I played this game when I was young, but cannot find it online. It is played on a checkers board (e.g. the black squares of a chess board) between two players P and T. The game goes as follows:

  1. P places 7 police within the bottom two rows of the board each at a distinct square.
  2. T places 2 thieves each at some (still) unoccupied square.
  3. P and T then take turns to move, starting from P. No piece is allowed to be moved onto the same square as another piece.
    (a) On P's turn, P must move exactly one police diagonally upwards by one step (like a normal checker piece).
    (b) On T's turn, T must move exactly one thief diagonally by one step (like a checker king). Whichever player cannot make a move loses.

So P wins if P can somehow trap both thieves. And T wins if T can somehow get at least one thief below the police (because police cannot move downwards). The question is, who wins under perfect play? I believe that P can win, but it is very easy to make a mistake. So is my belief correct, and what is the simplest strategy?

  • $\begingroup$ It is called "Fox and Hounds". You can read about it in wikipedia: en.wikipedia.org/wiki/Fox_games Usually there are 1 fox & 4 hounds. In this case hounds will win. $\endgroup$
    – Morris
    Nov 1 '21 at 11:38
  • $\begingroup$ @Morris: Thanks for finding the name, but of course having 2 thieves is very different from having only 1. =) $\endgroup$
    – user21820
    Nov 1 '21 at 14:59
  • $\begingroup$ I think that in the 4 to 1 the 1 wins. Here I think the 2 wins. $\endgroup$
    – Moti
    Nov 6 '21 at 20:34
  • $\begingroup$ @Moti: No, you did not read the wikipedia page that Morris linked. Also, it is very easy for the 4 police to catch the 1 thief... $\endgroup$
    – user21820
    Nov 7 '21 at 7:52

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