Expected Sum of Two Dice with Cancellation on Doubles
You roll two fair six-sided dice. If both dice show the same number, the outcome is "cancelled" and you record a sum of $0$. Otherwise, you record the sum of the two dice.
What is the expected value of the recorded sum?
Hints
- Think about decomposing the expectation into what happens when the dice match versus when they don't.
- Use the identity $E[R] = E[S] - E[S \cdot \mathbf{1}_{\text{doubles}}]$ where $S$ is the unrestricted sum.
- The doubles contribute $\frac{1}{36}(2 + 4 + 6 + 8 + 10 + 12) = \frac{42}{36} = \frac{7}{6}$ to the unconditional expectation of $S$.
Worked Solution
How to Think About It: This is a clean conditional expectation problem. Before doing any math, think about what's happening: five-sixths of the time you get a normal dice sum, and one-sixth of the time you get zero. So the answer should be roughly $(5/6) \times (\text{average sum when dice differ})$. The average sum of two dice is $7$, and the cancelled doubles ($2, 4, 6, 8, 10, 12$) are themselves symmetric about $7$, so throwing them out leaves the average of the surviving rolls at exactly $7$. Quick gut: the answer is exactly $(5/6) \times 7 \approx 5.83$.
Quick Estimate: There are 36 equally likely outcomes. Doubles occur 6 times (probability $1/6$), contributing 0 to the expectation. The remaining 30 outcomes contribute their sums. The total sum across all 36 outcomes is $36 \times 7 = 252$ (by symmetry, each die averages $3.5$). The doubles contribute $2 + 4 + 6 + 8 + 10 + 12 = 42$. So the non-doubles contribute $252 - 42 = 210$, giving expected recorded sum $= 210 / 36 = 35/6 \approx 5.833$.
Approach: Use the decomposition $E[\text{result}] = E[S] - E[S \cdot \mathbf{1}_{\text{same}}]$ where $S$ is the ordinary sum.
Formal Solution:
Let $S = X_1 + X_2$ be the sum of two fair dice, and let $D$ be the event that $X_1 = X_2$.
The recorded result is $R = S \cdot \mathbf{1}_{D^c}$, so:
$$E[R] = E[S] - E[S \cdot \mathbf{1}_D]$$
We know $E[S] = 7$.
For the doubles term:
$$E[S \cdot \mathbf{1}_D] = \sum_{k=1}^{6} P(X_1 = X_2 = k) \cdot 2k = \sum_{k=1}^{6} \frac{1}{36} \cdot 2k = \frac{2}{36} \sum_{k=1}^{6} k = \frac{2 \times 21}{36} = \frac{42}{36} = \frac{7}{6}$$
Therefore:
$$E[R] = 7 - \frac{7}{6} = \frac{35}{6}$$
Answer: The expected recorded sum is $\frac{35}{6} \approx 5.833$.
Intuition
This problem is really about the law of total expectation: split the world into mutually exclusive events (doubles vs. not doubles) and handle each piece separately. The slick move is computing $E[S \cdot \mathbf{1}_D]$ directly rather than first finding $E[S | D^c]$ and multiplying by $P(D^c)$ -- both work, but the indicator approach avoids an extra division.
In practice, this decomposition pattern shows up constantly in pricing. If a contract pays the sum of two correlated quantities but cancels under some condition (like a knock-out barrier), you compute the unconditional payoff and subtract the contribution from the knock-out region. It's the same $E[\text{payoff}] = E[\text{full payoff}] - E[\text{payoff} \cdot \mathbf{1}_{\text{knocked out}}]$ structure.