Find the probability the same color was used twice in a chess game given the player did not lose

Question

Here's the question and its solution:

I don't see how the solution to the problem is to compute:

$[1-P(W|L)]^2+[1-P(B|L)]^2$

i.e. I don't think the expression above reflects what the question is asking and I actually computed this:

$ \begin{align} P((W^1 \cap W^2) \cup (B^1 \cap B^2) \left| L_c^1 \cap L_c^2 \right.) &= \frac{P([(W^1 \cap W^2) \cup (B^1 \cap B^2)] \cap (L_c^1 \cap L_c^2))}{P(L_c^1 \cap L_c^2)} \\ &=\frac{P((W^1 \cap L_c^1) \cap (W^2 \cap L_c^2)) + P((B^1 \cap L_c^1) \cap (B^2 \cap L_c^2))}{P(L_c^1 \cap L_c^2)}\\ &=\frac{\left[P \left(L_c^1 \left| W^1\right.\right)P(W^1) \right]\left[ P \left(L_c^2 \left| W^2 \right. \right)P(W^2)\right] + \left[P \left(L_c^1 \left| B^1\right.\right)P(B^1) \right]\left[ P \left(L_c^2 \left| B^2 \right. \right)P(B^2)\right]}{P(L_c^1 \cap L_c^2)}\\ &=\frac{[(1-0.2)(0.5)][(1-0.2)(0.5)]+[(1-0.3)(0.5)][(1-0.3)(0.5)]}{(0.75)(0.75)}\\ &= \frac{0.4^2+0.35^2}{0.75^2} \\ &= \frac{113}{225} \end{align}$

Why is my attempted solution incorrect?

EDIT:

Where the superscripts denote the number of the game (1st game, 2nd game) and $L_c$ denote the event that the chess player does not lose i.e. win or draw.

Answer 1

The question is to find the probability that you played the two games with the same color knowing that you did not lose the games.

There are two situations here. You could have have played the two games as white, or the two games as black.

The probability that you played black knowing you did not lose for one game is $1-P(B|L)$ (simply the complementary probability of played black knowing you did lose the game). The probability that this happens two times in a a row is $[1-P(B|L)]^2$ since the outcome of the first game is independent to the ouctome of the second game (look up the multiplication rule and make a tree of possibilities if you don't get that).

It's the same thing for the probability that you played white two times in a row knowing you did not lose.

You add $[1-P(B|L)]^2$ and $[1-P(W|L)]^2$ because you are looking for the probability of either of the events I described above to be realised.

Your solution is not correct and is quite mysterious to me...