Expected value when rolling two dice and summing them

Question

Two fair dice are rolled. Let X be the sum of the numbers obtained. Find the expected value of X.

Solution — Step by Step

Each die has faces 1–6. When two dice are rolled, there are $6 \times 6 = 36$ equally likely outcomes. The sum X can range from $1+1=2$ to $6+6=12$ .

Let $X_1$ = outcome of die 1, $X_2$ = outcome of die 2. Then $X = X_1 + X_2$ .

By the linearity of expectation:

E(X) = E(X_1 + X_2) = E(X_1) + E(X_2)

For a single fair die:

E(X_1) = \frac{1+2+3+4+5+6}{6} = \frac{21}{6} = 3.5

Similarly, $E(X_2) = 3.5$ .

Therefore:

E(X) = 3.5 + 3.5 = \boxed{7}

Count outcomes for each sum:

Sum X	Number of ways	Probability P(X)
2	1 (1+1)	1/36
3	2 (1+2, 2+1)	2/36
4	3	3/36
5	4	4/36
6	5	5/36
7	6 (1+6, 2+5, 3+4, 4+3, 5+2, 6+1)	6/36
8	5	5/36
9	4	4/36
10	3	3/36
11	2	2/36
12	1	1/36

E(X) = \sum x \cdot P(X=x) = \frac{1}{36}[2(1)+3(2)+4(3)+5(4)+6(5)+7(6)+8(5)+9(4)+10(3)+11(2)+12(1)]

= \frac{1}{36}[2+6+12+20+30+42+40+36+30+22+12]

= \frac{252}{36} = 7 ✓

The expected sum of two dice is 7.

Notice that 7 is also the most likely single outcome (probability 6/36 = 1/6 — higher than any other sum). For a symmetric distribution centred at 7, the mean equals the mode and median — all are 7.

This is why in games like craps, 7 is such a significant number — it’s both the most likely outcome AND the expected value.

Why This Works

Linearity of expectation is one of the most powerful tools in probability:

E(X + Y) = E(X) + E(Y)

This holds regardless of whether X and Y are independent — no independence assumption required! It works because expectation is a linear operator.

This makes complex expected value problems easy: break the random variable into simpler components, find the expected value of each, and add them up.

For example: expected sum of 10 dice = $10 \times E(\text{one die}) = 10 \times 3.5 = 35$ .

Alternative Method — Symmetry Argument

The distribution of the sum X is symmetric about 7. Why? For every outcome (a, b) with sum $s$ , there is a corresponding outcome (7-a, 7-b)… actually, a cleaner symmetry: if sum = k is achieved by (a, b), then sum = $14-k$ is achieved by (7-a, 7-b). So the distribution is symmetric around 7.

By symmetry, the mean = the centre of symmetry = 7.

In CBSE Class 12 and JEE Main, the linearity of expectation approach is always faster for “sum of dice” or “sum of cards” problems. Instead of building the full probability table, just compute E(single die) and multiply by the number of dice. This saves significant time in exams.

Common Mistake

Students often build the full probability table (36 cases) for the sum of two dice and then laboriously compute E(X) from the table. While correct, this is unnecessarily time-consuming. The linearity trick — $E(X_1 + X_2) = E(X_1) + E(X_2) = 3.5 + 3.5 = 7$ — takes 30 seconds. Learn this approach for all “sum of multiple dice/random variables” problems.

Question

Solution — Step by Step

Why This Works

Alternative Method — Symmetry Argument

Common Mistake

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