probability - a chess problem The question is from: edx -> 6.0.41 -> unit2 -> solved problem -> A chess tournament problem,
as following:
A chess tournament problem. This year's Belmont chess champion is to be selected by the following procedure. 
Bo and Ci, the leading challengers, first play a two-game match. 
If one of them wins both games, he gets to play a two-game second round with Al, the current champion. 
Al retains his championship unless a second round is required and the challenger beats Al in both games. 
If Al wins the initial game of the second round, no more games are played.

Furthermore, we know the following:
*) The probability that Bo will beat Ci in any particular game is 0.6.
*) The probability that Al will beat Bo in any particular game is 0.5.
*) The probability that Al will beat Ci in any particular game is 0.7.

Assume no tie games are possible and all games are independent.

Part 1. Determine the a priori probabilities that
(a) the second round will be required.
(b) Bo will win the first round.
(c) Al will retain his championship this year.

Part 2. Given that the second round is required, determine the conditional probabilities that
(a) Bo is the surviving challenger.
(b) Al retains his championship.

My question is:
About Part 2 -> (b), according to my intuition, it should simply be 1 - (0.25 + 0.09) = 0.66, but according to P(A|B) = P(A∩B)/P(B), the result is 0.7992 which is also the answer from edx itself.
But still I feel my intuition is correct, can anyone tell me why?

Update - Summary:
I thought for a while, the correct answer by intuition could be:
9/13 * 0.75 + 4/13 * 0.91 = 0.7992

where, 9/13 = 0.36/(0.36+0.16) and 4/13 = 0.16/(0.36+0.16), is the possibility that Bo and Ci get to second round, then 0.75 = 1- (0.5*0.5) and 0.91 = 1-(0.3*0.3) is the possibility that AI win BO and Ci in second round.
 A: If a second round is required, the challenger playing against $A$ (=Al) could be either $B$ (=Bo) or $C$ (=Ci).  If it's $B$, he wins with probability $0.25$, so $A$ retains the championship with probability $0.75$.  If it's $C$, he wins with probability $0.09$, so $A$ retains the championship with probability $0.91$.  Those numbers appear in your answer, but the way you're combining them doesn't make sense, in part because $B$ and $C$ aren't equally likely to be the challenger.  The probability that $A$ retains the championship, given that a second round is required, is
$$
0.75 P({\text{challenger is }}B) + 0.91 P({\text{challenger is }}C)=0.91 - 0.16 P({\text{challenger is }}C).
$$
And the probability that the challenger is $B$ (still given that there is a second round, i.e., that either $B$ or $C$ won both of the first two games) is
$$
P({\text{challenger is }}B)=\frac{0.6^2}{0.4^2+0.6^2}=\frac{9}{13}
$$
(which you should have calculated in part 2a).  So the answer is
$$
0.91 - 0.16\cdot\frac{9}{13}\approx 0.7992.
$$
