Conundrum: Poker Game
Four poker friends played a hand of five-card stud. Each player was dealt one hole card face down, and then four additional cards face up. The cards were dealt, as in standard poker, one at a time around the table, from one regular poker deck. However, instead of betting each round, they decided to deal all twenty cards out in the beginning, and let winner take all!
1. As it turned out, any two consecutive cards dealt in this hand were either different color cards of the same rank or were consecutive ranks of the same suit, considering Aces as high cards only.
2. At least three of the four hole cards were Queens.
3. The last card dealt was a Heart.
4. At least one player was dealt more than one Ten. Nobody was ever dealt a Nine.
5. No Diamond was ever dealt immediately before or after a Spade.
6. Ron was dealt no Clubs, Lenny was dealt no Kings, Nick was dealt at least one Jack, and Frank’s hole card was a Spade.
Who won, and with what hand?
UPDATE: Puzzle solved by ArtVark. See comments for answer.
November 7th, 2007 at 1:48 pm
I think this question was on the LSAT. I got that one wrong, too.
January 11th, 2008 at 8:52 pm
Very nice puzzle.
The order of the people sitting are Lennie, Ron, Frank, and Nick (Lennie gets the first card). The cards are dealt in the following order: Qc,Qh,Qs,Ks,As,Ah,Kh,Kc,Ac,Ad,Kd,Qd,Jd,Td,Tc,Jc,Jh,Js,Ts,Th.
Lennie has: Qc, As, Ac, Jd, Jh — two pair Aces and Jacks
Ron has: Qh, Ah, Ad, Td, Js — pair of Aces
Frank has: Qs, Kh, Kd, Tc, Ts — two pair kings and tens
Nick has : Ks, Kc, Qd, Jc, Th — pair of Kings
So Lennie wins.
January 11th, 2008 at 9:18 pm
Welcome, ArtVark!
Your solution is correct. Very impressive work!
January 13th, 2008 at 5:57 am
Full solution:
We know that there are no nines. Since each new dealt card can only be one rank away from a previous one, and since there are cards higher than a nine already shown, there is no way a card below a nine can be dealt. Therefore all cards have to be one of T, J, Q, K, A. Since there are 20 cards in total dealt and 20 card higher than a nine, every one of the cards has to be used.
If all four hole cards are Q’s then if the next card is a K, it would be impossible to get any Js. If the next card is a J, it would be impossible to get any Ks. Therefore, there cannot be 4 Q’s for all the hole cards. After the third Q, the fourth Q would be the only way of getting from the cards above the Qs to the cards below the Qs (or vice-versa). The fourth Q can only be used once, so all of the cards above or below the Qs must be dealt first, followed by the fourth Q, followed by the cards opposite the previous set.
Based on the fact that if a card has a different rank than the previous card it has the same suit, then the following pattern of 3 dealt cards is impossible: X-Y-X (X’s represent the same rank).
Since there are 3 Q’s in the first four cards, the first four cards must be QQQx or xQQQ. (a pattern such as QQxQ wil violate the previous paragraph).
Now assume that the first five cards are JQQQJ. The next six cards after that have to be 4 T’s and 2 J’s. The seventh card after that has to be a Q, so the sixth cards has to be a J. The only patterns that can satisfy this are JTTTTJ or TTTTJJ (TTJTTJ, for example, would violate the X-Y-X rule). Now if you look at the JQQQJ pattern, you realize that the first two cards are the same suit and the last two cards are the same suit. The middle three Qs have to be red-black-red or black-red-black because two consecutive cards of the same rank have to be the same color. Therefore, the two J’s in the first five cards have to be the same color. The remaining J’s in the next six cards have to be the opposite color. This makes the TTTTJJ pattern impossible because the two J’s at the end have to be opposite colors. The four tens can either be red-black-red-black or black-red-black-red. So in the pattern JTTTTJ, the two J’s which have to be the same suit as the adjacent T’s also have to be opposite colors. Therefore the JTTTTJ pattern is impossible so the assumption that the first five cards are JQQQJ is incorrect.
A mirror image of the cards used in the above paragraph can be used to show that KQQQK is also impossible. Therefore the pattern xQQQx where the two x’s are the same rank cannot happen.
Now consider the case where 4 cards of the same rank are dealt in a row. This would mean that every person would get dealt one and only one card of that rank. This makes 4 T’s being dealt in a row impossible (someone had more than one T) and 4 K’s being dealt in a row impossible (Lennie had no K’s). Also note that if every K or T is either adjacent to the next card of the same rank or a multiple of 4 away froom the next card of the same rank, you will again get the situation where every player gets one card. So these situations cannot happen either.
So now let’s assume that the starting pattern is xQQQy (KQQQJ or JQQQK). After the initial KQQQ, for example, you will have 8 J’s and T’s followed by a Q and a K. There will be exactly 12 cards between the first two K’s so they will be dealt to consecutive players. That means that the following patterns for the seven last cards dealt are impossible because they will cause a K to be dealt to each player: KKKAAAA KKAAAAK KAAAAKK. The remaing possible seven card series of 3 K’s and 4 A’s are either KAAKKAA or KAAAKKA (other patterns would violate the X-Y-X rule and this pattern has to start with a K since the previous card was a Q). Now these two patterns can be shown as KA-AK-KA-A and KA-A-AK-KA where the dashes represent changes in suit. Since each suit change has to be to a suit of a different color, and since the last suit is hearts, the patterns have to be KA(black)-AK(diamonds)-KA(black)-A(heart) or KA(black)-A(diamonds)-AK(black)-KA(hearts). Any other combination would cause two cards of the same suit and rank to appear in the deck. Also the two black clusters have to be separate suits as well. This would force a diamond to be dealt adjacent to a spade, which cannot happen. Therefore the intial pattern of KQQQJ cannot happen.
Again, using the mirror image of the above paragraph, it can be demostrated that the pattern JQQQK cannot happen as well. Here the patterns with four T’s in a row will be illegal. Therefore the xQQQy starting pattern cannot happen. Therefore the cards dealt must start with QQQ. So the pattern of cards dealt will be QQQ followed by 8 J’s and T’s followed by a Q followed by 8 A’s and K’s, or QQQ followed by 8 A’s and K’s followed by a Q followed by 8 J’s and T’s.
So now let’s figure out the order of the 4th through 11th cards. The pattern of these cards has to be KxxxxxxK or JxxxxxxJ where the remaining six cards are the other cards on the same side of the deck (either A’s and K’s or J’s and T’s). The first and last cards of these sequences have to be adjacent in the deal to a Q, which is why they both start and end with J’s or K’s The four A’s or T’s in this case cannot be consecutive because that would cause one K or one T to be dealt to each player. Finally the X-Y-X rule cannot be violated, so the only possible sequence that works is KAAKKAAK (or JTTJJTTJ). Therefore the first twelve cards dealt are: QQQKAAKKAAKQ or QQQJTTJJTTJQ.
Now let’s consider the last 8 cards dealt. They would have to be in the pattern Kxxxxxxx or Jxxxxxxx (the location of the 20th card does not matter). The only possible patterns now (with dashes represent suit changes) are: KA-AK-KA-AK, K-KA-AK-KA-A, K-KA-A-AK-KA, KA-A-AK-K-KA, or KA-AK-K-KA-A. Since the last card must be a heart, it can be demostrated that all of the above patterns except for the last one will result in a spade being adjacent to a diamond. In the last case, the cards will be Kd,Ad,Ac,Kc,Kh,Ks,As,Ah. Therefore the suits of the last eight cards in order will be ddcchssh. Extrapolating further and applying the rules, the only possible suit order is chssshhccdddddcchssh.
So two possible solutions for the order of the deck are:
Qc,Qh,Qs,Js,Ts,Th,Jh,Jc,Tc,Td,Jd,Qd,Kd,Ad,Ac,Kc,Kh,Ks,As,Ah
or
Qc,Qh,Qs,Ks,As,Ah,Kh,Kc,Ac,Ad,Kd,Qd,Jd,Td,Tc,Jc,Jh,Js,Ts,Th
Dealing the first case into hands we get:
Hand one: Qc, Ts, Tc, Kd, Kh
Hand two: Qh, Th, Td, Ad, Ks
Hand three: Qs, Jh, Jd, Ac, As
Hand four: Js, Jc, Qd, Kc, Ah
In this case. Hand two has to be Ron (no clubs). Hand three has to be Lennie (no kings). Hand 4 has to be Frank (only hand left with a spade for a hole card). Hand 1 therefore is Nick’s. However, that hand must have a J and so this card order does not work.
So in the second case, the hands are:
Hand one: Qc, As, Ac, Jd, Jh
Hand two: Qh, Ah, Ad, Td, Js
Hand three: Qs, Kh, Kd, Tc, Ts
Hand four: Ks, Kc, Qd, Jc, Th
Here, Ron is hand two, Lennie is hand one, Nick is hand four, and Frank is hand three. This matches all the specifications of the problem and is the only unique deal that works.
January 13th, 2008 at 7:07 am
Wow.
ArtVark, I want to commend you for such an elegant explanation and to thank you for taking the time to share it with us.
I have an outline of how I thought someone might go about solving the puzzle, which is similiar to what ArtVark did, but slightly different. My solution also doesn’t go into nearly as much detail. It is not an exhaustive proof, as I leave many statements unsupported (if you do this, it leads to this) – though each statement could be backed up if challenged. I also don’t go into detail about how to arrange the cards, but ArtVark does a great job of that in his solution.
As ArtVark explained, the fourth Queen must be used as a bridge between the upper cards and lower cards, so only three of the hole cards can be Queens. Since Frank’s hole card is a Spade, it either is or is next to the Queen of Spades, making that one of the hole cards. The other two must therefore be the Queen of Hearts and the Queen of Clubs.
There are eight possibilities for the non-Queen hole card. It’s either a Jack or a King. It’s either a Spade or a Club. It’s either first or fourth. (For that last one, see ArtVark’s explanation of the X-Y-X pattern.) 2x2x2=8 possibilities.
If it’s a King, Spade, and first, it’s impossible to end on a Heart. The Ace of Spades has only two cards it can be next to. It cannot be next to any Diamond, and so the only cards it can be next to are the Ace of Hearts and the King of Spades. If the King of Spades is dealt first, that would leave the Ace of Spades with only one linking card, making it impossible to end on a heart. You’d have to end on the Ace of Spades.
It can’t be a Jack, Spade, and first for the same reason. You’d have to end on the Ten of Spades.
It can’t be a Jack, Club, and first, because that leads to everybody getting only one Ten, and someone needs to have more than one.
It can’t be a King, Club, and first, because that would lead to everybody having one King, and Lennie doesn’t have one.
Therefore, the first three cards must be Queens. The fourth card is a King or a Jack, and a Club or a Spade.
If the fourth card is the Jack of Clubs, it again leads to everyone getting a Ten. If it is the King of Clubs, it again leads to everyone having a King. Therefore, it is a Spade.
We are now left with only two possibilities. The fourth card is the Jack of Spades or the fourth card is the King of Spades. From there ArtVark’s solution matches mine.
January 14th, 2008 at 4:11 am
I just found a much more elegant way of describing the last eight cards dealt:
JD JC JH JS
TD TC TH TS
The 4×2 table above is a visual representation of the last eight cards. If any of these cards is dealt, the next card in the sequence has to be either to the left, right, above or below that card. The middle two columns touch both suits of the opposite color, so the middle suits cannot be Spades or Diamonds. So the solution will be a path through every card that starts at a J and ends on a heart by only moving left, right, up and down. It can be demonstrated that there is only one such path.
The advantage of this is that it can be also demonstrated that if you remove one of the J’s, the path cannot be found. Removing the JS or JD makes it impossible to get to the TS or TD and end on a heart. Removing the JH makes it impossible to get to from the Clubs to the Spades and also end on a heart. Removing the JC makes it impossible to end on a heart without getting 4 tens in a row. This would account for all the xQQQ deals that I demonstrated were impossible in the first part of my previous description.
January 14th, 2008 at 11:31 am
That is more elegant, and the table method holds up, but doesn’t that presuppose what the last eight cards are going to be?
I suppose you could make the same case with the eight cards above the Queens, by replacing Jacks with Kings, Tens with Aces, and “without getting 4 tens in a row” with “without dealing everyone a King”.
So this would be a visual way to prove xQQQy is impossible, in addition to the proof you provided above. Very nice.