There is a card game where one player gets three cards and only uses two (the third one is discarded without showing to an opponenet), and the second player gets also three cards and uses only two, discarding the third card in the similar way - without showing.(adsbygoogle = window.adsbygoogle || []).push({});

I am trying to enumerate all possible deals, but stuck with the logic. Do I account for the mucked (the third cards) or not?

If I have a 52-card standard deck, the first players hands number would be C(52,3) and the second players number of hands would be C(49,3)? Thus the total number of the deals would be C(52,3)xC(49,3).

What if I am one of the players and know the third discarded card of mine, but don't know the discarded card of my opponent, will it affect the above-mentioned formula?

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# I Two-player card game, combined probability

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