Conditional probability with liars

Question

Suppose an urn has $n$ balls of different colors and only one of them is white. Let there be two independent witnesses who speak the truth each with probability $.1$. A ball is selected at random from the urn and its colour is shown to the witnesses. Each asserts it is white. Then find the probability that it is truly white.

The solution is trivial using Bayes' formula but the problem arises when I am computing P(both assert it is white AND it is not). Shouldn't it be just P(it is not white)P(they both assert it is white|it is not)=$[(n-1)/n][0.9]^2$? The correct answer doesn't come using this.

Answer 1

Judging from the "correct" answer you posted in response gnasher729's answer, I think this is the issue: you are not asking the witnesses "is this ball white?" and getting a yes/no answer. You are asking the witnesses "what color is this ball?" and, if the witness doesn't tell the truth, you are getting one of the $n-1$ false answers (since there are $n$ different colors) with equal probability.

(If I'm actually correct here, then this is a really awful problem, because it tells you nothing about what it means for the witnesses to "lie" -- what I said above is just reasoning backwards from the "correct" answer. Since there's no information about how the liars lie, you really don't have enough information to solve this.)

Answer 2

Calculate the probability WT = "Both say it's white, and it is true", and WF = "Both say it's white, but it is false". The conditional probability that it is white when both say it is white is

$WT / (WT + WF)$.

WT and WF are easy enough to calculate. WT = "White ball and both say the truth", WF = "not white ball and both lie". So

$WT = (1/n) * 0.1^2$

$WF = ((n-1)/n) * 0.9^2$

This makes the probability roughly $1 / (81n)$ which is not very much.

Answer 3

Define $W$ as the event that the selected ball is white, $B$ as the selected ball is black, $W_1$ as "witness 1 says it is white" and $W_2$ accordingly. Then we have by Bayes'

$$ P(W\mid W_1, W_2) = \frac{P(W)P(W \cap W_1, W_2)}{P(W_1, W_2)} $$ To compute "white ball and both say it is white", we can condition on $W$: $$ \begin{align} P(W \cap W_1, W_2) &= P(W\cap W_1 \cap W_2)\\ &=P(W_1 \cap W_2 \mid W )P(W). \end{align} $$ From the wording of the question I think one can assume that $W_1$ and $W_2$ are conditionally independent, so \begin{align} P(W_1 \cap W_2 \mid W )P(W) &= P(W_1 \mid W)P(W_2 \mid W)P(W)\\ &= \frac{1}{10^2} P(W). \end{align} So putting it together we have $$ P(W\mid W_1, W_2) = \frac{P(W)^2 /10^2}{P(W_1, W_2)} = \frac{P(W)^2 /10^2}{P(W_1)^2} $$ the last equality coming from independence. By law of total probability, $$ P(W_1) = P(W_1 \mid W)P(W) + P(W_1 \mid B)P(B) = \frac{1}{n} \cdot \frac{1}{10}+\frac{n-1}{n} \cdot \frac{9}{10} = \frac{9n-8}{10n} $$ And the putting it all together, we get $$ P(W\mid W_1, W_2) = \frac{\frac{1}{(10n)^2}}{\left(\frac{9n-8}{10n}\right)^2}=(9n-8)^{-2} $$