 | i/six | 4/half-dozen |
| 5 | 1/6 | five/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 issue of 3.5 is not possible with a single die roll. However, if we whorl a half-dozen-sided die many times, the average outcome will exist three.five.
Equally an example, if we curl a 3 and so a four, the average of the 2 die rolls is (3 + iv) / 2 = vii / 2 = 3.5.
Of course, multiple dice rolls (or dice rolls for a dice with more than or less than 6 sides) volition requite u.s. different expected values. Permit'south start with 2 rolls of a 6-sided dice (or rolling 2 vi-sided dice) and taking the sum of the outcomes.
What Is The Expected Value Of 2 Dice Rolls?
The expected value of ii dice rolls is 7 (assuming we accept the sum of the two dice rolls).
When we curl 2 6-sided dice and sum the results (or curl one die twice and sum the results), we get a larger expected value than if we just rolled one dice one time.
If we call the outset die roll D1 and the second die coil D2, then the expected value of their sum D1 + D2 is:
- E(Di + D2) [accept the expected value of the sum]
- = E(D1) + E(Dtwo) [expected value of a sum is the sum of the expected values, since expected value is a linear operator]
- = 3.five + 3.5 [since the expected value of rolling a half dozen-sided dice is 3.five]
- = vii
The table below shows the outcome for both dice rolls and the sum of the two outcomes. The probability for each outcome in the tabular array is the same: one/36 (which comes from 1/six times 1/6).
Notice that the main diagonal (from bottom left to top right) has the value of 7, which is repeated half-dozen times. This value (7) is the most mutual value in the tabular array (way) and too the expected value (mean).
| 1st/second dice roll | one | 2 | 3 | 4 | v | 6 |
| 1 | 2 | 3 | 4 | 5 | vi | 7 |
| 2 | iii | 4 | v | 6 | vii | 8 |
| 3 | iv | 5 | six | seven | 8 | 9 |
| 4 | 5 | 6 | seven | 8 | 9 | 10 |
| 5 | 6 | 7 | 8 | nine | ten | 11 |
| six | 7 | 8 | 9 | ten | eleven | 12 |
rolls and the sum of the two outcomes. The
probability for each effect in the table is the
same: 1/36 (which comes from 1/6 times one/6).
What Is The Expected Value Of 3 Dice Rolls?
The expected value of 3 die rolls is 10.v (assuming nosotros have the sum of the 3 dice rolls).
When we roll 3 6-sided dice and sum the results (or roll one dice three times and sum the results), we go a larger expected value than if we just rolled i or two dice.
If we call the first die scroll Di, the second die scroll Dtwo, and the tertiary die curl D3, then the expected value of their sum D1 + D2 + Diii is:
- East(Done + D2 + D3) [have the expected value of the sum]
- = E(D1) + Eastward(Dii) + E(D3) [expected value of a sum is the sum of the expected values, since expected value is a linear operator]
- = iii.five + three.5 + 3.5 [since the expected value of rolling a 6-sided die is 3.5]
- = x.five
The "table" for rolling 3 dice would need to be three-dimensional, since in that location are 3 die rolls to account for. There are sums ranging from 3 (rolling a 1 on all iii die) to 18 (rolling a 6 on all three die).
The probabilities of these events vary. The most common sum is 10.5 (the expected value).
The probability of rolling a iii is 1/216 (which comes from i/vi times 1/half-dozen times 1/six). The same goes for rolling an eighteen.
What Is The Expected Value Of N Dice Rolls?
The expected value of N dice rolls is 3.5N (assuming we accept the sum of the N die rolls).
When we roll Northward 6-sided dice and sum the results (or roll one die Northward times and sum the results), we go a larger expected value than if we only rolled fewer dice.
If we call the first die roll D1, the 2nd dice roll D2, the tertiary dice gyre D3, etc., then the expected value of their sum Done + D2 + D3 + … + DN is:
- E(D1 + D2 + D3 + … + DN) [accept the expected value of the sum]
- = E(Done) + E(D2) + E(D3) + … + Eastward(DNorth) [expected value of a sum is the sum of the expected values, since expected value is a linear operator]
- = 3.5 + 3.5 + 3.v + … + 3.5 [since the expected value of rolling a half-dozen-sided die is three.5]
- = three.5N [since we added 3.5 N times]
The "table" for rolling North dice would need to be Due north-dimensional, since there are Northward die rolls to account for. There are sums ranging from N (rolling a 1 on all North dice) to 6N (rolling a vi on all Northward die).
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 one/6N (which comes from 1/half-dozen times 1/half-dozen times i/half-dozen …). The aforementioned goes for rolling a total of 6N.
What Is The Expected Value Of The Product Of Two Die Rolls?
The expected value of the product of 2 dice rolls is 12.25 for standard half dozen-sided dice.
When we take the production of two die rolls, we get different outcomes than if we took the sum of the two dice rolls. The tabular array below illustrates the possible products of ii dice rolls:
| 1st/second die roll | one | 2 | iii | 4 | 5 | 6 |
| 1 | 1 | ii | 3 | 4 | 5 | vi |
| ii | 2 | 4 | 6 | 8 | ten | 12 |
| 3 | 3 | 6 | 9 | 12 | 15 | 18 |
| four | 4 | viii | 12 | xvi | 20 | 24 |
| 5 | v | x | xv | 20 | 25 | 30 |
| half-dozen | 6 | 12 | xviii | 24 | 30 | 36 |
of ii dice rolls for 6-sided die.
As before, the probability of each product in the table is one/36 (which comes from 1/6 times 1/6). Taking the product of each table entry with one/36 (or adding up the entries and dividing by 36) gives the states the result of 12.25.
What Is The Expected Value Of The Minimum Of Ii Dice Rolls?
The expected value of the minimum of two die rolls is 91/36 (about 2.53) for standard 6-sided dice.
When we take the minimum of ii dice rolls, we get unlike outcomes than if we took the sum or product of the two dice rolls. The table below illustrates the possible minimums of 2 dice rolls:
| 1st/second dice ringlet | one | ii | 3 | four | 5 | half dozen |
| one | 1 | 1 | ane | 1 | one | i |
| 2 | 1 | 2 | 2 | 2 | 2 | two |
| 3 | 1 | 2 | 3 | three | 3 | 3 |
| 4 | 1 | 2 | iii | four | 4 | 4 |
| 5 | 1 | 2 | 3 | 4 | 5 | 5 |
| half-dozen | 1 | 2 | 3 | iv | v | 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/half-dozen times i/half dozen). Taking the production of each table entry with 1/36 (or adding up the entries and dividing by 36) gives us the effect of 91/36 or about ii.53.
What Is The Expected Value Of A four Sided Die?
The expected value of a dice roll is 2.5 for a standard 4-sided dice (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 four sides, the probability of each outcome is the same: 1/iv. The possible outcomes are the numbers 1 through 4: 1, ii, 3, and iv.
We accept the product of each consequence and its probability then add together upward those products to go the expected value (EV) of rolling a iv sided die:
- EV(roll 4-sided die) = 1(i/four) + 2(one/4) + 3(1/4) + 4(1/4) [sum of products: probability times outcome]
- = 1/iv + 2/4 + 3/4 + 4/iv
- = (1 + 2 + iii + 4) / 4
- = 10 / iv
- = 2.v
The table beneath illustrates the concept of expected value of a four-sided die roll in another way: with outcomes, probabilities, and their products.
| Die Coil | Proba bility | Production |
|---|---|---|
| one | 1/iv | 1/four |
| ii | 1/4 | 2/iv |
| 3 | 1/4 | 3/4 |
| 4 | 1/iv | 4/4 |
| Total | one | x/4 =two.v (EV) |
expected value of a iv-sided dice coil
in another way: with outcomes,
probabilities, and their products.
Note that an upshot of ii.v is non possible with a single dice curlicue. However, if we scroll a 4-sided die many times, the average outcome will be 2.5.
Every bit an case, if nosotros roll a 2 and then a 3, the average of the ii die rolls is (2 + 3) / ii = v / 2 = two.5.
What Is The Expected Value Of An 8 Sided Dice?
The expected value of a dice ringlet is 4.v for a standard 8-sided die (a die with each of the numbers 1 through 8 appearing on exactly one face of the dice).
In this instance, for a off-white die with eight sides, the probability of each outcome is the same: one/viii. The possible outcomes are the numbers 1 through viii: i, 2, 3, 4, five, 6, 7, and eight.
We take the product of each upshot and its probability and so add together up those products to become the expected value (EV) of rolling an 8 sided die:
- EV(roll 8-sided die) = 1(1/8) + 2(1/8) + 3(ane/eight) + 4(1/8) + 5(i/8) + 6(1/8) + 7(1/8) + 8(1/8) [sum of products: probability times outcome]
- = 1/viii + 2/8 + 3/8 + iv/8 + 5/8 + 6/eight + 7/8 + eight/8
- = (1 + 2 + 3 + iv + 5 + vi + vii + 8) / 8
- = 36 / eight
- = 4.5
The tabular array 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 | Production |
|---|---|---|
| one | 1/eight | 1/viii |
| 2 | 1/8 | 2/8 |
| 3 | 1/eight | iii/8 |
| 4 | i/viii | 4/8 |
| 5 | one/8 | 5/8 |
| half-dozen | 1/8 | 6/8 |
| 7 | 1/eight | 7/8 |
| eight | 1/viii | 8/eight |
| Full | 1 | 36/eight =4.5 (EV) |
expected value of a 8-sided die roll
in another way: with outcomes,
probabilities, and their products.
Annotation that an outcome of 4.v is not possible with a single die scroll. However, if nosotros roll an 8-sided die many times, the average effect volition be 4.v.
As an instance, if we roll a 4 and then a 5, the average of the ii dice rolls is (iv + five) / ii = 9 / 2 = 4.5.
What Is The Expected Value Of A 12 Sided Dice?
The expected value of a dice curl is vi.5 for a standard 12-sided die (a die with each of the numbers one through 12 appearing on exactly ane confront of the die).
In this instance, for a fair die with 12 sides, the probability of each outcome is the aforementioned: ane/12. The possible outcomes are the numbers 1 through 12: 1, 2, 3, 4, 5, 6, 7, 8, 9, ten, 11, and 12.
We have the product of each issue and its probability and and then add upwards those products to go the expected value (EV) of rolling a 12 sided die:
- EV(ringlet 12-sided die) = 1(1/12) + ii(1/12) + iii(1/12) + 4(ane/12) + v(1/12) + 6(ane/12) + 7(1/12) + viii(ane/12) + 9(one/12) + x(1/12) + xi(1/12) + 12(1/12) [sum of products: probability times effect]
- = 1/12 + 2/12 + iii/12 + 4/12 + 5/12 + 6/12 + 7/12 + 8/12 + ix/12 + x/12 + 11/12 + 12/12
- = (one + 2 + 3 + 4 + five + 6 + 7 + 8 + 9 + 10 + eleven + 12) / 12
- = 78 / 12
- = 6.5
The table beneath illustrates the concept of expected value of a 12-sided dice roll in another way: with outcomes, probabilities, and their products.
| Die Roll | Proba bility | Production |
|---|---|---|
| one | i/12 | 1/12 |
| 2 | 1/12 | ii/12 |
| 3 | one/12 | iii/12 |
| 4 | ane/12 | 4/12 |
| 5 | 1/12 | five/12 |
| vi | one/12 | vi/12 |
| 7 | 1/12 | 7/12 |
| 8 | 1/12 | 8/12 |
| nine | 1/12 | nine/12 |
| ten | i/12 | 10/12 |
| xi | ane/12 | 11/12 |
| 12 | 1/12 | 12/12 |
| Total | one | 78/12 =six.v (EV) |
expected value of a 12-sided dice curl
in another style: with outcomes,
probabilities, and their products.
Note that an outcome of half-dozen.5 is not possible with a unmarried dice coil. Notwithstanding, if nosotros roll a 12-sided die many times, the average outcome will be 4.5.
Equally an case, if we ringlet a 6 and and so a vii, the boilerplate of the 2 die rolls is (six + 7) / 2 = thirteen / 2 = 6.5.
What Is The Expected Value Of A 20 Sided Die?
The expected value of a dice roll is ten.5 for a standard 20-sided die (a die with each of the numbers 1 through 20 appearing on exactly one face up of the dice).
In this case, for a fair dice with 20 sides, the probability of each outcome is the same: 1/20. The possible outcomes are the numbers 1 through 20: i, two, 3, iv, 5, half dozen, vii, 8, ix, 10, 11, 12, 13, 14, 15, 16, 17, 18, nineteen, and 20.
Nosotros take the product of each consequence and its probability then add together upward those products to get the expected value (EV) of rolling a xx sided die:
- EV(curl xx-sided die) = 1(i/20) + 2(1/20) + 3(1/20) + 4(1/20) + 5(ane/twenty) + half-dozen(1/xx) + 7(i/20) + eight(1/20) + 9(1/20) + x(ane/twenty) + eleven(1/20) + 12(one/20) + 13(one/20) + 14(1/xx) + 15(1/twenty) + 16(1/20) + 17(1/20) + xviii(1/twenty) + 19(one/xx) + xx(1/xx) [sum of products: probability times outcome]
- = 1/20 + ii/20 + 3/20 + 4/20 + 5/20 + 6/twenty + 7/20 + viii/20 + 9/20 + 10/xx + xi/20 + 12/20 + xiii/xx + 14/20 + 15/20 + 16/20 + 17/20 + 18/20 + 19/xx + 20/xx
- = (one + 2 + 3 + iv + 5 + half-dozen + seven + 8 + 9 + 10 + 11 + 12 + 13 + 14 + 15 + 16 + 17 + 18 + 19 + twenty) / twenty
- = 210 / 20
- = x.5
The tabular array below illustrates the concept of expected value of a 20-sided dice roll in another way: with outcomes, probabilities, and their products.
| Dice Roll | Proba bility | Product |
|---|---|---|
| 1 | 1/twenty | 1/20 |
| two | 1/20 | 2/20 |
| 3 | 1/20 | 3/20 |
| 4 | one/20 | 4/20 |
| v | i/twenty | five/20 |
| six | one/20 | six/20 |
| seven | 1/20 | 7/twenty |
| viii | 1/20 | 8/20 |
| 9 | one/20 | 9/xx |
| x | i/xx | x/twenty |
| 11 | one/xx | 11/20 |
| 12 | ane/twenty | 12/xx |
| 13 | 1/twenty | thirteen/20 |
| 14 | one/20 | 14/20 |
| 15 | 1/20 | xv/20 |
| 16 | i/20 | 16/20 |
| 17 | 1/20 | 17/20 |
| 18 | 1/twenty | 18/xx |
| 19 | 1/20 | 19/xx |
| xx | 1/twenty | twenty/20 |
| Total | 1 | 210/20 =10.five (EV) |
expected value of a 20-sided dice roll
in another way: with outcomes,
probabilities, and their products.
Note that an upshot of 10.5 is not possible with a unmarried die ringlet. Yet, if we scroll a 20-sided dice many times, the boilerplate outcome volition be ten.v.
As an example, if nosotros curlicue a 10 and and then an 11, the average of the two die rolls is (ten + xi) / 2 = 21 / 2 = ten.5.
What Is The Expected Value Of An North Sided Die?
The expected value of a dice roll is (N + 1) / 2 for a standard N-sided die (a dice with each of the numbers 1 through N appearing on exactly one confront of the die).
In this case, for a fair die with Northward sides, the probability of each consequence is the same: ane/N. The possible outcomes are the numbers 1 through N: 1, ii, … , North.
We take the product of each issue and its probability and and then add together upward those products to get the expected value (EV) of rolling an N sided die:
- EV(roll N-sided die) = 1(1/N) + two(1/N) + … + N(one/N) [sum of products: probability times upshot]
- = 1/N + ii/N + … + N/N
- = (one + two + … + N / N
- (North(Due north+1)/2) / N [the sum of the numbers i to Due north is given by N(N+1) / two]
- N(Due north+1) / 2N
- (Due north+ane) / 2 [Due north cancels in numerator and denominator]
You tin test this formula for the expected values of each of the die rolls we looked at earlier.
For case, the expected value of a half dozen-sided die (Due north = 6) is iii.5, since (North+1)/2 = (6+1)/ii = vii/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 Die Roll |
|---|---|
| four | two.5 |
| 6 | 3.five |
| eight | 4.v |
| 10 | 5.five |
| 12 | 6.5 |
| xiv | 7.5 |
| sixteen | 8.5 |
| 18 | 9.v |
| xx | x.5 |
| Due north | (N+ane)/2 |
an Due north sided die for diverse values of N.
Conclusion
Now yous know more virtually the expected value of a die roll for diverse N-sided die. You too know how to do the calculations to verify the expected values.
You might also be interested to acquire more than about variance in my commodity here.
You tin can learn more about the uses of expected value in my article here.
I hope yous establish this article helpful. If and so, please share it with someone who tin use the information.
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~Jonathon
Source: https://jdmeducational.com/what-is-the-expected-value-of-a-dice-roll-11-common-questions/
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