Supercat's answer is correct. Below is a more rigorous proof of its correctness.
An example board is:
Ks Qs Js Kh Jh
On this board, the best possible 2-card holdings, in sorted order, are:
As Ts
(straight flush)
Ts 9s
(straight flush)
Kc Kd
(quad-kings)
Jc Jd
(quad-jacks)
K- Q-
(kings-full-of-queens)
K- J-
(kings-full-of-jacks)
Q- Q-
(queens-full-of-kings)
In order for Alice to make her declaration...
...she must must block all holdings of equal or better value. This means that exactly one of the following must be true:
Case 1: Alice has a straight flush.
Case 2: Alice blocks all straight flushes, and has quad-kings.
Case 3: Alice blocks all straight flushes, blocks quad-kings, and has quad-jacks.
Hands worse than quad-jacks must contain three queens in order to block kings-full-of-queens, but then would fail to block both quads, proving that the above list is exhaustive.
Alice's hand is:
As 9s Kc Kd
Alice's declaration is justified because...
...her hand fits Case 2.
Bob's hand is:
Qh Qd Jd Ts
Bob blocks all straight flushes as well with his Ts
. He knows then that it is either Case 2 or Case 3, which implies that Alice blocks all straight flushes; i.e., Alice must have As 9s
. It cannot be Case 3 because Bob has a J
. So Bob can deduce it is Case 2, and thus that Alice has exactly As 9s Kc Kd
. Referencing the sorted list of holdings above, his holding of queens-full-of-kings must beat Charlie's, as his Jd
blocks quad-jacks, and as Alice's holding blocks all other hands in the list.
So Bob’s declaration is justified.
Charlie's hand is:
Qc Jc - -
Charlie can rule out Case 1, because in that case, Bob would not be able to know all 4 of Alice's cards with certainty. He can also rule out Case 3 because he himself has a J
.
Charlie thus knows it is Case 2, and he also knows that Bob must have arrived at the same conclusion. Charlie thus knows that Bob was able to rule out Case 1, and thus that Alice and Bob mutually block each other's straight flushes. This means one player has the Ts
and the other has the As 9s
. If Bob had the As 9s
, then there is no way that Bob could have deduced all 4 of Alice's cards, as [Ts
+ two kings] only comprises 3 known cards. So, Alice must have the As 9s
, which uniquely constrains her cards to be As 9s Kc Kd
. And Bob must have the Ts
.
Charlie can reason further: in order for Bob to know his hand beats Charlie's, Bob must have known that he was not up against quad-jacks. The only way for Bob to know this is if he has the last J
, the Jd
. Holding just the Jd
and Ts
, then, how could Bob know he beats Charlie? Bob must have known he was not up against queens-full. And the only way for Bob to know this is if he himself has queens-full, which means that Bob has the last two queens in the deck, Qh Qd
.
Thus, Charlie can exactly deduce both Alice's holding of As 9s Kc Kd
and Bob's holding of Qh Qd Jd Ts
.
And so Charlie's declaration is justified.