For the second round, you must also include the term
$$
P(B_2|A_1^c, B_1, A_2^c)P(A_1^c,B_1,A_2^c) = 1\cdot P(A_1^c) = \frac16,
$$
which is not zero.
When this term is included, $P(B_2) = P(A_2) = \frac23.$
In the third round, your calculation of $P(A_3)$ should include the term
$$ P(A_3|A_1,B_1^c,A_2,B_2^c)P(A_1,B_1^c,A_2,B_2^c) =\frac16$$
while your calculation of $P(B_3)$ should include the terms
$$ P(B_3|A_1,B_1,A_2,B_2,A_3^c)P(A_1,B_1,A_2,B_2,A_3^c) =\frac16$$
and
$$ P(B_3|A_1,B_1,A_2,B_2,A_3^c)P(A_1,B_1,A_2,B_2,A_3^c) =\frac16.$$
When you include these missing terms, $P(A_3) = P(B_3) = \frac12,$
which is the correct probability that each player survives at the end.
Actually the probabilities "in real life" should be $P(A_3) = \frac13,$
$P(B_3) = \frac23,$ because if the first five chambers are empty then when $A$ hands the revolver to $B$ to try the final chamber, $B$ shoots $A$.
Avid readers of MAD Magazine would know this.
A variation on this theme was a plot element of The Deer Hunter.
Best Answer
As Win Vineeth stated, you are given $\mathsf P_B(b)$ for $b\in\{B_1, B_2, B_3\}$, and the tables is for $\mathsf P_{A\mid B}(a\mid b)$. Simply multiply as appropriate using: $$\mathsf P_{A,B}(a, b) = \mathsf P_{A\mid B}(a\mid b)~\mathsf P_B(b)$$
.