Player $A$ chooses two queens and an arbitrary finite number of bishops on $\infty \times \infty$ chessboard and places them wherever he/she wants. Then player $B$ chooses one knight and places him wherever he/she wants (but of course, knight cannot be placed on the fields which are under attack by $A$).
Then the game starts. First move is a move by player $A$, then by player $B$, and so on...
If $A$ succeeds in finding a trap for $B$ (check-mates him) the game is over and $A$ wins. If $B$ can avoid being check-mated indefinitely then $B$ wins .
Does $B$ always has a winning strategy?
There are two versions of this game:
1) Knight is not allowed to capture figures of $A$.
2) Knight is allowed to capture figures of $A$.
I would very like to see the solution of at least one of those two versions.
For the purposes of this question, suppose whatever version you want.
This is one of my problems, I like to create problems, especially simple ones.
Peter mentioned a very good question in a chat, namely the issue of a draw, so
*) If a knight is not under attack at some field but cannot move anywhere because all the fields where he can move are under attack then that is a draw.
So, $A$ wins if he/she checkmates knight, that is, if she/he attacks knight and knight does not have a field to move on because all are under attack, including the one at which he is at.
Notify me if we can improve this question.
Also, I think that there is an amount of bishops that guarantees the win of $A$, but do not know bounds on the number of bishops that guarantee the win.
And, if knight is not allowed to capture figures of $A$, then I think that two queens and three bishops always have a winning strategy.
Update: We have some strategies for $7$ bishops alone, which would mean that two queens and five bishops are enough, but with two queens the $5$ is too many bishops, Peter has the question on "are only two queens sufficient"? Also, now I think that two queens and two bishops are enough to always secure a winning strategy.