In a hand of 13 cards from a standard 52 card deck there will always be a certain number of singles, double, triples and quadruples. For example, I might have 6 singles, 2 doubles and a triple (for 13 cards in total) denoted by $a^6b^2c$, by which I mean that for precisely 1 face value (perhaps "King") I have 3 different cards, for precisely 2 more (perhaps "Jack" and "Eight") I have 2 different cards and for another 6 face values I have precisely one card. (My calculations suggest this is in fact the most likely combination, see the SageMath notebook here).
Now envisage a game of four-player President in which instead of playing the cards from a randomly selected hand, players pick up cards from the complete deck laid out on the table until they have a 13 card hand. Clearly, if we allow players to pick up any 13 cards the first player will always win (by first picking up 4 aces, then 4 kings, 4 queens and then 1 jack, at each point an unbeatable choice). Therefore, we must make an additional constraint on the hand that must be collected by specifying the cardinality pattern.
The problem is, for some cardinality patterns it is impossible for all four players to collect them at the same time, whilst for others it is possible but not guaranteed for all players to be able to collect them.
For example, when I tried the game with the constraint above ($a^6b^2c$) I arrived at
P1: QQQ9977K86542
P2: AAAJJ446532 left 32
P3: 000JJ99865432
P4: KKK8877AQ0653
as the hand of each player, player 2 being unable to finish. However, it is in theory possible for all players to finish with this rule:
P1: AAA0088J65432
P2: KKK0088A65432
P3: QQQ9977K65432
P4: JJJ9977Q65432
Clearly the trivial constraint $a^{13}$ (13 singles each) guarantees that all players will always be able to finish simultaneously, but this will give quite a boring game. This can be modified to $a^{10}c$ without issue because any face value for which one player takes a triple will have precisely a single remaining (the final split will be xxx|x||) thus reducing the problem to $a^9$ with 9 face values remaining once all players have taken precisely one triple and one single. (Edit: Not always winnable. See comments). However, doubles (which can split the four cards of a face value between the players either xx|x|x| or xx|xx||) create a lot of unpredictability.
Which other face value cardinalities are guaranteed to allow all 4 players to always finish successfully? Which among these allows for the most possible permutations of hands?