Conditional Geometric Probability Given Excluded Set
Let $X \sim \text{Geometric}(1/2)$ with support on the positive integers, so $P(X = k) = (1/2)^k$ for $k = 1, 2, 3, \ldots$
Let $B = \{1, 2, \ldots, 10\}$ be a set of excluded values. Given that $X \notin B$ (i.e., $X > 10$), compute: $P(X = 12 \mid X \notin B)$
Hints
- Apply the definition of conditional probability: $P(X = 12 \mid X \notin B) = P(X = 12) / P(X \notin B)$. The numerator simplifies immediately since 2 \notin B$.
- For the denominator, note that $\{X \notin B\} = \{X > 10\}$. For $X \sim \text{Geometric}(1/2)$, what is $P(X > 10)$?
- $P(X > 10) = (1/2)^{10}$ because $X > 10$ means the first 10 flips are all tails. The ratio $(1/2)^{12} / (1/2)^{10} = (1/2)^2 = 1/4$.
Worked Solution
How to Think About It: Conditional probability on a discrete random variable is just re-normalizing the PMF to the surviving probability mass. Here, conditioning on $X \notin B$ means we know $X > 10$ -- we are looking at the Geometric distribution truncated below 10. The conditional probability at $X = 12$ is simply the unconditional probability at 12 divided by the total probability mass that survives the conditioning.
Quick Estimate: $P(X = 12) = (1/2)^{12} = 1/4096$. The probability $X > 10$ is $P(X = 11) + P(X = 12) + \cdots = (1/2)^{10} + (1/2)^{11} + \cdots = (1/2)^{10} = 1/1024$ (geometric series). So the conditional probability is $(1/4096) / (1/1024) = 1024/4096 = 1/4$.
Approach: Apply the definition of conditional probability and use the memoryless property of the geometric distribution.
Formal Solution:
By definition of conditional probability: $P(X = 12 \mid X \notin B) = \frac{P(X = 12, \, X \notin B)}{P(X \notin B)}$
Since
2 \notin B$, the event $\{X = 12\}$ is already contained in $\{X \notin B\}$, so: $P(X = 12, \, X \notin B) = P(X = 12) = \left(\frac{1}{2}\right)^{12}$For the denominator, $\{X \notin B\} = \{X > 10\}$ since $B = \{1, \ldots, 10\}$. Thinking of $X$ as the number of flips until the first heads on a fair coin, $X > 10$ is the event that the first 10 flips are all tails: $P(X > 10) = \left(\frac{1}{2}\right)^{10}$
Therefore: $P(X = 12 \mid X \notin B) = \frac{(1/2)^{12}}{(1/2)^{10}} = \left(\frac{1}{2}\right)^{2} = \frac{1}{4}$
Answer: $\boxed{1/4}$.
Equivalently, this reflects the memoryless property: given $X > 10$, the remaining waiting time $(X - 10)$ is again Geometric$(1/2)$. So $P(X = 12 \mid X > 10) = P(X - 10 = 2) = P(X = 2) = (1/2)^2 = 1/4$.
Intuition
The elegant aspect of this problem is that it demonstrates the memoryless property of the geometric distribution in a slightly disguised form. Conditioning on $X > 10$ is equivalent to restarting the coin-flipping process from scratch -- the past 10 failures give you no information about when the first success will occur. So the conditional distribution of $X$ given $X > 10$ is just a shifted Geometric$(1/2)$, and the answer $(1/2)^2$ reflects the probability of waiting exactly 2 more steps.
More broadly, this illustrates the general formula for conditional PMFs: $P(X = i \mid X \notin B) = P(X = i) / P(X \notin B)$ for $i \notin B$. This is just rescaling the probability mass on the surviving set. In finance, this structure appears in conditional tail probabilities -- for example, computing the distribution of a loss given that it exceeds some threshold (the basis of Expected Shortfall / CVaR calculations).