Note: Knowledge of the poker game Texas Hold-em is (probably) needed in order to answer this question.

You and a friend (only two people are playing) are playing a standard game of Texas Hold-em but with the following rule changes:

  1. The players can see each other's cards.
  2. There is no betting other than an ante of 1 unit per hand.
  3. The dealer (which changes after each hand) gets to choose any two cards they want first, then the other player gets to choose any two cards they want with the exception that they cannot choose any cards of the same rank that the dealer chose. (For example, if the dealer chooses the King and Queen of clubs, the other player cannot choose any King or any Queen, but any of the other 44 cards left in the deck are allowed.)
  4. Once each player has chosen their cards, the remaining 48 cards are shuffled, then the 5 community cards are dealt out and the winner gets the pot, or the pot is split if the hands are tied.

Question 1: What is the optimal strategy for choosing cards when you are the dealer and when you are not the dealer?

Question 2: How would the optimal strategy change, if at all, if the following additional rule was added:

  1. If you choose two aces and you lose, you have to pay an additional 1 unit penalty.

2 Answers 2


Question 1.
Optimal strategy for the dealer:

Pick AA, see below

Probably not quite optimal, but close, strategy for the gambler:

Call the unordered suits: $\{$w,x,y,z$\}$, and
let N and M be ranks below an ace: 2,3,...,11,12,13$=$deuce,trey,...,J,Q,K

If the dealer picks AwAx choose either 6y5y or 6z5z
- these two hands have the highest showdown equity vs AwAx of $23.056\%$
this is because 65's straights are never beaten by a higher straight made by the AA (like JQKTN for T9 or 9TJQK for 78) and it always has a higher straight when the AA hits the wheel, they should choose a suited version that does not match either of the aces to maximise their chances of a flush.

If the dealer picks a suited hand without an A - NwMw, N$\gt$M, pick a pair one higher than her top card, with a minimum value of 5 with one of that same suit - $($max(N,4)$+1)$w$($max(N,4)$+1)\{$x,y,z$\}$
- block and counterfeit as many flushes and straights as possible, the reason to minimise your pair at $55$ is to reduce the double board pair counterfeits (you block the same straights).
worst case (I believe): $81.560\%$ equity (6w6x vs 5w4w - still better than AwAx @ 79.462%)

If the dealer picks an off-suit hand without an A - NwMx, N$\geq$M, do the same but chose one of each of the two suits - $($max(N,4)$+1)$w$($max(N,4)$+1)$x
worst case (I believe): $82.047\%$ (6x6y vs 5x5y)

If the dealer picks AwKw, pick TwTx
- block flushes and Broadway, while lessening the likelihood of being counterfeited (like choosing < TT for board double paired) while maximising your chance of an un-counterfeited straight (9876N, J987N, QJ97N with no T)
with $54.232\%$ equity
Note: JwJx and QwQx have slightly less at 54.206% and 54.117% respectively - while TT can be counterfeited (QQJJN with no T) the extra straights more than compensate.

If the dealer picks AwKx, pick TyTz
- here being able to hit a flush counts for more than blocking a card of either of the suits with $57.276\%$ equity

If the dealer picks AwNw, N$\lt13(\lt$K$)$, pick KwKx
worst case $66.943\%$ equity (KwKx vs Aw5w)

If the dealer picks AwNx, N$\lt13(\lt$K$)$, pick KxKy or KxKz
Once again being able to hit a flush counts for more than blocking the suit, w, of the Ace, but since N$\lt13$ it is best to be able to hit the higher flush in x
worst case $70.513\%$ equity (KxKy or KxKz vs Aw5x)

Note that

If the dealer picks AA (and the gambler acts optimally) she has $1-23.056\%=76.944\%$ equity which (even if there are some slight optimisations I missed) is better than choosing some other hand and letting the gambler act optimally; thus AA is the dealer's optimal choice.

Question 2.
Note: I assume here that, since it is "you and your friend", that the penalty of $1$ goes to the opponent rather than "the house".

Nothing changes (except that the penalties are paid)


If the strategy I outlined above for the gambler is optimal they are currently only choosing AA when the dealer chose KwNw, N$\lt$13($\lt$K$)$ or KwNx, N$\leq$13($\leq$K$)$, in such situations he is faced with choosing to either
(a) take his equity in a pot of $2$ with AA and forfeit $1$ every time he losses, or
(b) take his equity in a pot of $2$ with the next best equity hand he can choose.

There are always $$\prod_{i=0}^{4}{\frac{52-4-i}{i+1}}=1,712,304$$ possible board combinations that could appear, each with equal liklihood, so the question is whether the equity of the AA match up times the pot of $2$ minus the chance of forfeiting $1$ (times $1$) is less than the equity of the next best hand match up times the pot of $2$ - if it is they should play the next best hand instead. We can divide both sides of this inequality by $2$ to get: $$\text{equity of AA}-\frac{\text{losses of AA}}{3,424,608}\lt\text{equity of next best hand}$$

The next best hands are not always that obvious but the worst case for both $\text{equity of AA}$ and $\text{losses of AA}$ from this subset is when facing KwKx ($82.637\%$ and $292,660$) and the best case for the $\text{equity of next best hand}$ is when facing Kw2x, where the next best choice of QxQy or QxQz has an equity of $73.360\%$ and
$0.82637-\frac{292,660}{3,424,608}\approx0.74091 > 0.7336$

So the gambler should pick the flavour of AA suggested above in all such scenarios.

The only times the gambler could choose to play AA otherwise are in the other NwMw and NwMx cases. In those cases he chose other hands because they performed better than AA, so he has no incentive to play AA now with only a downside of a further penalty when he loses.

If the dealer picks AwAx the gambler could win more of the $1$ pots by picking 8y7y, 8z7z, 7y6y or 7z6z but not enough to up his total take from the pots than choosing 6y5y or 6z5z (87 takes 0.68916 , 76 takes 0.68937, 65 takes 0.68981) and the dealer puts up $2$ and takes out $2.31019$ in expectation.

If the dealer does not pick AA the gambler can always win more than half of the $2$ on the table making the choice a losing proposition (unless she knows that the gambler is not going to play optimally!).

So the dealer should still play AA.

Yes, I play a lot of poker - FWIW FL 2-7 triple draw and NL 2-7 single draw are better than NLHE IMO

  • $\begingroup$ This is very close to my line of thinking. I would double check your choice of suits (and I'll do the same). $\endgroup$
    – TTT
    Commented Jun 30, 2016 at 1:59
  • $\begingroup$ Edited for cases of dealer choosing AxKw and AxNw $\endgroup$ Commented Jun 30, 2016 at 2:24
  • $\begingroup$ Getting closer to my answer. I don't think there is ever a benefit to matching the dealer's suits if your card is lower. (You make more flushes with a different suit than you block by removing one possible flush card for the dealer's flush.) I'm referring to your defense against AKs. $\endgroup$
    – TTT
    Commented Jun 30, 2016 at 14:37
  • $\begingroup$ vs AcKc (win, lose, tie, equity): TdTh (919710, 786350, 6244, 53.894%); TcTd (924763, 779838, 7703, 54.232%), TcTd makes a flush ~1.7% of the time (and a bit less than half of those don't win) TdTh makes a flush ~2.25% of the time. The blocker value is more relevant in this one case that you suppose. (The Ak makes a flush 5.6% and 7.26% vs the blocker and no blocker hands.) $\endgroup$ Commented Jun 30, 2016 at 15:11
  • $\begingroup$ Wow- my apologies. My answer for that one case was wrong. Fantastic answer! $\endgroup$
    – TTT
    Commented Jun 30, 2016 at 16:30

Unless I'm mistaken, as the dealer ...

You'll always want to select AA as it will dominate any other hand.

When picking second, you'll want to select ...

9 and 10 of the same suit as this will give you the best chance of making either 2 pair, a flush, or filling out a straight that does not also make a higher one for the dealer's AA. Even then, though, your chances of winning are still only about 22%.

  • $\begingroup$ It's a good start, but note that your opponent may not necessarily be playing the optimal strategy... $\endgroup$
    – TTT
    Commented Jun 29, 2016 at 22:17

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