University of Virginia Computer Science
CS150: Computer Science, Fall 2005

Problem Set 2:
Procedurally Predicting Poker Probabilities
Due: Friday, 9 September
Beginning of class

Collaboration Policy - Read Carefully
For this problem set, you are required to work with an assigned partner. You will receive an email Thursday containing the partner assignments.

Before starting to work with your partner, you should go through questions 1 and 2 yourself on paper. When you meet with your partner, check if you made the same predicitions. If there are any discrepancies, try to decide which is correct before using DrScheme to evaluate the expressions.

You and your partner should work together on the rest of the assignment and turn in a single staped document containing both your answers to the questions 1 and 2, and one answer to the rest of the questions with both of your names on it. You should read the whole problem set yourself and think about the questions before beginning to work on them with your partner.

In addition to your partner, you may discuss this assignment with other students in the class and ask and provide help in useful ways. You may consult any outside resources you wish including books, papers, web sites and people except for materials from last year's CS200 course. If you use resources other than the class materials, indicate what you used along with your answer.

You are strongly encouraged to take advantage of the lab hours posted on the CS150 website.

Purpose

These robots are much better than the average player ... It would for sure make money online.
Phil Laak, Los Angeles Times, July 18, 2005.

In this problem set, you will develop procedures that can calculate odds for poker. Actually creating a poker bot involves making decisions based on incomplete information about what the other players have and how they behave. This is much harder than just calculating odds, but knowing the odds is important for any poker strategy.

In the game we will consider, Texas Hold 'Em (this is the poker variant that is used in most major tournaments) each player is dealt two hole cards. These are kept hidden from the other players and can only be used by the player who was dealt them. Then, five community cards are dealt. Every player may use these cards. There are betting rounds after the hold cards are dealt, and after the third, fourth, and final community cards.

At the end of the hand, each player makes their best five-card hand using as many of their own hole cards as they want and the remaining cards from the community cards. So, a player may make a hand using just the five community cards (and none of their hole cards), either of their hole cards and any four of the community cards, or both of their hold cards and any three of the community cards.

To calculate the odds a player will win a hand, we need to know all possible hands that player could get and how many of them beat the other player's hand. For example, when there is one community card left to be dealt, that means we need to consider how many of the remaining cards in the deck will allow the player to make a hand that is better than the other players hand.

Reading: Before going further, you should have finished reading Structure and Interpretation of Computer Programs, Section 2.2. You do not need to read Sections 2.2.3 or 2.2.4, although you may find the examples in those sections helpful and enjoy reading the footnote on page 127-8 about William Barton Rogers, a UVa professor who went on to found MIT. You can see Rogers' cabinet in the rotunda. Apparently, Professor Rogers was quite a teacher — here's a quote from one of his teaching evaluations when he was at UVa (from Francis H. Smith, History of the University of Virginia, 1819-1919, Vol. 2): Who can forget that stream of English undefiled, so smooth, so deep, and yet so clear, that passed from point to point with gentle touch, that commonly flowed along with the quiet of conscious power, yet sometimes became tumultuous with feeling, and then came the music of the cataract and the glory of the rainbow!

Becoming Pros at Cons

The simplest compound data structure is a pair: a way of grouping two things into one. In Scheme, we can make pairs using cons cells. A cons cell has two elements. For silly historical reasons, these are known as the car and cdr:
car elementcdr element
cons cell
The elements in a cons pair can be anything, even another cons pair. By using cons pairs inside cons pairs inside cons pairs we can build up complex data structures to represent any data we want.

Question 1: For this question you should not use the Scheme interpreter. For each fragment below, either:
  1. Explain why the fragment is not a valid Scheme expression; or,
  2. Predict what value the expression will evaluate to. If the expression evaluates to a compound data structure, draw a box-and-pointer picture showing the structure (for example, see Figure 2.3 in SICP).
  1. (cons 100 50)
  2. (cons (cons 1 2) 3)
  3. (cons 1 (cons 2 3))
  4. (car (cons (cons 1 2) null))
  5. (cdr (car (cons (cons 1 2) null)))
  6. (car (cdr (cons (cons 1 2) null)))

Nullifying Null

A list is a pair whose second element is either a list or the special value null (which is also considered a list). (Note that the SICP book uses nil instead of null.)

Scheme provides many useful procedures for manipulating lists. The ones we use in this assignment include:

In later classes we will see how do define these procedures; for now, it is enough to understand what they do so you can use them.

Question 2: For this question you should not use the Scheme interpreter. For each expression below, either:
  1. Explain why the fragment is not a valid Scheme expression; or,
  2. Predict what value the expression will evaluate to. If the expression evaluates to a compound data structure, draw a box-and-pointer picture showing the structure (for example, see Figure 2.3 in SICP).

Assume the following definition is evaluated before each expression below is evaluated: (define intsto3 (list 1 2 3))

  1. (list? (car intsto3))
  2. (list? (cdr intsto3))
  3. (length (car intsto3))
  4. (length (cdr (cdr (cdr intsto3))))
  5. (length (append intsto3 intsto3))
  6. (length (append intsto3 4))
  7. (length (append intsto3 null))
  8. (map (lambda (x) x) intsto3)
  9. (car (map (lambda (x) (> x 2)) (map (lambda (x) (+ x 3)) intsto3)))
  10. (length (apply append (map (lambda (x) (list x (+ 1 x) (+ 2 x))) intsto3))) Note: The procedure (apply Expression1 Expression2) expects the first operand to evaluate to a procedure and the second operand to evaluate to a list. It evaluates to the value Expression1 would evaluate to on operands that are the elements of the list Expression2 evaluates to. For example, (apply + (list 1 2 3)) means the same thing as (+ 1 2 3).
  11. (car (sort > intsto3))

After you have predicted how each expression evaluates, remember to meet with your partner and discuss your results. After you have done this, try evaluating them in DrScheme to check your predictions. If any evaluate differently than you expected, explain why DrScheme evaluates the express the way it does.

Download: Download ps2.zip to your machine and unzip it into your home directory J:\cs150\ps2 (it will take some time to download because it contains many tile images, so start the download and continue reading while it is finishing). See the Lab Guide document for instructions on unziping files and creating directories.

This file contains:

  • ps2.scm - A template for your answers. You should do the problem set by editing this file.
  • poker.scm - Provided code for this problem set. You should examine, but not need to modify, this file.

Representing Cards

To represent a card we need to keep track of both its rank (2 through Ace) and its suit (Hearts, Diamonds, Clubs, or Spades). We can do this using a cons cell: (these definitions are in poker.scm
   (define (make-card rank suit) (cons rank suit))
   (define card-rank car)
   (define card-suit cdr)
We use the numbers 2 through 10 to represent the normal card ranks, and 11 (Jack), 12 (Queen), 13 (King), and 14 (Ace) to represent the special cards. The definitions in poker.scm allow us to use the names Ace, King, Queen, and Jack for these ranks.

Question 3: Define a procedure higher-card? that takes two cards are operands and evaluates to #t if and only if the first card has a higher rank than the second card. Note that the card suit does not matter in this comparison.

If your higher-card? procedure is correct, you should get the following interactions:

> (higher-card? (make-card Ace Diamonds) (make-card King Spades))

#t

> (higher-card? (make-card 3 Clubs) (make-card 2 Clubs))

#t

> (higher-card? (make-card Ace Diamonds) (make-card Ace Spades))

#f

Sorting Hands

In order to evaluate a hand, it will be useful to sort the hand according to the card values.

Question 4: Define a procedure sort-hand that takes one operand that is a list of cards, and evaluates to the cards sorted by decreasing card value (the highest card should be first in the list).

If your sort-hand procedure is correct, you should get the following interactions (the display-cards procedure provided in poker.scm makes it easier to see the cards; the sample hands are also defined in poker.scm):

> (display-cards (sort-hand royal-flush))

"Ah Kh Qh Jh 10h"

> (display-cards (sort-hand ace-high))

"Ad 10s 8c 7c 3s"

Hint: you should not need more than one line for your definition.

Since the rankings of poker hands don't just depend on the card values, but on having pairs, triples, and quads of a given card, it will be more useful to sort the cards according to first the number of duplicates of each rank, and then by rank. For example, if the hand is Ac Js Jc 7d 4c the procedure sort-by-ranks should produce ((Js Jc) (Ac) (7d) (4c)) since the pair of Jacks are more important than the single A. Note that instead of being just a list of cards, the result is now a list of lists of cards.

The procedure sort-by-ranks is defined:

(define (sort-by-ranks cards)
  ;;; sorts cards into lists of cards of each rank, ordered by most
  ;;; cards and highest cards within group
  (sort (lambda (r1 r2) 
          (if (= (length r1) (length r2))
              (higher-card? (car r1) (car r2))
	    (> (length r1) (length r2))))
        (combine-adjacent-matches  ;; combine them into lists of matching rank
	 same-rank?	   
	 (sort-hand cards))))      ;; cards sorted by rank

Ranking Hands

There are two ways that a hand could beat other hands:
  1. It is in a better category of hands (e.g., a flush beats a straight)
  2. It is in the same category of hands, but with higher cards
The ranking of the hand categories is shown next.
CategoryDescriptionExample
Straight FlushFive cards in sequence all of the same suitKs Qs Js 10s 9s
Four-of-a-Kind ("Quads")Four cards of the same rank3h 3d 3c 3s Jd
Full HouseThree cards of matching rank and two different cards of matching rankAc As Ah 7d 7h
FlushFive cards of the same suitQh 10h 8h 3h 2h
StraightFive cards in sequence9d 8h 7c 6h 5s
Wheel StraightA-5 straight5d 4h 3c 2h As
Three-of-a-Kind ("Trips")Three cards of matching rank5d 5h 5c Kh Js
Two PairTwo different pairs of matching rankJd Jh 5c 5h As
PairTwo cards of matching rank8d 8h Ac 7h 3s
High CardAnything elseKd 9h 7c 5h 2s

(Note: we've listed wheel straight as a separate hand category, even though it is usually listed just as a straight. It is the only straight that can use A as a low card, and the lowest possible straight.)

We have provided the higher-hand? procedure that defines the poker rules:

(define (higher-hand? hand1 hand2)
  (cond
    ((straight-flush? hand1)     (or (not (straight-flush? hand2)) 
                                     (and (straight-flush? hand2) 
					  (higher-similar-hand? hand1 hand2))))
    ((four-of-a-kind? hand1)     (or (and (not (straight-flush? hand2)) 
					  (not (four-of-a-kind? hand2)))
                                     (and (four-of-a-kind? hand2)
					  (higher-similar-hand? hand1 hand2))))
    ((full-house? hand1)         (or (and (not (straight-flush? hand2)) 
					  (not (four-of-a-kind? hand2)) 
					  (not (full-house? hand2)))
                                     (and (full-house? hand2) 
					  (higher-similar-hand? hand1 hand2))))
    ((flush? hand1)              (and (not (beats-flush? hand2))
                                      (or (and (flush? hand2) 
					       (higher-similar-hand? hand1 hand2))
                                          (not (flush? hand2)))))
    ((wheel-straight? hand1)     (not (or (any-straight? hand2) (beats-straight? hand2))))
    ((three-of-a-kind? hand1)    (and (not (beats-trips? hand2))
                                      (or (and (three-of-a-kind? hand2)
					       (higher-similar-hand? hand1 hand2))
                                          (not (three-of-a-kind? hand2)))))
    ((three-of-a-kind? hand1)    (and (not (beats-trips? hand2))
                                      (or (and (three-of-a-kind? hand2) 
					       (higher-similar-hand? hand1 hand2))
                                          (not (three-of-a-kind? hand2)))))
    ((two-pair? hand1)           (and (not (beats-two-pair? hand2))
                                      (or (and (two-pair? hand2) 
					       (higher-similar-hand? hand1 hand2))
                                          (not (two-pair? hand2)))))
    ((pair? hand1)               (and (not (beats-pair? hand2))
                                      (or (and (pair? hand2) 
					       (higher-similar-hand? hand1 hand2))
                                          (not (pair? hand2)))))
    (#t                          (and (not (beats-high-card? hand2))
                                      (higher-similar-hand? hand1 hand2)))))
Your job is to define the higher-similar-hand? procedure it uses to compare two hands in the same category.

The rules for comparing poker hands of the same category specify that the most important part of the hand should be compared first. The most important part is the highest card with the highest number of duplicates. If the most important parts are equal, than the next most important part of the hand determines the higher hand. Note that the sort-by-ranks procedure we defined sorts the cards in a hand according to importance, so you can determine the higher hand by considering each element of the list produced by sort-by-ranks in order until you find one that is unequal.

Here are a few examples:

Question 5: Define the higher-similar-hand? procedure, completing the template we have provided in ps2.scm. It should take two poker hands as operands, and can assume the hands are of the same category. It should evaluate to #t if and only if the first hand beats the second hand.

If your procedure is correct, you should get the following interactions:

> (higher-hand? pair-jacks pair-kings)

#f

> (higher-hand? pair-kings pair-jacks)

#t

> (higher-hand? queens-up queens-up)

#f

> (higher-hand? ace-high ace-higher)

#f

> (higher-hand? ace-higher ace-high)

#t

> (higher-hand? kings-full-of-aces kings-full-of-jacks)

#t

Finding the Monster

To find the best possible hand, we will need to consider all possibilities and use the higher-hand? procedure to identify the best hand among them. There are five community cards and two hole cards, so a player can make a hand by using both hole cards and choosing any three of the community cards, by choosing one of the hole cards and any four of the community cards, or by just using the five community cards.

To find possible hands, you will find this procedure (defined in poker.scm useful:

(define (choose-n n lst)
  ;; operands: a number n and a list (of at least n elements)
  ;; result: evaluates to a list of all possible was of choosing n elements from lst
  (if (= n 0) 
      (list null) 
      (if (= (length lst) n) 
          (list lst) ; must use all elements
          (append 
	   (choose-n n (cdr lst)) ;; all possibilities not using the first element
	   (map (lambda (clst) (cons (car lst) clst)) 
		(choose-n (- n 1) ;;; all possibilities using the first element
			  (cdr lst))))))) 
Question 6: Define the possible-hands procedure, completing the template we have provided in ps2.scm. It should take two operands: the first is a list of two cards representing a player's hole cards; the second is a list of 5 cards representing the community cards. It should evaluate to a list of all possible hands that could be made using 0, 1, or 2 of the player's hole cards and enough of the community cards to make a five-card hand.

Note that your possible-hands procedure doesn't depend on the elements of the operands being cards. You may find it easier to test by using scalar values instead. For example, (possible-hands (list 1 2) (list 'a 'b 'c 'd 'e)) should evaluate to a list containing these elements (in any order):

((a b c d e) 
 (1 b c d e) (1 a c d e) (1 a b d e) (1 a b c e) (1 a b c d)
 (2 b c d e) (2 a c d e) (2 a b d e) (2 a b c e) (2 a b c d)
 (1 2 c d e) (1 2 b d e) (1 2 b c e) (1 2 b c d) (1 2 a d e)
 (1 2 a c e) (1 2 a c d) (1 2 a b e) (1 2 a b d) (1 2 a b c))

Now that we have a list of all possible hands, we can find the best hand using the higher-hand? procedure.

Question 7: Define the find-best-hand procedure, completing the template we have provided in ps2.scm. It should take two operands: the first is a list of two cards representing a player's hole cards; the second is a list of 5 cards representing the community cards. It should evaluate the best possible hands that could be made by the player using 0, 1, or 2 of the player's hole cards and enough of the community cards to make a five-card hand.

If your procedure is correct, you should get the following interactions:

> (display-cards (find-best-hand aces-in-hole trip-nines))

"9d 9c 9s Ah Ad"

> (display-cards (find-best-hand big-slick community-clubs4))

"Ac Qc Jc 7c 5c"

> (display-cards (find-best-hand big-slick royal-flush))

"Ah Kh Qh Jh 10h"

For the first hand, both hole cards are used with the three nines to make a full house. For the second hand, the Ace of clubs hole card is used with the four club community cards to make a flush. For the third hand, no hole cards are used and the five community cards make a straight flush.

Hint: as in Question 4, you should be able to define this using only one line. Don't worry about efficiency in your definition (for now).

After the Turn

After the fourth community card (known as turn) has been dealt, there is one more community card to come. To make good decisions, a player needs to know the likelihood that she will have the winning hand after the final card is dealt, but doesn't know what the final card is.

For simplicity, we will assume the player is playing against only one opponent, and has good enough card-reading skills to know the other player's exact hand. (Of course, no real poker player is that good, but in many cases players do have a reasonable guess what cards the other players are holding.)

To analyze a hand, we determine how many of the possible river cards would allow the player to win or draw. The analyze-turn-situation procedure is defined below (and in poker.scm):

(define (accumulate-outs lst)
  ; lst is a list of triples representing  cards
  (if (null? lst) (list null null null)
      (let ((rest-outs (accumulate-outs (cdr lst))))
        (list (append (car (car lst)) 
		      (car rest-outs))
              (append (car (cdr (car lst))) 
		      (car (cdr rest-outs)))
              (append (car (cdr (cdr (car lst))))
		      (car (cdr (cdr rest-outs))))))))
                 
(define (analyze-turn-situation hole1 hole2 community)
  ;; remove all known cards from the deck
  (let ((current-deck (remove-cards (append hole1 hole2 community) 
				    full-deck)))
    ;; we want to find out how many of the remaining cards produce each
    result
    (accumulate-outs
     (map (lambda (river-card) 
            (let ((outcome 
		   (compare-hands?
		    (find-best-hand hole1 
				    (cons river-card community))
		    (find-best-hand hole2 
				    (cons river-card community)))))
              (if (eq? outcome 'higher)
                  (list (list river-card) null null)
                  (if (eq? outcome 'equal)
                      (list null (list river-card) null) ; chop
                      (list null null (list river-card))))))
          current-deck))))
The analyze-turn-situation procedure determines the cards left in the deck by removing the known hole cards and community cards as current-deck. Then, it uses map to try each possible river card from the current deck. If the river card would allow player 1 to produce a better hand than player 2, it puts that card as the first element in the list; if the hands would be equal, it puts that card as the second element; if player 2 would win, it puts that card as the third element. The accumulate-outs procedure combines all the sublists to form a list of all the winning and chopping outs.

Here's an example:

> (show-analysis (analyze-turn-situation connect67 aces-in-hole straight-draw4))

Winning outs (15): 2c 3c 5h 5d 5c 5s 9c 10h 10d 10c 10s Jc Qc Kc Ac
Chopping outs (0):
Losers (29): 2h 2d 2s 3h 3d 3s 4h 4d 4s 6h 6d 6s 7h 7d 7s 8h 8d 8s 9h 9d
Jd Js Qh Qd Qs Kh Kd Ks As three-clubs))

To consider the situation after the flop (the first three community cards have been dealt), we need to look at all possibilities for both the fourth and fifth card. The analyze-flop-situation does this:
(define (analyze-flop-situation hole1 hole2 community)
  ;; operands: hole cards for player 1 and play 2 and community cards
  ;;     there must be 2 cards in each players hole cards and 3 community cards
  ;; result: a list of three elements (winning-outs, chopping-outs, loser) showing
  ;;     the turn and river cards that will lead for the each outcome for player 1.
  (let ((current-deck (remove-cards (append hole1 hole2 community) full-deck)))
    ;; we want to find out how many of the remaining cards produce each
    result
    (map (lambda (turn-card) 
           (analyze-turn-situation hole1 hole2 (cons turn-card
	   community)))
         current-deck)))

Question 8: Predict how long it will take to evaluate an application of analyze-flop-situation (for example, (analyze-flop-situation connect67 aces-in-hole straight-draw3)). You should start by timing analyze-turn-situation. You can do this using the time procedure: (time (show-analysis (analyze-turn-situation connect67 aces-in-hole straight-draw4))). This will print out something like cpu time: 3523 real time: 4823 gc time: 203 (the numbers here are made up for the example). The number after real time gives the number of milliseconds it took to do the evaluation (in this case 4.823 seconds). If you want to try evaluation and timing analyze-flop-situation you can, but you should predict how long it will take first.

Question 9: This implementation is obviously too slow for many practical uses, including building a poker bot. Suggest some approaches you would use if you wanted to make a faster implementation.

Credits: This problem set was developed for UVA CS200 Fall 2005 by David Evans
and tested and improved by Dan Upton.

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CS 150: Computer Science
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