The Poker Game, solved (I think)

[ralphmerridew]

original puzzle

First, I suggest a slight change to the rules: As compensation for going second, the second player wins if both players draw royal flushes.

With that change, if I haven't made an error, the second player can win against any play by the first. (Otherwise, the game will be a simple draw.)

The first player must prevent the second player from drawing a royal flush. To do that, he must draw at least one card from TJQKA in each suit. That leaves one kicker; WLOG say it is in diamonds.

My basic strategy is as follows: Draw 5 cards of different rank, such that I threaten two straight flushes that share only cards I hold, the higher straight flush beats any SF he can draw to, any draw that blocks my higher straight flush can not beat my lower straight flush, and any draw that blocks both straight flushes has at least three different ranks without being a SF (is at best a flush).

Note: If I draw 5 different ranks, I can get a full house on my second draw against any defense. (First player would have to block ten different cards to do so, and can block at most nine, the five he held initially, and up to four on his second turn.

Call a suit strong if the first player can potentially get a SF to 9 or better on his second turn. (Alternately, the first player either has the ace of that suit or two cards within five of each other.) I block a strong suit by preventing him from drawing to SF to 9 or higher.

First, if all suits are strong, and his doubleton is QJ, JT, J9, or T9, draw KD, QS, JC, 9TH (threatening SF to KH, SF to TH), while he can, at best, draw SF to QD. To block the SF to KH, he must draw KH, QH, or JH and can no longer get an SF. To block the SF to TH, he must draw the 6H, 7H, or 8H. In addition, he must hold an ace or the doubleton (at most one of which matches the card to block the high SF, and neither matches the card to block the low SF).
If all suits are strong, and the doubleton is Jx or Tx, play as above, but take the 9D instead of the KD. While I have a duplicated rank, he would still have to block seven cards to keep me from getting a full house, and he can only block 6. The only full houses he can get while blocking both my straight flushes are jacks full and (kicker)s full; in either case, I can beat that with queens full.

Otherwise:
For the doubleton suit:
If his doubleton is AK, AQ, AT, Ax, KQ, KT, Kx, QT, Qx, I draw the jack of that suit.
If his doubleton is AJ or KJ, I draw the queen of that suit.
If his doubleton is QJ, JT, Jx, or Tx, he has at most three strong suits (or I would have used one of the above strategies. For QJ and JT, draw K8. For J8 or T8, Q7. For J(2-7 or 9) or T(2-7 or 9), draw Q8.

If he has any other strong suits, draw 9T in one, and draw J Q or K in any other strong suits, without matching the cards drawn in diamonds. (If there were two cards drawn in diamonds, then there were at most three strong suits, so this would take at most five cards. Otherwise, there was one card in diamonds, 9T in another suit, and at most 2 other cards.) If you have drawn fewer than five cards, draw any other cards of distinct rank to fill the hand. (This threatens SF to K, SF to T, and full house.)

If he has no other strong suits, draw 56S and enough cards of distinct rank to fill the hand. This threatens SF to 9S, SF to 6S and full house.

I think that covers every case. Can anybody see a flaw?

[Trojan Horse]

I see a flaw, though it's one that I think can be easily plugged:

Say I'm the first player, and you're the second. I take AD, KD, AS, KC, KH. Since my doubleton is an AK, you take the JD. Since I also have spades as a strong suit, you take the TS and the 9S. You now take any other two cards of distinct rank.

Thing is, if you don't choose those two cards carefully, I'll toss my aces and take the KS and the 8S. I've blocked your straight flush, and I have an unbeatable four-of-a-kind. You lose.

Of course, you've got me if, say, you take the AC as one of the two extra cards. I've gotta take the 2C, 3C, 4C, or 5C to block your club straight flush, and I've got 3 cards of different ranks. I'm sunk.

Just cover for cases like that, and I think you've got it.

[ralphmerridew]

I wanted to leave the weak suits out of this as much as possible; alternate blocks more consistent with my approach would be: draw the KS as one kicker, or draw the 8S instead of TS.

Still, the weak point is that I'd glossed over the "must include at least three ranks" part of the proof.

[ralphmerridew]

Wait... drawing the 8s instead of the ten won't work because the other person could then get a royal flush. "For each weak suit, draw the card in my flush suit matching his card in the weak suit" should block it, but might require six cards; "... in the weak suit IF HE HAS AT LEAST TWO CARDS OF THAT RANK." should only take five cards, but might not block all threats.

[Zag]

I was struggling to come up with the counterattack to the original solution, but I got it as I was writing that I couldn't get it.

Here it is:

First player: As Ah Ad Ac Ts
Second player: Td 9d Jc Jh Js

The first player can no longer make a straight flush, and he cannot block the second player's straight flush in diamonds (which could go king-high or 10-high, as needed) without dropping down to three aces. If he does this, the second player makes jacks-full to win.


When I originally read your rule change (that second player wins ties) I had thought it self-evident that the second player would win. It was only after reading your description that I realized it wasn't obvious.

[ralphmerridew]

My rule change was just that the second player wins tied royal flushes. Any tie on a lower hand is a tie.