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A crooked gambler has nine dice in her coat pocket. Three are fair and six are not. The biased ones are loaded in such a way that the probability of rolling a 6 is 1/2. She takes out one die at random and rolls it twice. Let A be the event “6 appears on first roll” and let B be the event “6 appears on the second roll.” Are A and B independent? Our intuition here would probably answer “yes”: How can two rolls of a die not be independent? For every dice problem we have encountered so far, they have been. But this is not a typical dice problem. Repeated throws of a die do qualify as independent eventsif the probabilities associated with the different faces are known. In this situation, though, those probabilities are not known and depend in a random way on which die the gambler draws from her pocket. To see what effect not knowing which die is being tossed has on the relationship between A and B requires an application of Theorem 2.4.1. Let F and L denote the events “fair die is selected” and “loaded die is selected”, respectively. Then P (A ∩ B) = P (6 on first roll ∩ 6 on second roll) = P (A ∩ B | F) P (F) + P (A ∩ B | L) P (L) Conditional on either F or L, A and B are independent, so P(A ∩ B) = (1/6)(1/6)(3/9) + (1/2)(1/2)(6/9) = 19/108 Similarly, P (A) = P (A | F) P (F) + P (A | L) P (L) = (1/6) (3/9) + (1/2) (6/9) = 7/18 = P (B) But note that P (A ∩ B) = 19/108 = 57/324 ,which is greater than P (A) · P (B) = (7/18) (7/18) = 49/324 proving that A and B are not independent.

So that chunk I copied from a book. I was wondering since A and B are not independent, how the occurrence of one of them has an impact on the other. Computationally, it does convince me that they are not independent, but I couldn't convince myself from the nature of independence. It seems to me that the occurrence of A does not influence the occurrence of B. Any clues would be appreciated.

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This is how I look at it - if A occurs (a 6), then the probability of the die being loaded increases. $$ P(L|A) = \frac{P(L \cap A)}{P(A)} = \frac{P(A)\cdot P(L)}{P(A|L)\cdot P(L)+P(A|F)\cdot P(F)}$$ $$P(L|A) =\frac{ \frac{2}{3} \cdot \frac{1}{2} }{ \frac{2}{3}\cdot\frac{1}{2}+\frac{1}{3}\cdot\frac{1}{6} } = \frac{6}{7}$$ So the odds of having pulled a loaded die went up from an a priori $\frac{2}{3}$ to a revised $\frac{6}{7}$ This increases your odds of the 2nd 6 from $\frac{7}{18}$ to $\frac{19}{42}$

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