# Mind Reading in Three-Handed Pot Limit Omaha

Alice, Bob, and Charlie are playing a game of Pot Limit Omaha. After the river card is dealt, but before the players' hands are revealed, the following conversation takes place:

Alice: "My hand beats Bob’s."

Bob: "Then my hand must beat Charlie’s. Also, I now know your exact 4 cards, Alice."

Charlie: "I now know both Alice's and Bob's exact 4 cards."

What's an example of a situation where this conversation could take place?

(In the game of Pot Limit Omaha, each player is dealt 4 private hole cards, and 5 public community cards are dealt, using a standard 52 card deck. Each player must use exactly two of their hole cards and exactly three of the community cards to make a five card poker hand.)

• Interesting question! Just to clarify, when Charlie says he knows both players' exact hole cards, does that mean he knows the exact cards down to their suits?
– Jafe
Feb 14 at 3:00
• @Jafe Yes, exact means down to the suits. Feb 14 at 4:22
• @Jafe Thank you for the reputation bounty! That was extremely generous! Feb 18 at 16:25

Suppose the board is:

sKQJ hKJ

with Charlie (last to speak) holding

cQJ32

To see how Charlie could infer the contents of Alice and Bob's hands, consider that...

...if Alice were to hold a straight flush, at most two of her cards could be relevant and there would be no way Bob could know the others. If any of her cards were a queen or jack, there would be no way Bob could know which queen or jack she held, since Charlie holds the other.

As a consequence, Alice's winning hand must be

sA9 dK cK

Since Alice can't have a jack, she can't block four jacks and her hand must therefore be better than that. If she held the ten of spades along with the two kings, there would be no way Bob could know her last card.

Bob, meanwhile, must have:

sT hQ dQJ

Charlie can see that no four of a kind is possible, other than the four kings which Bob must know that Alice has. The only hand Bob could have that would beat Charlie's hand is queens full of kings, but Bob couldn't know that was the best hand unless he had both the straight flush and four jacks blocked.

This is far from the only possible arrangement of cards, but if the board and Charlie's hand are as indicated, the remaining cards would be forced.

• Couple nitpicks. You mention quad queens a few times - is this a typo? Also, in a straight flush hand, only two of the cards are relevant. Feb 16 at 2:57
• @dshin: This is a refinement of my first solution; I guess I didn't fix all the prose to match the tweak. Feb 16 at 8:33
• @dshin: I'll have to think about how best to describe the need to have four kings. Feb 16 at 8:40
• I've accepted your answer, as it is correct. I also posted my own version of your solution, with slightly more rigorous reasoning. Feb 16 at 14:46

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.

• Nice write-up. Thanks. I wonder how many meaningfully different variations there are of the solution. I think the prospective straight flush could be of any rank if the board cards had a gap (e.g. 3-4-6-3-6), but perhaps there substantively different arrangements. Feb 16 at 17:46
• I have a rough proof sketch that all solutions look similar (a board with a prospective straight flush, a paired top card, and a paired bottom card). But it's quite complicated, and I'm not keen on trying to write it up. It might be best to validate the hypothesis by computer program. Feb 16 at 19:41
• nitpick: queens full of kings Feb 16 at 23:50
• @DanielMathias thanks, fixed Feb 17 at 0:34

The community cards must:

Hint at four possible high hands, all of which require two specific cards to complete. We can almost get there with two aces, and a 7, 10, J which share a suit with one of the aces. This would hint at the best hand being a royal flush, second best being a 7,8,9,10,J straight flush, third best being four of a kind aces, and fourth best being aces-over-jacks full house. This specific setup leads to a contradiction later, unfortunately.
Even though it's flawed and will eventually collapse, I'll use it as an example.

Suppose the community cards are

7, 10, J, A of Spades and the A of Diamonds.

Alice must have:

...the third best possible hand, and must know that the best and second best hands are impossible.
In our example setup, this would mean that she has the aces of Hearts and Clubs, plus either the Q or K of Spades and either the 8 or 9 of Spades, so she knows that both straight flushes are impossible.

Bob must have:

...the fourth best possible hand, and must know that the best and second best hands are impossible.
In our example setup, this means that Bob has two Jacks, and whichever of Q/K and 8/9 Spades Alice didn't have.

Charlie must:

...be able to determine Bob's hand, and who is holding which disproof cards.
We can get Bob's hand by giving Charlie the other Jack.

Unfortunately, this setup fails because...

No matter what Charlie is holding, he can't determine which of Alice/Bob has the 8 and who has the 9, and similarly for Q and K. He can determine what Alice and Bob's scoring hands are, and what the four unused pocket cards are, but not who has which ones exactly.