Let $X1,X2,X3$ be i.i.d. taking values in a finite set, and not constant. Is it necessarily true that $P(X3=X2|X2≠X1)≤P(X3=X2)$? Give a proof or a counterexample.
Since the two events $A=\{X3=X2\}$ and $B=\{X2≠X1\}$ are dependent, I believe that it is possible that $A$ is favorable to $B$. However I cannot come up with a counterexample. Any idea? Appreciate in advance!
Keeping the definiton of the events $A$ and $B$ and applying Bayes' Theorem twice $$ P(A | B) = \frac{P(A\cap B)}{P(B)} = \frac{P(B|A)P(A)}{P(B)}. $$ I am answering the question: is it possible to have $P(A|B) > P(A)$ ? From the equation above, that is indeed possible if $$ P(B|A) > P(B) $$ or, in terms of the original random variables, if $$ P(X_2 \neq X_1| X_3 = X_2) > P(X_2 \neq X_1). $$
Example Let us consider 3 doors, with only one door hiding something behind. Let us say that we open the three doors one after another and that $X_1$ is a binomial r.v. modeling the output of first door to be opened (find something / not finding anything), $X_2$ the output of the second try and $X_3$ the final attempt. It seems to me that if we know that the second and third attempt gave the same result (it can only be failure), than $$ P(X_2 \neq X_1| X_3 = X_2) = 1 > P(X_2 \neq X_1) $$