This is a question from my actuary exam p manual, and I don't quite understand the solution. The question is:
An inspector has been informed that a certain gambling casino uses a "fixed" deck of cards one-quarter of the time in its blackjack games. With a fair deck, the probability of the casino winning a particular hand is 0.52, but with a fixed deck the probability of winning a particular hand is 0.75. The inspector visits the casino and plays 3 games of blackjack (from the same deck of cards), losing all of them. Find the conditional probability that the deck was fixed given that the inspector lost all three games.
The solution states: A=The deck is fixed X=Number of games lost out of 3 games. We are looking for $P(A|X=3)=\frac{P(X=3|A)P(A)}{P(X=3|A)P(A)+P(X=3|A')P(A')}$, and we are given that P(A)=0.25. Then is deck is fixed, p=0.75, and if deck is fair, then p=0.52.
What I don't understand is why does P(X=3|A)=0.75, and P(X=3|A')=0.52?
Since X is defined to be the event where we lose, shouldn't P(X=3|A)=1-0.75 =0.25 , and P(X=3|A')=1-0.52 = 0.48 Because, if our deck is fair, we have a 52% chance of winning, therefore a 48% chance of losing, and if the deck is fixed, we have a 75% chance of winning, therefore a 25% chance of losing.
$P(X=a)$ measures the number of games that the inspector lost, so you want to use the probability that the house wins. But, in fact, $P(X=3|A)=(0.75)^3$ and $P(X=3|A')=(0.52)^3$, because the probabilities were for individual hands, not three hands. The fact that the inspector lost all three hands is far more persuasive evidence that the deck was marked than losing just a single hand.