ON THE CARDS

let's reprise

Do you mean, "What is the likelihood that only 3 of those 4 players hold hands containing at least 4 cards of any suit?"

If so, it is zero. You can't deal a hand of 13 cards in 4 suits without having at least 4 cards in at least 1 suit because 4x3=12.

Otherwise, I don't know what you mean by "the one" suit.

Are you asking what the probability is of having one of the suits break 4-4-4-1 or 5-4-4-0?

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