Re: Update #1
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Here's an example from the current run: B's hand is 2s3c3d5h6h6s7c7s8d9cJcJhAs. The hands and arrangements in A's latest strategy that don't contain one of these cards are:
9h9sKcKhKs 2d3h4h6d7h ThQdAh
9d9hAcAdAh 2d4s5d6c9s TsJsKh
5c5dTcThTs 2h3s4c6c8h KdKsAc
8h9dTdJsQd 2d4d5d8sTs 4hAdAh
TdThQcQhQs 2h3h4c5c9d 3sKhKs
4d6dQdKdAd 2c4h5s6c7h 9sTsJs
5c5d5s9d9s 2c3h4c6d8h ThKhAc
QcQdKcKdKs 2d4d5dJsAc 9d9h9s
2c4c6cTcKc 2d3h4s5d8c 3sThKd
6cJdJsKcKh 2h4d5d6d9s QdAcAh
TcThQdQhQs 2c3s5d6c9d 6d9hKd
TcJdQcKhAc 2d3h4d7h8h 9dThTs
4d6dJdQdAd 2h5c6c8s9h 4h4sTh
TcThTsKcKd 3s5d8hJdQh 4d4h4s
4dTcTdThTs 2c2d4h8h9s QdQhAc
4c4d4sTdTh 2d3s8c9dQh 9sAcAh
4dJdQhQsKh 2d3s4s5c9h 2h8c8s
8hTsJdKcKd 2c3h5c6d7d QcQhAc
7dJdQdKdAd 3h5c8h9hTs 5sKcKh
etc. (113 possible hands out of 10000) All are treated as equally likely.
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As we proceed through iterations of A1,A2,A3... and B1,B2,B3..., does B keep playing this same hand against the same 113 hands being played by A?
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