A fair die is rolled repeatedly. Let [math]X[/math] be the number of rolls needed to obtain a 5 and [math]Y[/math] the number of rolls needed to obtain a 6.

Calculate [math]\operatorname{E}(X | Y = 2). [/math]

• 5.0
• 5.2
• 6.0
• 6.6
• 6.8

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.

Step 1: Understanding the Random Variables

The problem describes two independent random variables, [math]X[/math] and [math]Y[/math], which represent the number of rolls needed to obtain a specific outcome from a fair die.

• [math]X[/math]: Number of rolls to obtain a 5.
• [math]Y[/math]: Number of rolls to obtain a 6.

Both [math]X[/math] and [math]Y[/math] follow a geometric distribution with success probability [math]p = \frac{1}{6}[/math], since the die is fair and each roll is independent. The expected value of a geometric distribution is [math]\frac{1}{p}[/math].

Step 2: Analyzing the Condition [math]Y=2[/math]

The condition [math]Y=2[/math] means that the first time a 6 appears is on the second roll. This implies two specific outcomes for the first two rolls:

• The first roll was not a 6.
• The second roll was a 6.

Let's denote the outcome of the [math]k[/math]-th roll as [math]R_k[/math]. So, [math]Y=2[/math] means [math]R_1 \ne 6[/math] and [math]R_2 = 6[/math].

Step 3: Decomposing [math]\operatorname{E}(X | Y = 2)[/math] based on the First Roll

We want to calculate the expected number of rolls to get a 5, given that the first 6 appeared on the second roll. We can use the law of total expectation, considering the outcome of the first roll for [math]X[/math]. Given [math]Y=2[/math] (i.e., [math]R_1 \ne 6[/math] and [math]R_2 = 6[/math]), [math]X[/math] cannot be 2 (because [math]R_2=6 \ne 5[/math]). Therefore, we consider two mutually exclusive scenarios for [math]X[/math]:

• Scenario 1: A 5 is obtained on the first roll ([math]X=1[/math]).
• For this to happen, [math]R_1 = 5[/math].
• This is consistent with [math]R_1 \ne 6[/math] from the condition [math]Y=2[/math].
• The probability of [math]R_1 = 5[/math] given [math]R_1 \ne 6[/math] is:

[[math]]P(X=1 | Y=2) = P(R_1=5 | R_1 \ne 6, R_2=6)[[/math]]

Since rolls are independent, and [math]R_1=5[/math] is compatible with [math]R_1 \ne 6[/math], this simplifies to:

[[math]]P(R_1=5 | R_1 \ne 6)[[/math]]

Given that [math]R_1 \ne 6[/math], the possible outcomes for [math]R_1[/math] are [math]\{1, 2, 3, 4, 5\}[/math]. Each of these outcomes has a probability of [math]\frac{1}{5}[/math]. Therefore, [math]P(X=1 | Y=2) = \frac{1}{5} = 0.2[/math]. In this scenario, [math]\operatorname{E}(X | X=1, Y=2) = 1[/math].

• Scenario 2: A 5 is obtained on the third roll or later ([math]X \ge 3[/math]).
• This occurs if the first roll is not a 5.
• The probability of this scenario is [math]1 - P(X=1 | Y=2)[/math]:

[[math]]P(X \ge 3 | Y=2) = 1 - 0.2 = 0.8[[/math]]

Alternatively, given [math]R_1 \ne 6[/math], the first roll must be [math]R_1 \in \{1, 2, 3, 4\}[/math]. The probability of this is [math]\frac{4}{5} = 0.8[/math].

Step 4: Calculating Conditional Expectation for [math]X \ge 3[/math]

We need to find [math]\operatorname{E}(X | X \ge 3, Y=2)[/math]. In this scenario:

• [math]R_1 \ne 5[/math] (since [math]X \ge 3[/math]) and [math]R_1 \ne 6[/math] (from [math]Y=2[/math]).
• [math]R_2 = 6[/math] (from [math]Y=2[/math]), which also implies [math]R_2 \ne 5[/math].

This means that neither of the first two rolls was a 5. Due to the memoryless property of the geometric distribution, the number of *additional* rolls needed to get a 5, starting from the third roll, is still distributed as a geometric random variable with [math]p=\frac{1}{6}[/math]. The expected number of additional rolls is [math]\frac{1}{p} = \frac{1}{1/6} = 6[/math]. Since two rolls have already occurred, the total number of rolls [math]X[/math] in this case is [math]2 + 6 = 8[/math]. Thus, [math]\operatorname{E}(X | X \ge 3, Y=2) = 8[/math].

Step 5: Final Calculation of [math]\operatorname{E}(X | Y = 2)[/math]

Using the law of total expectation, we combine the results from the two scenarios:

[[math]]\operatorname{E}(X | Y=2) = P(X=1 | Y=2) \times \operatorname{E}(X | X=1, Y=2) + P(X \ge 3 | Y=2) \times \operatorname{E}(X | X \ge 3, Y=2)[[/math]]

Plugging in the values we calculated:

[[math]]\operatorname{E}(X | Y=2) = (0.2 \times 1) + (0.8 \times 8)[[/math]]

[[math]]\operatorname{E}(X | Y=2) = 0.2 + 6.4[[/math]]

[[math]]\operatorname{E}(X | Y=2) = 6.6[[/math]]

Key Insights

• Conditional expectation problems often benefit from decomposing the event space into simpler, mutually exclusive scenarios.
• The memoryless property of the geometric distribution is crucial for calculating conditional expectations like [math]\operatorname{E}(X | X \gt k)[/math]. If [math]X[/math] is geometric, then [math]\operatorname{E}(X | X \gt k) = k + \operatorname{E}(X) = k + 1/p[/math].
• When conditioning on an event that restricts previous outcomes (e.g., [math]Y=2[/math] implies [math]R_1 \ne 6[/math]), probabilities for subsequent events (e.g., [math]R_1=5[/math]) must be calculated conditionally based on the restricted sample space.
• Understanding how the conditioning event [math]Y=2[/math] impacts the possible values of [math]X[/math] (e.g., ruling out [math]X=2[/math]) is essential for setting up the total expectation correctly.

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Solution: D

[math]X[/math] follows a geometric distribution with [math]p = \frac{1}{6}[/math]. [math]Y = 2[/math] implies the first roll is not a 6 and the second roll is a 6. This means a 5 is obtained or the first time on the first roll (probability = 20%) or a 5 is obtained for the first time on the third or later roll (probability = 80%).

[[math]] \operatorname{E}[X | X \geq 3] = \frac{1}{p} + 2 = 6 + 2 = 8, [[/math]]

so [math]\operatorname{E}[X | Y = 2] [/math] equals

0.2(1) + 0.8(8) = 6.6. 

Copyright 2023. The Society of Actuaries, Schaumburg, Illinois. Reproduced with permission.