# Theory of Expectation :: Problems on Throwing/Rolling Dice : Probability Distribution

 Problem Back to Problems Page
Two unbiased dice are throws together at random. Find the expected value of the total number of points shown up.
(Or) Calculate the expected value of "x", the sum of the scores when two dice are rolled.

 [Expectation: 7 ; Variance: 5.83 ; Standard Deviation: +2.412]

 Solution

Let "x" indicate the sum of the points on a die (which is nothing but the number of points on the die)

The sum/number of points on the dice would be

• 1 if the dice shows up ONE
• 2 if the dice shows up TWO
• 3 if the dice shows up THREE
• 4 if the dice shows up FOUR
• 5 if the dice shows up FIVE
• 6 if the dice shows up SIX

⇒ The values carried by the variable ("x") would be either 1, 2, 3, 4, 5 or 6
⇒ "X" is a discrete random variable with range = {1, 2, 3, 4, 5, 6}

"X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x"

In the experiment of tossing a dice, there are six possible elementary events, the events of the die showing up either ONE, TWO, THREE, FOUR, FIVE or SIX all of which are mutually exclusive, equally likely and exhaustive.

Therefore the probability of occurance of each elementary event is 1/6 Probabilty that the dice would show up

• ONE ⇒ P(ONE) =
 1 6
• TWO ⇒ P(TWO) =
 1 6
• THREE ⇒ P(THREE) =
 1 6
• FOUR ⇒ P(FOUR) =
 1 6
• FIVE ⇒ P(FIVE) =
 1 6
• SIX ⇒ P(SIX) =
 1 6

Probability for the sum/number of points on the dice to be

• 1 ⇒ P(X = 1) =
 1 6
• 2 ⇒ P(X = 2) =
 1 6
• 3 ⇒ P(X = 3) =
 1 6
• 4 ⇒ P(X = 4) =
 1 6
• 5 ⇒ P(X = 5) =
 1 6
• 6 ⇒ P(X = 6) =
 1 6

The probabilty distribution of "x" would be
x 1 2 3 4 5 6
P(X = x)
 1 6
 1 6
 1 6
 1 6
 1 6
 1 6

Calculations for Mean and Variance

x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
1
 1 6
 1 6
1
 1 6
2
 1 6
 2 6
4
 4 6
3
 1 6
 3 6
9
 9 6
4
 1 6
 4 6
16
 16 6
5
 1 6
 5 6
25
 25 6
6
 1 6
 6 6
36
 36 6
Total 1
 21 6
 91 6
= 3.5 = + 15.17

Expectation of the sum/number of points on the dice

⇒ Expectation of "x"
 ⇒ E (x) = Σ px = 3.50
Variance of the sum/number of points on the die
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 15.17 − (3.5)2 = 15.17 − 12.25 = 2.92

Let x1, x2 represent the sum of the points on the first die and the second dice respectivley

Expected sum/number of the points

• On the First Dice
⇒ E (x1) = 3.5
• On the Second Dice
⇒ E (x2) = 3.5

Each trial (throwing of the dice) is identical and therefore the expected sum/number of points on the dice in each trial would be the same

Therefore, expected sum of the points on the two dice

= E (x1) + E (x2)
= 3.5 + 3.5
= 2 × 3.5
= 7

Variance of the sum/number of points

• On the First Dice
⇒ var (x1) = 2.92
• On the Second Dice
⇒ var (x2) = 2.92

Each trial (throwing of the dice) is identical and therefore the variance of the sum/number of points on the dice in each trial would be the same

Variance of the sum of the points on the two dice

= var (x1) + var (x2)
= 2.92 + 2.92
= 2 × 2.92
= 5.84

## Alternative

Where all the trials are identical

• The expected sum of the points is given by

Number of Trials × Expectation in each trial

⇒ Expected sum of the points on two dice

= No. of dice thrown × Expected sum of the points per dice
= n × E (x)
= 2 × 3.5
= 7

• The variance of the sum of the points is given by

Number of Trials × variance in each trial

⇒ variance of the sum of the points on two dice

= No. of dice thrown × variance of the sum of the points on the two dice
= n × var (x)
= 2 × 2.92
= 5.84

### Alternative

Let "x" indicate the sum of the points on the two dice

[Since you are required to find the expected value the sum of the scores on the two dice, the variable would represent the sum of the points on the dice]

The sum of the points on the two dice would be

• 2 if the dice show up {(1,1)}
• 3 if the dice show up {(1,2), (2,1)}
• 4 if the dice show up {(1,3), (2,2), (3,1)}
• 5 if the dice show up {(1,4), (2,3), (3,2), (4,1)}
• 6 if the dice show up {(1,5), (2,4), (3,3), (4,2), (5,1)}
• 7 if the dice show up {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)}
• 8 if the dice show up {(2,6), (3,5), (4,4), (5,3), (6,2)}
• 9 if the dice show up {(3,6), (4,5), (5,4), (6,3)}
• 10 if the dice show up {(4,6), (5,5), (6,4)}
• 11 if the dice show up {(5,6), (6,5)}
• 12 if the dice show up {(6,6)}

⇒ The values carried by the variable ("x") would be either 1, 2, 3, ... , 12
⇒ "X" is a discrete random variable with range = {1, 2, 3, ... , 12}

"X" represents the random variable and P(X = x) represents the probability that the value within the range of the random variable is a specified value of "x"

In the experiment of tossing 2 dice

 Total no. of possible choices = 6 × 6 = 36
Probabilty that the two dice would show up

• {(1,1)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 2) =
 1 36

• {(1,2), (2,1)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 3) =
 2 36

• {(1,3), (2,2), (3,1)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 4) =
 3 36

• {(1,4), (2,3), (3,2), (4,1)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 5) =
 4 36

• {(1,5), (2,4), (3,3), (4,2), (5,1)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 6) =
 5 36

• {(1,6), (2,5), (3,4), (4,3), (5,2), (6,1)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 7) =
 6 36

• {(2,6), (3,5), (4,4), (5,3), (6,2)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 8) =
 5 36

• {(3,6), (4,5), (5,4), (6,3)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 9) =
 4 36

• {(4,6), (5,5), (6,4)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 10) =
 3 36

• {(5,6), (6,5)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 11) =
 2 36

• {(6,6)} =
 No. of favorable choices Total no. of possible choices

⇒ P(Sum= 12) =
 1 36

Probability that the sum of the numbers on the two dice would be

• 2 ⇒ P(X = 2) =
 1 36
• 3 ⇒ P(X = 3) =
 2 36
• 4 ⇒ P(X = 4) =
 3 36
• 5 ⇒ P(X = 5) =
 4 36
• 6 ⇒ P(X = 6) =
 5 36
• 7 ⇒ P(X = 7) =
 6 36
• 8 ⇒ P(X = 8) =
 5 36
• 9 ⇒ P(X = 9) =
 4 36
• 10 ⇒ P(X = 10) =
 3 36
• 11 ⇒ P(X = 11) =
 2 36
• 12 ⇒ P(X = 12) =
 1 36

The probabilty distribution of "x" would be
x 2 3 4 5 6 7 8 9 10 11 12
P(X = x)
 1 36
 2 36
 3 36
 4 36
 5 36
 6 36
 5 36
 4 36
 3 36
 2 36
 1 36

Calculations for Mean and Variance

x P (X = x) px
[x × P (X = x)]
x2 px2
[x2 × P (X = x)]
2
 1 36
 2 36
4
 4 6
3
 2 36
 6 36
9
 18 36
4
 3 36
 12 36
16
 48 36
5
 4 36
 20 36
25
 100 36
6
 5 36
 30 36
36
 180 36
7
 6 36
 42 36
49
 294 36
8
 5 36
 40 36
64
 320 36
9
 4 36
 36 36
81
 324 36
10
 3 36
 30 36
100
 300 36
11
 2 36
 22 36
121
 242 36
12
 1 36
 12 36
144
 144 36
Total 1
 252 36
 1974 36
= 7 = + 54.83

Expectation of the sum of the points on the dice

⇒ Expectation of "x"
 ⇒ E (x) = Σ px = 7
Variance of the sum of the points on the two dice
 ⇒ var (x) = E (x2) − (E(x))2 ⇒ var (x) = Σ px2 − (Σ px)2 = 54.83 − (7)2 = 54.83 − 49 = 5.83
Standard Deviation of the sum of the points on the dice
 ⇒ SD (x) = + √ Var (x) = + √ 5.83 = + 2.412

 Credit : Vijayalakshmi Desu