Probability of Truth Given Corroboration
Alice and Bob are witnesses to a crime. When asked questions, Alice tells the truth with probability $p_A$ and lies with probability $1 - p_A$. Bob tells the truth with probability $p_B$ and lies with probability $1 - p_B$. Their truthfulness is independent.
The detective asks Alice a yes/no question. She gives an answer, and Bob then states that Alice's answer is truthful.
Given that $p_A = 1/3$ and $p_B = 3/5$, find the probability that Alice actually told the truth.
Hints
- Bob saying "Alice told the truth" is consistent with two scenarios: both are truthful, or both are lying. What are the probabilities of each?
- Use Bayes' theorem: $P(T \mid E) = P(\text{both truthful}) / P(\text{Bob confirms})$, where the denominator sums over both consistent scenarios.
- Compute $p_A p_B = (1/3)(3/5) = 1/5$ and $(1-p_A)(1-p_B) = (2/3)(2/5) = 4/15$. The answer is $(1/5) / (1/5 + 4/15)$.
Worked Solution
How to Think About It: Bob says "Alice told the truth." But Bob might be lying about that. There are two scenarios consistent with Bob's statement: either Alice really did tell the truth AND Bob is honestly reporting that, or Alice lied AND Bob is also lying (falsely claiming she told the truth). We need the conditional probability that we are in the first scenario.
Quick Estimate: With $p_A = 1/3$ and $p_B = 3/5$: both tell truth has probability $(1/3)(3/5) = 1/5$. Both lie has probability $(2/3)(2/5) = 4/15$. The confirmation event has total probability $1/5 + 4/15 = 3/15 + 4/15 = 7/15$. So the answer is $(1/5)/(7/15) = (3/15)/(7/15) = 3/7 \approx 0.429$.
Approach: Apply Bayes' theorem with the two scenarios that produce Bob's confirmation.
Formal Solution:
Let $T$ = "Alice told the truth" and $E$ = "Bob confirms Alice's answer is truthful."
Bob confirms Alice if either: - Alice told the truth and Bob tells the truth: probability $p_A \cdot p_B$ - Alice lied and Bob lies (falsely says she was truthful): probability $(1 - p_A)(1 - p_B)$
By Bayes' theorem:
$$P(T \mid E) = \frac{P(E \mid T) P(T)}{P(E)} = \frac{p_A p_B}{p_A p_B + (1 - p_A)(1 - p_B)}$$
Plugging in $p_A = 1/3$, $p_B = 3/5$:
$$P(T \mid E) = \frac{\frac{1}{3} \cdot \frac{3}{5}}{\frac{1}{3} \cdot \frac{3}{5} + \frac{2}{3} \cdot \frac{2}{5}} = \frac{\frac{3}{15}}{\frac{3}{15} + \frac{4}{15}} = \frac{3}{7}$$
Answer: $P(\text{Alice told truth} \mid \text{Bob confirms}) = \dfrac{3}{7} \approx 0.4286$.
Intuition
This is a classic Bayesian reasoning problem that tests whether you can identify all scenarios consistent with an observation. The trap is thinking Bob's confirmation simply multiplies Alice's credibility -- it does not. Bob's confirmation is ambiguous: it happens both when everyone is honest and when everyone is lying. The confirmation actually makes Alice's truthfulness LESS certain than her raw $p_A$ if the "both lying" scenario is likely enough.
Notice the result: even though Bob claims Alice is truthful, the posterior probability of truth is only $3/7 \approx 43\%$, which is actually higher than Alice's prior $1/3 \approx 33\%$. Bob's confirmation helps, but not as much as you might hope, because Bob is also unreliable. This has direct parallels in trading: corroborating signals from unreliable sources (noisy indicators) improve your estimate but less than independent reliable signals would. The formula $p_A p_B / [p_A p_B + (1-p_A)(1-p_B)]$ is just Bayes in odds form -- prior odds $p_A/(1-p_A)$ times the likelihood ratio $p_B/(1-p_B)$ -- for one independent binary report, and it appears in information aggregation models throughout market microstructure.