Poker Riddle
Posted by Rob Herman at October 18th, 2006
From the XKCD forums. Recast here for clarity and because I am not confident that the poster posted with enough clarity. I think I have the solution but, unlike most days, I am not 100% confident.
For those who are unfamiliar, Texas Hold’em is a particular kind of poker that has become very popular in recent years. Here’s how it works: A single deck is used. Each player gets two cards face-down. There is a round of betting. Three cards are flipped face-up, and there is another round of betting. Another face-up card, another round of betting, then a final face-up card, and a final round of betting. Everyone who is left makes their best 5-card poker hand from their own two hidden cards plus the five common cards on the table, and the best hand wins. You are under no obligation to use any cards from your own hand; if the board contains AKQJT of one suit, all players who have not folded will tie with a royal flush.
To simplify the problem, since betting is not relevant to this riddle: In Texas Hold’em, the goal is to make the best poker hand possible out of your two hidden cards plus five that are face up and common to everyone. High pairs, especially aces and kings, are regarded as excellent hands.
The riddle is this: One player has a pair of kings–a strong hand! Adding as few other players as possible, construct a setup where that player has no chance to win any of the pot. That is, no matter what 5 cards end up being common to all players, he cannot hope to even tie for the best hand.
I think I have the solution that requires 6 other players, which is what the poster alludes to.
I’m answering this riddle… but I don’t have to like it.
Taking betting out of poker is like taking white out of vanilla.
You can just imagine how I felt the first time I opened my mothers cupboard.
I have 4 other players if I know the suits… and five if I don’t.
In poker the hands rank as:
Straight Flush
4 kind
Full-house
Flush
Straight
3 kind
2 pair
pair
IF his hand is:
KS, KD
then the others must have:
AS, AD
QS, QD
KH, KC
QH, QC
Straight flush is out, both Ace & Queens of same suit are taken.
4 of a kind is out since the other kings are taken.
Any full house will be beaten by the pair of Aces.
Same with the Flushes.
the K’s can’t have a straight since A’s & Q’s are taken.
3 of a kind are out due to kings being taken.
any 2 - pair that is picked up can be used by the pair of A’s.
IF I’m not allowed to know what his suit is for some bizzare reason then they must have:
K?, K?
then
A?, A?
A?, A?
k?, k?
Q?, Q?
Q?, Q?
This is the same as above, but I take all 4 A’s and all 4 Q’s
MJB: what if the cards in the Common 5 are
2H, 3H, 4H, 5H, 6H
Then every one has a straight flush 6 high for their highest possible hand and everyone including the KS KD hand ties.
Here is the solution I came up with.
I got it for 6 other players. I am not certain that is optimal.
So you have 2 kings, in poker the suits are symetric so lets say you have KH and KD. If garbage is turned up in the 5 common cards the only thing that will beat you is a pair of aces. So I will give that to player 1 (P1)(you are P0)
P0:KH KD
P1:A A
So how can you beat or tie a pair of aces? I will look at each of the poker hands to make sure P0 can not win.
Pair: well P0 and P1 both have pairs and P1 has better one. So far so good
Two pair: Well to get two pair P0 needs a pair in the common 5 so then P1 will also have the same pair and a pair of aces. Still good
Three of a kind: Ok here is our first issue, if a king is in the common 5 then P0’s three kings will beat P1’s 2 aces. So I will give some other players the other 2 kings so that they can not be in the common 5
P0: KH KD
P1: A A
PX: KS KC
Straight: Another issue. If there is a straight in the common 5 then P0 can win or tie. I will start with the highest straights and work down. To get an ace high straight AQJT must be in the common 5. I will eliminate this possiblity by giving the Aces to other players. To get a King high straight QJT9 must be in the common 5. I will pair one of the Aces with one of the Kings and give it to P2. That way P2s Ace high straight will beat P0s King high straight. A queen high straight becomes a King high straight for P0 and that case has already been dealt with. For a P0 to get a Jack high straight JT987 would need to be in the common 5 so to elimnate this possiblity I will give all the sevens to other players. This also eliminates the possiblity of 7-T high straights. A 6 high straight is the lowest possible with out using an ace and the players with sevens will be able to make a seven high straight to beat P0s six high straight.
P0: KH KD
P1: A A
P2: A KS
PX: A KC 7H 7D 7S 7C
Flush: Not a problem. To get a Flush P0 needs either 4 Hearts, 4 Diamonds, or 5 of any other suit in the common 5. In any of those cases the player with the Ace in that suit will make a better flush.
Full house: Since P2 has the other two kings any full house P0 has would need three of a kind in the common 5 which means P1 can use the same three of a kind with his aces to make a higher full house
Four of a kind: The Kings and Aces are all given to other players so any four of a kind P0 makes will be 2-Q with a king kicker which means any player with an Ace can make the same four of a kind with an Ace kicker to beat P0
Straight flush: The highest Straight flush that P0 can have is King high. So to eliminate this possiblty I will give 9H and 9D to a different player so P0 can not have KQJT9 of the same suit. If there is a Queen high straight flush in spades or Clubs then the Player with the King of that suit will be able to make a higher straight flush. Since all the sevens and Aces are taken the only other straight flush P0 can get is 65432, In which case the pLayer with the appropriate seven will be able to make a higher straight flush.
P0: KH KD
P1: A A
P2: A KS
PX: A KC 9H 9D 7H 7D 7S 7C
The solution (after remaining arbitrary assignment)
P0: KH KD
P1: AH AD
P2: AS KS
P3: AC KC
P4: 9H 9D
P5: 7H 7D
P6: 7C 7S
I was about to concur with your solution, but on further review, it must be noted:
a setup where that player has no chance to win any of the pot. That is, no matter what 5 cards end up being common to all players, he cannot hope to even tie for the best hand.
In this solution, T9876 on the board will tie all with the straight.
*Spoiler - complete solution (I think)*
Add two hands of 66 to break up any relavent low straights (the JT987 straight will be split by the two QQ hands with higher valued straights at QJT98).
We’re up to five other hands. The stated solution value is six. Any pairs or threes or full houses similarly lose to the bullets. The only possible straights will lose to the QQ or 66 hands. As far as a flush goes, any flush which utilizes one of his kings loses to the ace pair hand, and any flush which does not loses to the king pair hand (even if it includes the ace of that suit, it still loses on second card substituting K for low card showing). Any straight flush showing loses as per a straight (all four Q and 6 in hands). As it turns out, we do need the other aces out of the deck. Otherwise JJJJA showing will tie all. Now we’re up to the stated six, and this appears to be complete. Any four of a kind showing is split by the ace pair hands on kicker.
AA
AA
KK
QQ
QQ
66
66
Optional note: J/5, T/4 or 9/3 work as above in place of Q/6. T/5, 9/5, 9/4 also work by breaking all straights so they cannot at all appear instead of topping them to break ties.
Alatar, If I understand your solution you only take to kings out of the deck (for the person who is supposed to lose).
Thus
AA
AA
KK
QQ
QQ
66
66
being 7 hands includes the hand that is supposed to lose.
what if the common 5 cards are K 2 4 7 9. Wouldn’t the “losing” player win with three of a kind?
Hrm. I miscounted something somewhere. Bother. Lost track of hand count, I did mean for the KK hand to be the other pair of kings. So now I’m suboptimal. I think the key is going to be splitting AA, KK into AK, AK so that it isn’t necessary to break up high straights except in the target player’s suits, which appears to be in yours.
I blame random revision midway through. Bah.
A more difficult problem is when the losing player has a pair of aces. I have a solution for that which uses 12 other players (a very large game indeed) It does not feel optimal though can anyone do better?
I think I have it solved at 11 opponents. Still quite large, but it really can’t get much smaller, if at all. I’ve been twiddling around trying for 10, but I’m reasonably convinced that it cannot be done. The threat of a four-card flush to one of the aces forces a lot of cards out of the deck.
Unless specified by “x” or “y”, suits of cards do not matter.
Kx, Ky draws dead to:
A, A
A, K
and the following cards distributed in any order:
A, K, Qx, Qy, 7, 7, 7, 7
no?