Class Notes for Probability & Statistics I (Math 3350)

R. Sinn North Georgia College & State University August 25, 2009

Abstract A poker-based introduction to probability and statistics. Almost all the examples are real poker problems from No Limit Texas Holdem (NLHE) whose answers interest winning players.

1 Basic Probability

1.1 Organizing Principals Professional gamblers and poker players actually use mathematics to earn a living. Career paths for professional gamblers include running casinos and betting parlors in places where such things are legal. They o¤er odds for betting on athletic events and various races. They design and perfect games of chance for the casino ‡oor and verify that the games are fair. Well, some do. Professional poker players are currently the most high pro…le examples of professional gamblers. Ten years ago, few people outside Vegas knew of the legendary Doyle Brunson. Today, many of us can name poker celebrities like Phil Ivey, the "Poker Brat" Phil Hellmuth or Annie Duke having seen them playing high stakes poker on TV. What matematics do they use? They use probability theory to predict the future. They use statistics to analyzes the past. With an understanding of both, they manufacture hundreds of "small edges" over their competition to earn money.

1 A roulette ball falling into its slot and a card drawn from a well-shu• ed deck are two example of probability experiments. A probability experiment is any event whose outcomes are perfectly random and exactly pre- dicted by probabilities. The probability space (or sample space) is the collection of all possible outcomes of the experiment. A simple random sample (SRS) is a collection of outcomes drawn at random from the probability space S. An event is set of outcomes. We use set notation: an event A is a subset of S. Determining probabilities typically devolves into a counting excercise. We simply count the number of outcomes in event A and divide by the number of outcomes in S. To simplify our work, we will use the cardinality operator n(A) to indicate "the number of elements (outcomes) in set (event) A." This leads us to the only formula in discrete probability theory:

n(A) P (A) = n(S)

Example 1 Suppose we draw a single card at random from a well-shu• ed standard deck of 52 cards. What is the probability of drawing a heat?

Let the event (set) H represent the 13 "hearts" outcomes. Then

n(H) 13 1 P (A) = n(S) = 52 = 4

Generally, in probability caclulations, the compound event "A or B" means A B, while "A and B" indicates A B. Independence of the events causes[ a vital divergence. Sets A and B\ are independent provided A B = ?. \

1. For indepedent sets: P (A B) = P (A) + P (B). [ 2. For dependent sets: P (A B) = P (A) + P (B) P (A B). [ \

2 Note that when A B = ?, then by de…nition P (A B) = 0 since \ \ n(?) = 0. The cardinality of set with no elements is zero. We could about the probability of drawing one card and getting both an Ace and an 8. The events are certainly independent. But they can never occur simultaneously. For dependent sets, we typically have to …nd a creative way to count n(A B). Strategically, our counting job is much easier if we can partition the event\ into independent sets. When this is not possible, the formula is based on an idea called the "principal of inclusion-exclusion." The basic problem is that, when we sum, the outcomes in the "overlap" or intersection are counted twice. So we exclude them by subtracting. For three depedenet sets A, B and C, we have:

P (A B C) = P (A) + P (B) + P (C) P (A B) [ P ([A C) P (B C) + P (A B C) \ \ \ \ \ We count all the two-way overlaps twice, and throw them out. In the process we "exclude" the three-way overlap three times - it was added three times - so now it has be included again. The same idea works regardless of how many sets are involved, but the results are tedious and in the long tradition of math texts are therefore left to the reader. It is often far easier to count the outcomes in the complement of event A, e.g. the probability that event A does not occur. We denote set complement as Ac. The formula is:

c P (Ac) = n(A ) = 1 P (A) n(S) In probability calculations, we often …nd the notion of "throwing away" the overlap between two sets useful. This helps us identify independent sets. Given two events A and B such that A B = ?, the event "A but not B" is given by A Bc which we denote as A\ B6. \ n Example 2 Hero holds A J , and ‡op comes J 9 3 . Hero feels he will win with any club| (‡ush),| Jack (set), or Ace~ (two| pair),| but feels otherwise his hand is worse than his opponents. He believes he will lose unless his hand improves. What is his probability of making a winning hand on the next card?

3 Let A represent the event of a turning an Ace, and let C represent the event of turning a club. Since Hero holds the A , A C = ?, and thus P (A C) = P (A) + P (C). Hero can see four clubs,| the\ two in his hand and the two[ on the board. So n(C) = 9. There are three Aces left, so n(A) = 3. We know where …ve the 52 cards in the deck are, so:

P (A C) = n(A)+n(C) = 3+9 = 12 [ n(S) 47 47

Example 3 Hero holds J T , and ‡op comes K Q 3 . Hero feels he will win with a straight, but~ worries} that Villain holds~ two| clubs| (‡ushes beat straights). What is his probability of making the winning hand on the next card?

Let T be the event Hero turns a straight card (any A or 9), and let F be the event a ‡ush card hits (any club). Here, the winning event for Hero is T F , and n(T F ) = 8 2, that is the 8 straight cards from which we "thrown away" the An and the 9 . The solution is 6 . | | 47

Example 4 Hero faces one villain heading to the ‡op where Hero bets last. It is known that the probability of "hitting the ‡op" with any two unpaird cards is approximately 1/3. Hero decides to blu¤ his opponent. What is the probability his opponent "missed" the ‡op and will therefore be likely to fold?

Let H be the event of "hitting the ‡op," e.g. pairing one (or more) of the board cards. The important probability is knowing that P (not H) = 1 2 1 3 = 3 . Aggressive poker players use this likelihood of "unimproved" cards in villain’shand to bet (blu¤) on the ‡op. This is so common it has a name: a continuation bet, or cbet.

4 1.1.1 Odds One of the more confusing aspects of Vegas-style betting is that gamblers tend to state probabilities in terms of odds rather than percentages. A weatherman would say "there’s a 70% chance of rain tomorrow." A gam- ber would lay 7 to 3 odds for rain tomorrow. A percentage probability is 7 based on the fraction, here, 10 . In fractional probabilities, we give the "number of successes" divided by the "total outcomes." For odds, we give a ratio of "number of successes" to "number of failures." The order is important: success to failure. Poker players know a host of odds. For example, Hero faces odds of roughly 7.5 to 1 against ‡opping set when starting with a poket pair (pp). The "against" turns the odds around, meaning there is likely to be "1" success for each "7.5" failures. To convert to a probability, we …nd the denominator by …rst adding successes and failures. Let FS be the event of ‡opping a set given that Hero starts with a pp. Then:

1 1 2 P (FS) = 7:5+1 = 8:5 = 17 = 0:117 65

Suppose Hero holds TT while his opponent holds 55. Hero is about a 4 to 1 favorite. This means that a hand like TT will beat a hand like 55 about 4 times in 5, or 80% of the time. Poker players tend to use both percentage chance of winning (called equity) and odds, depending upon the situation. When betting, knowing one’s odds is more useful since it allows us to match our bet-to-pot ratio and compare it to our successes-to-failure ratio. For example, Hero estimates his chances of winning the pot are 1 in 3, or 2 to 1 against. Villain bets $50 into a $100 pot. Hero must call $50 for a chance to win $150. Hero has 2 to 1 odds (against) making his hand, but the pot is "laying him" 3 to 1 odds ($150 to $50). In this case, it is best for Hero to call, since the odds against him making his hand are lower than the pot odds.

1.1.2 Equity Once you understand the ratio vs. fraction concept, converting odds to probabilities is pretty simple and can usually be at least estimated in your

5 head. Going from percentages to odds can be a bit more tricky. Poker players tend to use various odds calculators like PokerStove (highly recommended, free – google "Poker Stove"). Suppose in a tournament, two players bet all-in pre‡op, Hero has QQ, Villain has AQs. Using "the stove," we …nd that Hero has a 65.725% chance of winning. Rounding a bit, we …nd that Hero’s odds are 65.8 to 34.2, but we simplify so that the …nal number is a 65:8 "1." Dividing both numbers in the ratio by 34.2, we …nd that 34:2 = 1:924 0. Gamblers and poker players would say he had odds of "slightly worse than 2 to 1," call it 1.9 to 1, or round to "2 to 1" for simplicity. Generally, odds are only quoted with only the accuracy needed for whatever bet is being made. Suppose Hero has 22% equity in a hand. What are his odds of winning if he were to go all-in and be called? The answer is that Hero’sodds are 78 78 39 to 22, so we divide both numbers by 22: 22 = 11 = 3: 545 5, or about 3.5 to 1 against. The is also a notion of "fold equity," which is the percentage chance Villain will fold if Hero bets or raises. Of course, fold equity must be estimated, like implied odds must be. Good players add fold equity to their percentage chance of winning the hand when deciding whether to bet or raise. Equity is the percentage of the pot Hero "deserves" if all cards were known to all players, and is equal to Hero’s percentage chance of winning times the size of the pot.

1.2 Poker Primer Understanding the examples in this document requires a basic understanding of poker. For examples in this text, we will use a game poker players call 200nl which means "No Limit Texas Holdem, $200 max buy-in," a game where the small blind (SB) is $1 and the big blind "BB" is $2. All bets must be at least as big as the BB, or $2. This will keep the math simpler, though the calculations apply equally well to all poker games. It is highly recommended that the student wishing to perform well in the probability units for this class spend at least two hours playing free poker online. You should play NLHE and learn the basic mechanics and language of the game. If you have a good deal of experience playing poker, you may skim the following information. However, pay close attention to sections with an asterisk (*) as they are vital to understanding later examples.

6 1.2.1 Cards Poker is played with a standard deck of 52 cards (no jokers or wild cards). The deck has four suits: spades ( ),( ),( ), and ( ). Spades and clubs are black. Hearts and diamonds are red.• The~ cards} have| 13 di¤erent values, from strongest to weakest: A K Q J T 9 8 7 6 5 4 3 2. Aces can be counted as low (A = 1) for the purposes of straights. Most poker books and articles use shorthand for the cards. A typical variation is to indicate the card strength in capital letters or numbers followed by the suit in lowercase letters. The Ace of spades is As, the 8 of hearts is 8h, and so forth. There are times when the suit indicator letter is left o¤ which means no ‡ushes or ‡ush draws come into play. Pre‡op, we often write AK to indicate an Ace and King of di¤erent suits. AKs indicates they are of the same suited, or "suited."

1.2.2 Hands The winning hands in poker are based on …ve card hands, though rarely is poker played with only …ve cards available for each player in modern times. Most often, the player uses seven cards and makes his best 5-card hand from them. The ranking of hands is as follows, from best to worst:

1. Straight Flush (J T 9 8 7 of same suit)

2. Quads or 4-of-a-kind

3. Full House (8 8 8 2 2)

4. Flush (all …ve cards same suit)

5. Straight (all …ve cards "in a row," 9 8 7 6 5)

6. 3-of-a-kind (called set when Hero starts with a pocket pair, trips otherwise)

7. 2 Pair

8. 1 Pair

9. Ace high

7 Ties result when players have identical hands, and the pot is divided equally among the two (or more) tied players. Unlike Bridge, there is no ranking of suits. A 7-high heart ‡ush ties with a 7-high club ‡ush.

1.2.3 Betting In most common poker games, the …rst betting round occurs after the …rst group cards are dealt to each player (2 each in NLHE). Most modern poker games have "forced bets." In stud games, each player antes, and a "bring-in bet" is required of half the size of a standard bet. An ante is a small amount required of each player to participate in the hand. In "holdem" poker games like Texas Holdem and Omaha, forced bets are called "blinds" and are based on your position. The …rst player to the dealer’s left pays the small blind (SB) and the second player to the dealer’sleft pays the big blind (BB). The SB is usually half the BB. Two main types of betting are allowed, one called "limit poker," the other called "no limit poker." In limit poker games, bets and raises are …xed. In early rounds, the …xed bet is equal to the BB. In later rounds, the …xed bet doubles. In no limit poker, a bet of any amount is allowed at any time, leading to the exciting "I’mall in!" moments shown on televised poker tournaments. A raise must be at least as large as the bet that preceeds it. When betting …rst in a round, the bet must be at least as large as the BB. Opening bets have several names. Since the SB and BB are forced, any bet larget than the BB is actually a "raise." When a player just "completes" or matches the BB, he is said to be "limping in" or limping. When a player is said to have opened, we mean he has raised unless an "open limp" is speci…ed. Oddly, a pre‡op reraise (a second unforced bet that raises the player who opened) is called a 3-bet despite being either the 2nd "real" bet or the 4th bet total. A re-reraise is called a 4-bet. In limit poker, a round of betting is usually limitted to 4 total bets, e.g. a bet and 3 raises.

Texas Holdem There are four betting rounds in holdem: pre‡op, ‡op, turn and river. Pre‡op, the blinds "post" or make the forced bets, and two cards are dealt face down to each player. A round of betting begins with the …rst player to the left of the BB. The next …ve cards dealt are all dealt face up in the center of the table. They are community cards and can be

8 used by any player along with 0, 1, or 2 of his own cards to make the best possible 5-card poker hand. The ‡op is dealt: three cards. A betting round follows starting with the SB or, if he has folded, the …rst player to the left of the dealer still active in the hand. The turn is the fourth face up card dealt in the middle, and river is …nal card. Each is followed by a betting round starting with the …rst active player to the dealer’sleft.

Stud The players all post their antes, and the hand begins with two face down cards and one face up card dealt to each player. A betting round begins with a "bring-in" of half the standard bet posted by the player with the highest face up card showing. If two (or more) players have an A, the "best" hand is determined by suit rank (similar to Bridge: spades, hearts, diamonds, clubs in descending order of strength). This is only time suit rank matters in poker. After the bring-in bet, betting proceeds clockwise starting at the left of the bring-in bet. The remaing four cards are dealt one at a time, with a betting round that begins at the strongest hand that’sshowing face up. The …nal (seventh) card is dealt face down. The players who make it to this point have three "hole" cards (face down) and four cards face up. Stud is almost invariably played as limit poker with …xed bets. The …rst three betting rounds have a …xed bet amount typically equal to double the ante. For the last two betting rounds, the standard bet doubles.

Omaha This game is played exactly like Texas Holdem, with two excep- tions. Pre‡op, each player is dealt four cards face down instead of only two. And, at showdown, a player must use exactly two cards from his own hand and exactly three on the board. This leads to many "misread" hands, especially by novices. The betting rounds are identical.

1.2.4 Games * The most popular poker game in the world right now is NLHE, or "no limit" (Texas) holdem. LHE, or limit holdem, is much less popular than it used to be. In between "limit" and "no limit" poker is a variation called "pot limit" where any bet up to the size of the pot is allowed. This allows for huge pots,

9 especially in late betting rounds, like no limit games. In fact, pot limit games play much like no limit, so this document will not explore the di¤erences. Some poker games also include a Hi/Lo option, where the pot is split in half between the player holding the best "high card" hand (according to typical rules) and one holding the best "low card" hand. To qualify a low card hand, a player must have 5 unpaired cards that are 8 or lower, including Aces. Straights (and ‡ushes) do not count. The lowest or "best" low card hand is A 2 3 4 5. This is also a straight, and a player holding it often "scoops" the pot, or wins both halves, the high and the low. Often, no player is able to "qualify" a low card hand. Then the best high card hand wins the entire pot. Omaha and Stud have common Hi/Lo variants. PLO refers to "pot limit Omaha" and is a high card only game. O8B refers to "Omaha 8 or better," meaning it is a Hi/Lo version. Most commonly, this game is limit, but a pot limit version PLO8 is also played. Stud refers to the common high card game. Stud E ("E" means 8 or better) refers to the Hi/Lo variant. The most common …ve games are played in alternating rounds, a practice begun in the "Big Game," a legendary poker game played for the highest stakes by the best poker players in the world and few billionaires who en- joy gambling with them. The game is called HORSE and rotates between Holdem, O8B, Razz, Stud and Stud E. Razz is a low-card only game, with no straights or ‡ushes, where all hands qualify as low. The best low card wins the entire pot in Razz, which is dealt like Stud. The entire array of poker games make for interesting probability questions. The sitation in each will be explained in enough detail to solve the problem. However, the person with no experience playing any poker game at all will be at a disadvantage. Go play some FREE poker online!!

1.2.5 Strategy Several considerations a¤ect every poker decision. Oddly, rather far down the list of important considerations is the actual cards Hero holds. More important are considerations like position, e¤ective stack size and "reads" or an understanding of how opponents play their holdings. In poker, the player who bets the most wins. The cards only break ties. The probability concept of expected value is vital. Poker players know that luck is involved because cards are drawn randomly from a well shu• ed

10 deck. However, when the odds of having cards "hit" the board favor Hero winning more than he has to bet, he is said to have positive expected value. Poker players try to invest heavilty when they have good expected value, and they try to "get away from" hands when they are facing negative expected value situations. Since Hero’s opponents actual cards are only guessed at, Hero can make big mistakes and lose all his chips (called "getting stacked").

Position * Position is so important that many abbreviations have developed to help poker players quickly discuss hands. In casinos, the dealer position rotates, the position marked by a "button" with a "D" on it. So the dealer’sposition is called the button (BTN). The position to the dealer’sleft is the small blind (SB), next is big blind (BB). Those who bet …rst are said to be in early position (EP). Later bettors are middle position (MP) while those betting last are late position (LP). A standard NLHE poker game is played with a table of 9 or 10 players. In a 9-player game, the names of the positions are as follows, starting immediately to the dealer’sleft:

SB - small blind BB - big blind UTG - under the gun (…rst to act in …rst betting round) UTG+1 MP1 - 1st middle position player MP2 - 2nd middle position player MP3 - also called "hijack" seat CO - cuto¤ seat BTN - button or dealer For a 10-handed game, we add UTG+2. A popular variant especially online (where dealer’sdon’tearn wages) is called 6max, where only 6 players play at a table. There, the seats would be labeled as follows:

11 SB BB UTG MP CO BTN

Example 5 A typical "no limit" hand with jargon explained - PLEASE READ!! Hero is in the CO for a hand of 6max 200nl. The SB places his $1 forced bet, the BB his $2. Two cards are dealt to each player. Hero receives TT. UTG folds. MP limps (e.g. "calls" the $2 BB). Hero opens for a standard raise, typically about 4BB + 1BB/limper, meaning Hero tends to open raise $8 (4 x $2 BB) plus an additional BB or $2 for each person who has limped. In this case, Hero bets $10. It folds around to the BB, who calls. He has already posted his $2 BB, so he adds $8 more to "call." The limper folds. The pot is $23, $10 each from Hero and the BB plus the $1 SB and $2 from MP who limped and then folded. The ‡op comes Q T 5. Villain bets $12, and Hero raises to $30 (matches the $12 bet and raises $18 more). Villain decides to call, which means he must add $18 to his $12 to make $30. The turn is a 7, and Villain checks. Hero makes a PSB, or a "pot-sized bet." Since pre‡op $23 was bet with $60 added on the ‡op, the turn came with $83 in the pot. Since both players started (we assume, unless speci…ed othewise) with 100BB or $200, they each have almost $160. Hero wagers about half of his remaining chips, matching the $83 in the pot. This gives Villain 2 to 1 "pot odds," meaning he must put up $83 for a chance to win $166 more. Villain folds. Hero is returned his $83 bet and rakes in the $83 in the pot.

Example 6 A typical "limit" hand with jargon explained - PLEASE READ!! Hero is in the CO for a hand of 6max 200 LHE or limit holdem. The SB places his $1 forced bet, the BB his $2. Two cards are dealt to each player. Hero receives TT. UTG folds. MP limps (e.g. "calls" the $2 BB). Hero opens for a standard raise: 1 BB or $2 more - he puts $4 in the pot. It folds around to the BB, who calls $2 more. The limper calls. The pot is

12 $13, $4 each from Hero, the BB and MP plus the $1 SB. The ‡op comes Q T 5. BB checks (does not bet), MP bets $2, and Hero raises to $4. BB folds, but MP calls $2. The pot after the ‡op is $21.The turn is a 7, and Villain checks. Hero bets, but the standard amount doubles on the turn and river. So Hero bets $4, Villain raises $4 to $8, and Hero reraises $4 to $12. Villain calls, so both players have added $12 to the pot on the turn. Heading the river, there is $45 in the pot. The river is an Ace. Villain checks, Hero bets $4. This gives Villain better than 11 to 1 "pot odds," meaning he must put up only $4 for a chance to win $45 more. Villain calls and shows AQ for top two pair. Hero shows his set of tens, and rakes in the pot.

Stack Sizes Unlike old movies and TV shows, players cannot be "raised out" of a hand.

Example 7 Suppose Hero has a full buy-in (BI) of $200 playing 200nl, but all the other players have only 50 BB, or $100 stacks. Hero holds AA on the button (OTB) and an EP player opens for a standard raise of $8 showing signi…cant strength. Hero 3-bets to $20, everyone else folds except EP who 4-bets to $50. Hero goes "all in," meaning he shoves all $200 worth of his chips into the center of the table. But the EP villain only had $100 to start the hand, and he bet $8 to open and 4-bet $42 more putting a total of $50 in the middle. He only has $50 remaining. So Hero’s "all-in" bet is actually just a $50 bet. If villain calls, Hero’s remaining chips will be counted, and Hero will be returned any chips which Villain cannot match.

The example leads to the notion of "e¤ective stack size." Since a deep stacked player with lots of chips to wager can only bet as much as his opponents have in front of them, "e¤ective stack size" means the smallest amount of chips any player has who is left in hand. The reason stack sizes are so important is due to a phenomenon known as "implied pot odds." Quite possibly the most valuable NLHE hand is the "set," where Hero holds a pocket pair (pp) like the TT’s in the two examples above, then catches a third T on the ‡op - the classic "‡opping a set" that every NLHE player dreams about. Typically, Villain has open raised for something like $10 and Hero decides to call with 22. If villain has $150 in his stack (after opening), Hero is getting 15 to 1 "set odds" on his hand,

13 meaning he stands to win $150 more if the ‡op comes something like A Q 2. Hero’s"implied odds" are good for two reasons. The EP open raise indicates a strong hand for Villain, and the ‡op is likely to have helped Villain’shand much more than Hero’s. So we would expect Villain to continue betting, especially with AK or AQ type hands. Knowing one’simplied odds is impossible, but they are greater whenever stacks are deep, when Villain has a good but second-best hand, and when Villain is aggressive (meaning he prefers to bet and raise rather than call - a sure-…re way to get lots of chips in the pot). In general, Hero needs at least 15 to 1 and sometimes 20 to 1 set odds to call a pre‡op bet with a hand like 33. And hands like 8h 7h (called suited connectors) hit two pair or better on the ‡op once in twenty times. So most poker players suggest at least 30 to 1 e¤ective stack sizes to call a bet with this type of hand.

2 Probability Calculations

We are …nally ready to do some computations. The formula for determining any (discrete) probability is simply:

n(A) P (A) = n(S)

Determining the cardinality of the sets involved - when they are very large - can be di¢ cult. So mathematicians have developed some helpful machinery.

2.1 Repeated Trials Most often in poker, we pick more than card at a time: Hero’s two pocket cards, the three ‡op cards, and so on. These picks of multiple items at once are examples of repeated trials. We repeat the probability experiment a certain number of times and determine the likelihood of some series of events. The repetead trials may be either indepedent or dependent.

14 De…nition: Independent Trials. Let A1,A2, An represent the idea of event A occuring during one of the n repeated trials of a probability experiment S. The experiment is said to be independent provided that for any event in S and for any whole number n, P (A1) = P (A2) = = P (An).

If the probability of events remain …xed for each trial, the events are said to be independent, and draws of samples are said to be conducted with replacement. An example would be shu• ing the deck and drawing a single card. If the card were replaced into the deck before another was drawn, the probability of drawing any card would remain unchanged. For depedent trials, the probability in later trials is a¤ected by earlier draws. Sampling is conducted without replacement. This is the common sampling type for card games, where cards are dealt and remain with the players or on the board until the hand is …nished. Thus, if Hero is a dealt an Ace, the probability of any other Ace appearing is reduced due to the fact that only three - not four - remain in the deck.

2.1.1 Order of Outcomes Suppose we draw three cards without replacement from the thirteen clubs. One outcome is A K Q. In poker, order does not matter, so the event A K Q is identical to the event K Q A. Unordered events are called combinations. In this case, we would say there are "13 choose 3" combinations.

De…nition: Combinations. Suppose we draw a sample of k objects from n objects. The total number of possible combinations or unordered events is given by:

n n! k = k!(n k)! On the other hand, if the events are ordered, then there is a di¤erence between the following six permutations of A K Q:

AKQ = KQA = QAK = AQK = KA Q = QKA 6 6 6 6 6

15 In the case of ordered events, we would say there are "13 permute 3" possibilities.

De…nition: Permutations. Suppose we draw a sample of k objects from n objects. The total number of possible permutations or ordered events is given by:

n! (n k)!

I will conclude with a notational issue. On the graphing calculator, combinations are noted as 13C3 while permutations are noted as 13P3. This is not correct matematical notation. Combinations are related to the binomial n coe¢ cient k , read "n choose k," and should be written in binomial nota- n! tion. For permutations, we will simply write them as (n k)! or refer to them as "n permute k."

2.2 Counting The following series of examples will introduce the way in which these ideas interelate. They are not given in any particular order on purpose,with hopes that you will stop before reading the solution and try to identify whether the samples are drawn with or without replacement and whether the events are ordered or unordered. Performing this simple exercise will help immensely on later tests and quizzes. Also note that, for each example, the event set is explicitly de…ned. This is an important habit to develop so your test and quiz responses are clear to the grader.

Example 8 How many starting hands exist in Texas Holdem?

Let T be the event of drawing two cards from 52. This is sampling without replacement where events are unordered (e.g. AK is equivalent to KA). Then we have "52 cards choose 2" possibilities:

52 52! 52 51 50 49 2 1 52 51 n(T ) = 2 = 2!50! = (2 1) (50 4921) = 2 = 1326 16 Example 9 Hero sits down at a tournament table with 9 other players of skill level ranked 1 through 9, with 9 being best. In how many di¤erent orders of skill level can the players be seated around the table relative to Hero’s position?

Let R be the event of seating all 9 players at random at the table. Once a player has been seated, he remains seated. Therefore, this is sampling without replacement, but here the event is ordered. In poker, money moves from right to left around the table (the power of position). So Hero generally prefers the best players to be to his right. Here, an order of 1 2 3 4 5 6 7 8 9 from Hero’s immediate left to Hero’s immediate right is preferred. We …nd there are "9 permute 9" possibilities:

9! 9 8 7 2 1 n(R) = (9 9)! = 0! = 9 8 7 2 1 = 362; 880

Example 10 How many total ‡ops are possible given that we know Hero’s two pocket cards?

Let F be the event of drawing 3 cards for the ‡op. This is sampling without replacement, and the event is unordered. We know Hero’scards, so the deck has only 50 cards. There are "50 choose 3" possibilities for the ‡op:

50 50! 50 49 48 n(F ) = 3 = 3!47! = 3 2 1 = 19; 600 Example 11 Hero holds ATs. What is the probability that he ‡ops a ‡ush, e.g. that all three ‡op cards are of the same suit as his A and T?

Let F be the event of all 3 cards being the same suit as Hero’s cards. This is sampling without replacement, and the event is unordered. We know two of the cards, so the deck has only 50 cards. Since there are 11 of the 13 suited cards Hero needs (13 minus the 2 in his hand), we …nd there are "11 choose 3" possibilities for him to draw only cards in his suit:

17 11 n(F ) ( 3 ) 33 P (F ) = n(S) = 50 = 3920 0:008418 4 ( 3 )

Example 12 What is the probability of drawing …ve cards and having a ‡ush, e.g. all of cards of the same suit?

Let HF be the event of all 5 cards being hearts. We will …nd the probability of a heart ‡ush, then multiply the result by four since there are four di¤erent suits. (There are other more elegant ways to proceed, which we will cover in the next section.) This is sampling without replacement, and the event is unordered. There are "52 choose 5" possibilities in the probability space. There are "13 choose 5" ways for him to draw only hearts:

13 n(HF ) ( 5 ) 33 4 P (HF ) = n(S) = 52 = 66 640 4:952 0 10 ( 5 )

Then,

4 33 33 3 P (F ) = 4 P (HF ) = = 1:980 8 10 66 640 16 660

Example 13 What is the probability of drawing seven cards and having a 5-card ‡ush, e.g. 5 cards of the same suit?

This is the more typical example in modern poker. There are actually three seperate probability quesions here. Hero has a ‡ush as long as at least …ve cards are of the same suit. As rare as it might be, Hero can have six or seven cards of the same suit and make a ‡ush with the best …ve. We will create three events, and combine them.This is sampling without replacement, and the event is unordered. Let HF 5 be the event of having …ve hearts and any two other cards. Let HF 6 be the event of having six hearts and any other card. Let HF 7 be the event of having all seven cards be hearts. We are trying to …nd

18 P (HF 5 OR HF 6 OR HF 7) = P (HF 5 HF 6 HF 7). [ [

Again, we will …nd the probability of any heart ‡ush, then multiply by four. There are "52 choose 7" possibilities in the probability space. There are "13 choose 5" ways for him to draw …ve hearts and "39 choose 2" ways to draw any other two cards:

13 39 n(HF 5) ( 5 )( 2 ) 73;359 P (HF 5) = n(S) = 52 = 10;291;120 ( 7 )

We proceed the same way for a 6-‡ush and a 7-‡ush:

13 39 n(HF 6) ( 6 )( 1 ) 1;287 P (HF 6) = n(S) = 52 = 2;572;780 ( 7 )

13 39 n(HF 7) ( 7 )( 0 ) 33 P (HF 7) = n(S) = 52 = 2;572;780 ( 7 )

Because it is impossible to have exactly …ve hearts and exactly six (or seven) hearts simulatneoulsy, the events HF 5, HF 6 and HF 7 are mutually independent. Therefore:

P (HF 5 HF 6 HF 7) = P (HF 5) + P (HF 6) + P (HF 7) [ [ = 73;359 + 1;287+33 = 78;639 0:0076414 10;291;120 2;572;780 10;291;120

Now that we have the probability of drawing a heart ‡ush with seven cards, we multiply:

4 78;639 78;639 P (F ) = 4 P (HF ) = = 0:030566 10;291;120 2;572;780

Comparing this result to the previous example, we …nd that ‡ushes are about ten times more likely with seven cards than with …ve, a main reason most modern poker games use (at least) seven cards.

19 2.3 Partitions We can use the idea of combinations to partition the deck into groups of cards, or subsets. Consider the problem of drawing …ve cards and having exactly one pair. Let R be the event of having exactly two cards of the same rank. One way to perform the calculation is as follows:

13 4 12 4 4 4 ( 1 )(2)( 3 )(1)(1)(1) 352 P (R) = 52 = 833 = 0:42257 ( 5 )

The terms in the numerator arise from the following reasoning:

13 1 There are "13 choose 1" ways to pick the rank or value of the pair, e.g. QQ, for example. 4 There are "4 choose 2" ways to pick two of that rank. 2 12 3 There are "12 choose 3" ways to draw cards of three di¤erent ranks from the remaining 12 insuring exactly one pair. 4 1 For each of the three values, there are "4 choose 1" ways to pick 3 one card of that value. We generally write this as 4 , but we listed 1 all three individually for clarity.

Example 14 What is the probability of drawing two pair into a …ve card hand?

Let 2P be the event of having exactly two pair. Then:

13 4 2 11 4 ( 2 )(2) ( 1 )(1) 198 P (2P ) = 52 = 4;165 = 0:047539 ( 5 )

The terms in the numerator arise from the following reasoning:

20 13 2 There are "13 choose 2" ways to pick the rank or value of the two pairs, e.g. QQ and 99, for example. 4 2 There are "4 choose 2" ways to pick two of each rank. We square this term, since we’re picking two cards from each rank. 11 1 There are "11 choose 1" ways to draw a card of a di¤erent rank from the pairs. 4 There are "4 choose 1" ways to pick one card of that value. 1 The …rst term in the numerator partitions the deck into a group of two values (Q’sand 9’s, for example) and group of the 11 other values. We then pick the necesary cards from those two groups. The third term partitions the remaining 11 values into a group of 1 and a group of 10. From the group of 1 value, the …nal numerator term selects the …fth card. Once you’ve learned how to use partitions e¤ectively, probability calculations for most poker situations become very simple. Also, when you know multiple ways to calculate a probability, you can use the various approaches together. If you get the same answer in two or more ways, you have more con…dence you are correct. The following example will demonstrate multiple ways to calculate the probability of drawing 3-of-a-kind.

Example 15 What is the probability of drawing a 3-of-a-kind into a …ve card hand?

Let 3K be the event of having 3-of-a-kind. Then:

13 4 48 ( 1 )(3)( 2 ) 94 P (3K) = 52 = 4165 0:022569 ( 5 )

The terms in the numerator arise from the following reasoning:

13 1 There are "13 choose 1" ways to pick the rank or value of the 3-of-a-kind. 21 4 There are "4 choose 3" ways to pick three of that rank. 3 48 There are "48 choose 2" ways to draw any two others cards. 3 There is a slight ‡aw in this reasoning. We have calculated the probability of drawing a hand that at least as good as 3-of-kind. A full houses are possible if the two remaining cards have the same rank. We can calculate the probability of a full house, then subtract all of them out. Let B be the event of drawing a "boat" or full house into a …ve card hand. Then:

13 4 12 4 ( 1 )(3)( 1 )(2) 6 P (B) = 52 = 4165 0:0014406 ( 5 )

The terms in the numerator arise from the following reasoning:

13 1 There are "13 choose 1" ways to pick the rank or value of the 3-of-a-kind. 4 There are "4 choose 3" ways to pick three of that rank. 3 12 There are "12 choose 1" ways to pick another value for the pair. 1 4 There are "4 choose 2" ways to pick two of that rank. 2 Then the probability of exactly 3-of-a-kind is given by:

P (3K B) = 94 6 = 88 0:021128 n 4165 4165 4165

22 2.4 Exercises Solve each of the following using choose notation (combinations) or factorial notation (permutations). Be sure to de…ne your event set(s) clearly. You should provide an exact probability as well as a numeric estimate for each problem.

1. What is the total number of pocket pairs in Texas Holdem? In other words, in how many ways can Hero be dealt two pocket or hole cards of the same rank or value?

2. Stack sizes matter. Hero generally prefers deep stacks to his right and small stacks to his left. At a 6max table (e.g. Hero and …ve opponents) with …ve distinctly di¤erent opponent stack sizes, what is the probability that they are arranged from smallest stack immediately to Hero’sleft to largest immediately to Hero’sright?

3. What is the total number of unpaired pocket holdings Hero can have in Texas Holdem?

4. What is the total number of "suited" holdings Hero can have in Texas Holdem, e.g. hands like A Q where both cards are of the same suit? • • 5. What is the probability of Hero being dealt "big slick" pre‡op, e.g. AK?

6. Given that we know Hero’stwo pocket cards, what is the total number of ‡ops possible in Holdem?

7. Given that we know Hero’stwo pocket cards, what is the total number of monotone ‡ops possible in Holdem, e.g. all three ‡op cards of the same suit?

8. What is the probability of a two-tone ‡op in Holdem, e.g. two cards of one suit and one card of another?

9. What is the probability of Hero drawing a broadway hand pre‡op, e.g. any two cards of di¤erent rank both T or higher?

10. What is the probability of Hero drawing "pocket rockets" pre‡op, e.g. AA?

23 11. What is the probability of Hero drawing "quads" or 4-of-a-kind into a seven card hand?

2.5 Challenge Problems 1. What is the probability of drawing a full house if Hero is dealt seven cards? (Note that we must rule out two 3-of-a-kinds and quads + pairs/3-of-a-kind.)

2. Hero is seated at the …nal table of a large MTT (multi-table tournament). Seat positions are determined by random draw. There are 9 opponents plus Hero. There are 2 small stacks, 4 medium stacks (about the same as Hero’s stack), and the rest are big stacks. How many di¤erent con…gurations of stacks are possible?

3. Hero is seated at the …nal table of a large MTT with nine opponents: three professional players and six amateurs. Hero is worried about the professionals, and he’d like them to be on his right. What is the probability if, when seat positions are determined by random draw, that all three professionals are in the four seats immediately to Hero’s right?

3 Binomial Theorem

The examples in Chapter 2 all related to repeated draws without replacement. How do we handle repeated draws with replacement? The answer is connected to Pascal’sTriangle and binomial expansions. We will start with a simple example of ‡ipping a coin.

3.1 Pascal’sTriangle Example 16 What is the probability of ‡ipping a coin three times and getting exactly two "heads"?

24 Let H be the event of ‡ipping a coin and having it come up heads, and T for tails. One way to answer the question is to list all the possibilities to …nd n(S), and then count those which have exactly two H’s, e.g. n(HH). The best option to create an ordered list to insure that we cover all the permutations, for example:

HHH HHT * HTH * HTT THH * THT TTH TTT

I placed an asterisk after each permutation that has two H’s, so the probability is:

n(HH) 3 P (HH) = n(S) = 8

At this point, it is interesting to list all the probabilities and compare them with the third row of Pascal’sTriangle:

1 P (HHH) = 8 3 P (HH) = 8 3 P (H) = 8 1 P (no H) = 8

Pascal’sTriangle follows:

1 1 1 1 2 1 1 3 3 1 1 4 6 4 1 1 5 10 10 5 1

25 The entries in row three sum to 8 ( 1 + 3 + 3 + 1 ), the number of possible permutations of H’sand T’sinto three slots. The entries also correspond to the numerators in the probabilities we listed above. We can use the …fth row of Pascal’sTriangle to compute the following:

Example 17 What is the probability of ‡ipping a coin three times and getting (a) exactly two "heads"? (b) exactly 3 heads, and (c) at least two heads.

The sum of the entries in the …fth row is 25 = 32. So for this probability experiment, n(S) = 32. The probabilities can be read o¤ now.

1 P (HHHHH) = 32 = P (no H) 5 P (HHHH) = 32 = P (H) 10 P (HHH) = 32 = P (HH)

This answers the …rst two parts of the question. Let 2H be the event of ‡ipping a coin …ve times and getting at least 2 heads. The events of exactly 3 heads and exactly 4 heads are independent, so we have:

P (2H) = P (HH) + P (HHH) + P (HHHH) + P (HHHHH) 5+10+10+5+1 = 32 = 0:96875

The fourth row (nth row) of Pascal’striangle are also the coe¢ cients in the expansion of a fourth degree (nth degree) binomial:

(x + y)4 = x4 + 4x3y + 6x2y2 + 4xy3 + y4

Pascal’s Triangle can also be written in "choose" notation using combinations as follows:

26 1 1 1 2 2 2 0 2 2 3 3 3 3 0 1 2 3 4 4 4 4 4 0 1 2 3 4 5 5 5 5 5 5 0 1 2 3 4 5 The Binomial Theorem. The expansion of (x+y)n is given by

n n n n n 1 n n 2 2 n n 1 n n (x + y) = 0 x + 1 x y + 2 x y + + n 1 xy + n y or, in summation notation:

n n n k n k (x + y) = k x y . k=0 X

3.2 Bernouli Trials Bernouli trials are repetitions of the same probability experiment with probability of success x and probability of failure y = 1 x. Notice that, since x+y = 1, then (x+y)n = 1 for any whole number n. This pleasant turn of events allow us to use the Binomial Theorem to calculate the probability of k successes and n k failures in n trials.

Example 18 Suppose the chance of rain on any given June day in Dahlonega, GA, is known to be 40%. We’re currently experiencing a drought, so we’ll consider a rainy day a "success." For any …ve-day period, what is the probability of (a) exactly two days of rain, (b) …ve days of rain, and (c) at least two days of rain?

Let R be the event that, on any one day, there is signi…cant rainfall. Then a successful outcome is P (R) = 0:4 = x. Of course, failure is given by y = 1 x = 0:6. (a) The probability of exactly two days of rain (e.g. two successes in …ve trials) is given by the x2y3 term of the …th degree binonial expansion. There

27 are "5 choose 2" ways to distribute two successfully rainy days into …ve days total, so we have:

P (2R) = 5 (:4)2(:6)3 0:3456 2 (b) The probability of …ve rainy days (e.g. …ve successes in …ve trials) is given by the x5 term:

P (5R) = 5 (:4)5 0:01024 5 (c) If we de…ne the event RR to mean at least two rainy days in …ve, we can simply add terms from the binomial expansion:

P (RR) = P (2R) + P (3R) + P (4R) + P (5R)

5 5 k 5 k = (:4) (:6) 0:66304 k k=2 X Example 19 The Top 5 hands in Holdem are AA, KK, AK, QQ and JJ. Late in an MTT, when the SB and BB are large compared to Hero’sstack of chips, these hands are extremely valuable. Hero’s stack has dwindled to the point where he feels he can only a¤ord to wait 20 hands before desperation will force him to go all-in with just about any two cards. What is the probability that Hero gets a Top 5 hand in the next 20?

Let T 5 be the event of being dealt a Top 5 hand pre‡op. We must …rst calculate the probability of the event T 5 on any one hand of poker, then we can apply the Binomial Theorem. There are "4 choose 2" ways to be dealt 4 4 any given pocket pair, or 6 total. There are 1 1 or 16 ways to be dealt any two given unpaired cards. So we have:

6+6+16+6+6 20 P (T 5) = 52 = 663 0:030166 ( 2 )

28 Since P (T 5) 0:03 = x, we use the 20th degree binomial expansion:

P (T 5) = 20 (0:03)(0:97)19 0:33637 1 We can calculate the probability of getting at least one T 5 hand by using set complements:

P (1 or more T 5) = 1 P (no T 5) = 20 (0:97)20 0:54379 0 So Hero has hope of getting a huge hand before time runs out. We can compute the probability of Hero getting 3 or more Top 5 hands by summing directly (if we have good technology):

20 20 k 20 k P (3 or more T 5) = (:03) (:97) 0:02101 k k=3 X

We can proceed using set complements (with poorer technology options) and thus have fewer terms to evaluate:

P (3 or more T 5) = 1 P (2 or fewer T 5) 2 20 k 20 k = 1 (:03) (:97) 1 0:97899 = 0:02101 k k=0 X

29 3.3 Exercises 1. What is the probability of Hero being dealt AA exactly twice in …ve hands?

2. With a hand like AK, Hero’schance of "hitting the ‡op" (e.g. ‡opping 1 a pair or better) is roughly 3 . In the next four times Hero is dealt AK, what is the probability that he hits the ‡op (a) exactly twice, (b) all four times, (c) two or more times, and (d) never?

3. With pocket pairs (e.g. JJ), Hero’s odds of "‡opping a set or better" is about 7.5 to 1 against. (A set is 3-of-a-kind, e.g. one of the ‡op cards is the same value as his pair.) In the next ten times Hero is dealt a pocket pair, what is the probability that he ‡ops a set (a) exactly twice, (b) four times or fewer, (c) two or more times, and (d) never?

4. A "rotation" in poker is playing the same number of hands as there are players at the table. In Holdem, this means that each player plays each position exactly once. At a table of ten players, what is the probability that, in one rotation, Hero is dealt a Top 10% hand (a) exactly three times, (b) …ve times or fewer, (c) at least once, and (d) all ten times?

4 Assorted Probability Problems

Most probability questions are rather simple - as long as you know whether to use combinations, permutations or the binomial theorem. Having di¤erent problem-types mixed altogether is more realistic. Having several types of problems inside the same question is challenging, but realistic. Not every probability question we ask ourselves is a discrete, easy-to-classify problem. We often have to invent our own events and sub-events, working and rework- ing them to verify mutual exclusivity. The following section tackles some complex (and some simple) poker probabilities. The ideas are all mixed together haphazardly. You should try very hard not to look ahead before you’ve attempted the problems. You’ll get more out of them, that way.

30 4.1 Multinomial Partitions

Before doing problems, let’s introduce another, even more elegant, partitioning method. Consider the problem of drawing exactly 3-of-a-kind into a 7-card hand. We might proceed as follows using binomial partitions:

13 4 12 4 4 ( 1 )(3)( 4 )(1) 6336 P (3K) = 52 = 128 639 ( 7 )

In this problem, we …rst select the 3-of-a-kind value, then select 4 others values from the remaining 12. We can do the partitioning all at once with the multinomial coe¢ cient. I will de…ne it completely. A few examples will more than su¢ ce. In this example, we want to partition the 13 card values into a group of 1 (the 3-of-a-kind value), a group of 4 (the other values in the hand), and a group of 8 (the values not used in our hand). The multinomial coe¢ cient is given by:

13 13! 1;4;8 = 1!4!8! = 6435 Note that this is exactly the same as:

13 12 1 4 = 6435 When using the multinomial coe¢ cient, it is customary to list all the groups in the lower part of the "choose notation." Why? They must …rst sum to the number in the upper portion. This also clari…es what can be a murky situation. With the binomial coe¢ cient, the second number is "left out" since no ambiguity exists. There is another problem in both the calculations above. Generally, when we say 3-of-a-kind, we mean 3-of-a-kind but not better. With a 7-card hand, we have counted several straights and ‡ushes in the above calculation, in which case the …nal hand wouldn’tbe evaluated as 3-of-a-kind.

31 4.2 Assorted Calculations

Example 20 For 5-card poker hands, calculate the probability of holding a pair, two pair, 3-of-a-kind, straight, ‡ush, full house, quads and straight ‡ush.

The probability of exactly one pair ("13 choose 1" ways to pick the value of the pair; "4 choose 2" ways to pick a pair; "12 choose 3" ways to pick three more values to eliminate 2 pair/trips/boats/quads as possibilties; and "4 choose 1" ways to pick cards from each of those values):

13 4 12 4 3 ( 1 )(2)( 3 )(1) 352 P (Pr) = 52 = 833 ( 5 )

The probability of exactly two pair ("13 choose 2" ways to pick the values of the 2 pair; "4 choose 2" ways to pick the pairs; "12 choose 1" way to pick three more values to eliminate trips/boats/quads as possibilties; and "4 choose 1" ways to pick cards from each of those values):

13 4 2 11 4 ( 2 )(2) ( 1 )(1) 198 P (2Pr) = 52 = 4165 ( 5 )

The probability of exactly 3-of-a-kind ("13 choose 1" ways to pick the values of the trips; "4 choose 3" ways to make trips; "12 choose 2" ways to pick two other values to eliminate boats/quads as possibilties; and "4 choose 1" ways to pick cards from each of those values):

13 4 12 4 2 ( 1 )(3)( 2 )(1) 88 P (3K) = 52 = 4165 ( 5 )

The exact number of straight ‡ushes is 40. We will count them here and subtract them out of straights and ‡ushes in later calculations ("4 choose 1" ways to pick a suit; there are 10 straights ranging from Ace-high to 5-high - no "wrap arounds" alowed; thus "10 choose 1" ways to pick the value of the straight):

32 4 10 n(SF) = 1 1 = 40 The probability of exactly a straight ("10 choose 1" ways to pick the values of the straight and "4 choose 1" ways to pick the suit of each card value; subtract out straight ‡ushes):

5 (10)(4) 40 1 1 5 P (Str) = 52 = 1274 ( 5 )

The probability of a ‡ush ("4 choose 1" ways to pick the suit and "13 choose 5" ways to pick 5 cards from it; subtract out straight ‡ushes):

4 13 (1)( 5 ) 33 P (FL) = 52 = 16 660 ( 5 )

The probability of a full house ("13 choose 1" ways to pick the value of the trips; "4 choose 3" ways to make trips; "12 choose 1" ways to pick the pair; and "4 choose 2" ways make a pair):

13 4 12 4 ( 1 )(3)( 1 )(2) 6 P (FH) = 52 = 4165 ( 5 )

The probability of 4-of-a-kind ("13 choose 1" ways to pick the value of the quads; "4 choose 4" ways to make quads; "48 choose 1" ways to pick any other card):

13 4 48 ( 1 )(4)( 1 ) 1 P (4K) = 52 = 4165 ( 5 )

The probability of a straight ‡ush:

4 10 (1)( 1 ) 1 P (SF) = 52 = 64 974 ( 5 )

33 You should compare the probability expressions for 2 Pair and a Full 13 House. For 2 Pair, we use 2 to pick two values since the hand T T 8 8 2 is identical to 8 8 T T 2. For the Full House calculation, the order of selection 13 matters. If we pick two values with 2 , then we must select which of the two values is trips: 2 . You can verify that 1 13 2 = 13 12 = 13 2 1 1 1 6 2 We pick two values, then multiply by the 2! permutations those values can be put in the trips place or the pair place. This is a technical detail that I often manage to confuse when running through these calculations quickly in classroom examples. The hand T T T 8 8 is not the same as T T 8 8 8. So selection order is important here, when choosing which value is trips.

Example 21 For 7-card poker hands, calculate the probability of holding a pair, two pair, 3-of-a-kind, straight, ‡ush, full house, quads and straight ‡ush.

Solution possibly forthcoming...note that it’spossible to hold a ‡ush while simultaneously holding 3-of-a-kind, a pair, 2 pair, or even a straight that’s not a straight ‡ush - yikes!! It’s also possible to have 6- and 7-card ‡ushes and straights, which makes double-counting all the possibilities extremely easy to do but very di¢ cult to ferret out.

Example 22 In Holdem, what is the probability that Shero ‡ops a set or trips?

In poker, we call it a "set" (event S) when Shero holds a pocket pair (event "pp") and ‡ops a card of that value. We call it "trips" (event T) when Shero holds unpaired cards (event "up") and ‡ops two cards of one of those values. Examples: Shero holds 88 and ‡ops T 8 2 for a "set of 8’s"; Shero hold T8 and ‡ops 882 for "trip 8’s". We are concerned with event S and event T, which are disjoint. Event S can only happen if Hero’scards are paired pre‡op. Conversely, event T can only happen with unpaired pre‡op holdings. So we calculate:

34 13 4 ( 1 )(2) 1 P (pp) = 52 = 17 ( 2 ) ) P (up) = 1 P (pp) = 16 17

In the case of sets, we must match one value exactly from two remaining cards in that suit, yet rule out full houses and quads. The following expanded probability expression calculates the 7.51 to 1 odds against ‡opping a set or better while holding a pocket pair:

2 2 12 4 + 12 4 + 2 48 (1) ( 2 )(1) ( 1 )(2) (2)( 1 ) 144 P (S+) = 50 = 1225 ( 3 )

12 4 The 1 2 term generates all the full houses, adding a pair on the board to the set. The …nal term 2 48 generates quads. The odds of simply having 2 1 a set are about 8.28 to 1 against:

2 12 4 2 (1)( 2 )(1) 132 P (S) = 50 = 1225 ( 3 )

For the trips, we must have a pair on the board that exactly matches one of Shero’scards, but no other matches to her hand. The odds are about 73.2 to 1 against:

2 3 44 (1)(2)( 1 ) 33 P (T) = 50 = 2450 ( 3 )

For the overall probability of Shero having exactly 3-of-a-kind after the ‡op, we caclulate as follows:

P (3K) = P (pp)P (S) + P (up)P (T) = ) 1 132 16 33 396 P (3K) = 17 1225 + 17 2450 = 20 825 = 0:019016

35 Example 23 Suppose Hero de…nes "Premium" hand as AA, KK, QQ, JJ, AK and AQ. What is the probability that, in the next 20 rounds of play, Hero is dealt a premium hand 4 times? never? at least 3 times? 5 times or fewer?

Let M be the event that, in a given trial, a premium hand was dealt. This event is unodered, and draws are with replacement. These are Bernouli trials, so we must calculate the probability of a success for a single trial:

4 4 2 4(2)+2(1) 28 P (M) = 52 = 663 ( 2 )

The probability of 4 premium hands in the next 20:

20 28 4 635 16 3 P (M=4) = 7: 727 4 10 4 663 663 The probability of no premium hands in the next 20:

P (M=0) = 20 28 0 635 20 0:421 89 0 663 663 The probability of at least three:

20 20 k 20 k 28 635 P (M 3) = k 663 663 0:05018 4 k=3 P or, with less computational work:

2 20 k 20 k 28 635 P (M 3) = 1 P (M<3) = 1 k 663 663 k=0 P The probability of …ve or fewer:

36 5 20 k 20 k 28 635 P (M 5) = k 663 663 0:999 87 k=0 P Example 24 In a typical "Sit n Go" tournament online, a 10-player tournament begins as soon as 10 players sit down at the table. If the buy-in is $21 per player, $20 goes into the prize pool while $1 goes to the casino. A typical payout structure pays …rst place $100, second place $60 and third place $40. If all players are about equal in ability, the …nal order of …nish is more or less random. For a random order of …nish, what is Hero’schance of …nishing "in the money" or ITM e.g. in the top 3?

Let N be the event that Hero …nishes in the money (unordered, without replacement). The event N is not ordered, but all the orders of …nish are. This problem requires both combinations and permutations. There are 10! orders of …nish, and the result is as expected even if the calculations are knotty:

9 (2)3!7! 3 P (N) = 10! = 10

The terms in the numerator arise as follows:

9 2 There are "9 choose 2" ways to pick Hero plus two opponents to …nish ITM. 3! There are "3 permute 3" ways for these 3 players to …nish 1st, 2nd or 3rd.

7! There are "7 permute 7" ways for the other players to …nish 4th through 10th.

37 4.3 Exercises

The following exercises each include the odds against the event happening for reference. You should compute the probablitiy, convert to odds against, and compare. These odds come from tables in the Appendices of Doyle Brunson’s famous Super System: A Course in Poker Poker and were calculated by "Crazy Mike" Caro, an expert poker player who is well-versed in game theory, probability theory and statistics.

1. The odds against drawing AQ or AJ suited prelfop are approximately 165 to 1. Odds against drawing AQ or AJ unsuited are approximately 54.3 to 1.

2. Odds against drawing "bad" Aces pre‡op, e.g. AT or worse, are 35.8 to 1 (suited) and 11.3 to 1 (unsuited).

3. Hero holds AK. Odds against ‡opping at least an Ace or King are 2.08 to 1.

4. Hero holds AK. Odds against ‡opping either AAK or AKK (full house) are 1088 to 1.

5. Hero holds AKs. Odds against ‡opping QJT not all in Hero’s suit (a straight, but not a ‡ush) are 310 to 1.

6. Hero holds KK. Odds against ‡opping 1 Ace and other 2 cards 2 - Q (including pairs) are 4.18 to 1.

7. Hero hold QJo. Odds against ‡opping any straight are 101 to 1.

8. Hero hold QJo. Odds against ‡opping QQQ or JJJ are 9799 to 1.

9. Hero’sstarting hand will have no Ace 85.07% of the time.

10. If Hero holds exactly one Ace pre‡op, then the probability of no other player holding an Ace is 50.41% for a 6-player table, 30.53% for a 9- player table, and 25.31% for a 10-player table.

38 11. Hero holds KK pre‡op (and would hate for an opponent to have Ax pre‡op), the probability of no other player holding an Ace is 39.68% for a 6-player table, 20.14% for a 9-player table, and 15.61% for a 10-player table. 12. Hero holds 65o pre‡op. Odds against ‡opping a straight are 75.6 to 1. 13. Hero holds 75o pre‡op. Odds against ‡opping a straight are 101 to 1. 14. Hero holds 85o pre‡op. Odds against ‡opping a straight are 152 to 1. 15. Hero holds 95o pre‡op. Odds against ‡opping a straight are 305 to 1. 16. Hero holds 98s pre‡op. Odds against ‡opping a straight ‡ush are 4899 to 1. 17. In a 10-handed game, the odds against no player being dealt an Ace or King pre‡op are 70.5 to 1. 18. Hero will not hold an Ace or Pair pre‡op for the next 50 hands, odds against are 87,897 to 1.

5 Expected Value

Poker players often talk about the "expected value" (EV) of their playing decisions. Positive EV is good. It means that, on average, playing that hand that way will win money. Negative EV is bad - it’s means the play loses money on average. Even simple poker scenarios often lead di¢ cult EV calculations. A simple EV calculation relates to the Sit n Go example above. Sit n Go’sare also called SnG’sor STT’sfor "single table tournaments." Suppose that, based on experience of playing 100 $21 SnG’s, Hero expects to win 10% of the time, …nish 2nd 20% of the time and place 3rd 10% of the time. This is statistics - analyzing past performance and converting the results to probabilities or expectation of the future. Note that a sample size of 100 is "short term" in poker, so this would be a rough estimate.

Example 25 What is Hero’s Expected Value for playing $21 SnG’s?

39 De…ne x1 to be the event of Hero …nishing in 1st place, x2 …nishing 2nd, and so forth. Then x1 = 100 21 = 79, Hero’s winning minus his buy-in, x2 = 39, and x3 = 19. Any other …nishing place is out of the money, or OTM, so xOTM = 21.

De…nition. The expected value E(X) of a random variable X is given by:

E(X) = p1x1 + p2x2 + ::: + pnxn, or

E(X) = pkxk k=1 X

where pi is the probability event xi occurs.

In this example, we have:

E(X) = 79(0:1) + 39(0:2) + 19(0:1) 21(0:6) = 5

This means that, for each SnG Hero plays, he expects a return of $5 on his $21 investment. Over the course of the next 10 SnG’s, Hero would expect to win about $50. Expected value estimates only match reality over large samples, and then only when the underlying probabilities are extremely accurate.

Example 26 Playing a 200nl cash game with 100BB e¤ective stacks, Hero is UTG (under the gun) with KK and opens for a standard raise of $8. MP opponent reraises (3-bets) for 12 BB or $24. The action folds around to Hero. Hero knows Villain 3-bets with only four hands: AA, KK, QQ and AK. What is the expected value for Hero if he just shoves all-in, given prior experience against Villain suggesting Villain will likely fold QQ but call with AK, KK and AA?

40 We use Poker Stove to …nd the relevent probabilities of winning an all-in.

KK vs. AA - Hero has 18.054% equity (chance of winning) KK vs. KK - Hero has 50% equity KK vs. AK - Hero has 68.884% equity

Next, we need to …nd the probability that Villain is holding these various hands. We can ignore most of the 1326 possible starting combinations since Villain 3-bet. There are "4 choose 2" ways to have a pocket pair, so there are 6 combinations of AA and QQ. There are 42 = 16 ways to have AK. And there is only way Villain could hold KK since Hero has two of them in his hand. So Villain has one of 6 + 6 + 16 + 1 = 29 combinations. We think of these outcomes as the possible values of a random variable X, with the outcomes listed (if we needed to) as X = x1; x2; : : : ; x29 . Each outcome is equally likely to occur. Summarizing, Villainf has a: g

6 0:20690 probability of holding AA 29 1 0:03448 probability of holding KK 29 6 0:20690 probability of holding QQ 29 16 0:55172 probability of holding AK 29

At the time Hero considers the $192 all-in bet, he has already commited $8 to the pot. Villain has committed $24. There is $3 posted by the SB and BB. The pot is $35. If Villain folds, Hero wins $35. If Villain calls, Hero must win a showdown after all …ve community cards are dealt. If Hero wins the showdown, he wins the $35 plus Villain’scall ($200 $24 =) $176, or $211. If Hero loses the showdown, he loses $192. The value of having QQ is $35, since Villain folds. For the other hands, we have to multiply the probability of winning times the amount won and subtract the product of the probability of losing times the amount to be lost:

41 Value of AA 0:18054(211) (1 0:18054)(192) = 119:24, which occurs 6 times in 29.

Value of KK 0:5(211) (0:5)(192) = 9:5, which occurs 1 time in 29. Value of AK 0:68884(211) (1 0:68884)(192) 85:60, which occurs 16 times in 29.

Value of QQ 35, which occurs 6 times in 29.

The expected value of Hero’sall-in shove is:

E(X) = 6 ( 119:24) + 1 (9:5) + 16 (85:60) + 6 (35) = 30:126 29 29 29 29

On average, Hero expects to win $30.13 with the shove. This does not mean that Hero is guaranteed to win or lose any given exchange. It means rather that Hero will win an average of $30.13 per hand if the same hand were played the same way hundreds of thousands of times. Poker players are not simply interested in positive expected value. They want the maximum possible EV which often means choosing between several positive EV plays. For example, Hero could just call the $24 3-bet. Hero would keep QQ holdings in the hand for the ‡op and be "way ahead" of both QQ and AK, but Hero would be but "way behind" AA. He could avoid committing his stack on ‡ops that contained A’s, but Villain might also not wish to commit a large amount of chips with AK or QQ. The EV for folding is -$8, so Hero can rule that action out. However, it would take more calculations and assumptions to …nd the EV for calling.

6 Solutions to Selected Exercises

Problems 2.4. Brief solutions to odd questions below. Challenge problems excluded (otherwise, they wouldn’tbe much of a challenge, now would they?)

1. What is the total number of pocket pairs in Texas Holdem?

42 13 4 Solution: n(pp) = 1 2 = 78 3. What is the total number of unpaired pocket holdings??

13 4 2 Solution: n(up) = 2 1 = 1248 5. What is the probability of Hero being dealt AK?

4 2 (1) 8 Solution: P (AK) = 52 = 663 ( 2 ) 7. Given that we know Hero’s two pocket cards, what is the total number of monotone ‡ops possible in Holdem, e.g. all three ‡op cards of the same suit?

Solution: Hero can have (a) suited cards or (b) unsuited. 4 13 (a) n(sc) = 1 2 = 312 (b) n(uc) = n(scc) = 52 312 = 1014 2 11 3 13 ( 3 )+(1)( 3 ) 1023 (a) P (M) = 50 = 19 600 ( 3 ) 2 12 2 13 (1)( 3 )+(1)( 3 ) 1012 (b) P (M) = 50 = 19 600 ( 3 ) The solution is a weighted average of the two probabilities: 312 1023 1014 1012 22 P (M) = 1326 19 600 + 1326 19 600 = 425 Compare: Ignoring Hero’sholdings, we would have: 4 13 (1)( 3 ) 22 P (M) = 52 = 425 ( 3 ) 9. What is the probability of Hero drawing a broadway hand?

(20) (5)(4) 2 1 2 80 Solution: P (B) = 52 = 663 ( 2 )

11. What is the probability of Hero drawing "quads" (7-card hand)?

13 4 48 ( 1 )(4)( 3 ) 1 Solution: P (Q) = 52 = 595 ( 7 )

43 Problems 3.3 Brief solutions to odd questions below.

1. What is the probability of Hero being dealt AA twice in …ve hands?

4 (2) 1 Solution: Let x = P (AA) = 52 = 221 : ( 2 ) Then let A2 be the event of being dealt AA twice in …ve trials (unordered event, with replacement) = 5 1 2 220)3 106 480 000 4 P (A2) = = 2: 019 8 10 2 221 221 527 182 965 101 3. With pocket pairs (e.g. JJ), Hero’s odds of "‡opping a set or better" is about 7.5 to 1 against. In the next ten times Hero is dealt a pocket pair, what is the probability that he ‡ops a set (a) exactly twice, (b) four times or fewer, (c) two or more times, and (d) never?

1 2 Solution: Let x = 7:5+1 = 17 :

(a) P (S = 2) = 10 2 2 15 8 0:228 83 2 17 17 4 10 k 10 k 2 15 (b) P (S 4) = k 17 17 0:996 60 k=0 P 1 10 k 10 k 2 15 (c) P (S 2) = 1 k 17 17 0:332 58 k=0 P (c) P (S = 0) = 10 2 0 15 10 0:286 04 0 17 17 Problems 4.3 Brief solutions to odd questions below.

1. The odds against drawing AQ or AJ suited prelfop are approximately 165 to 1. Odds against drawing AQ or AJ unsuited are approximately 54.3 to 1.

4 4 Solution: n(AQ) = 1 1 = 16, of which 4 are suited, 12 unsuited. We double these values to include n(AJ). 8 24 For suited, we have P = 1326 , and for unsuited, P = 1326 . 1318 The odds (suited) are 1318 to 8, or 8 = 164: 75 to 1.

44 3. Hero holds AK. Odds against ‡opping at least an Ace or King are 2.08 to 1.

Solution: Approach this "backwards": 44 ( 3 ) 227 700 227 1 P (no A or K) = 1 50 = 700 227 = 2: 083 7 to 1. ( 3 ) ) 5. Hero holds AKs. Odds against ‡opping a straight (but not a ‡ush) are 310 to 1.

4 3 63 Solution: n(QJT) = 1 = 64, but 1 is a straight ‡ush P = 19600 19600 63 2791 ) Odds: = = 310: 11 to1. 63 9 7. Hero hold QJo. Odds against ‡opping any straight are 101 to 1.

Solution: The possible straights are built with AKT, KT9, or T98. 4 3 Each straight combination can be made in 1 = 64 ways 3(64) 12 1213 ) P = 50 = 1225 12 = 101: 08 to 1. ( 3 ) ) 9. Hero’sstarting hand will have no Ace 85.07% of the time.

48 ( 2 ) 188 Solution: P (no A) = 52 = 221 = 0:850 68 ( 2 )

11. Hero holds KK pre‡op (and would hate for an opponent to have Ax pre‡op), the probability of no other player holding is 39.68% for a 6-player table, 20.14% for a 9-player table, and 15.61% for a 10-player table.

46 (10) 9139 Solution (6-players): P (no Aces) = 50 = 23 030 = 0:396 83 (10) 46 (16) 11 594 (9-players): P (no Aces) = 50 = 57 575 = 0:201 37 (16)

46 (18) 1798 (10-players): P (no Aces) = 50 = 11 515 = 0:156 14 (18) 13. Hero holds 75o pre‡op. Odds against ‡opping a straight are 101 to 1.

Solution: The possible straights are built with 986, 864, or 643 (see #7 above).

45 15. Hero holds 95o pre‡op. Odds against ‡opping a straight are 305 to 1.

64 Solution: The only possible straight is with 876, so P = 19600 = 4 1225 . 17. In a 10-handed game, the odds against no player being dealt an Ace or King pre‡op are 70.5 to 1.

44 (20) 1798 128639 1798 Solution: P (no A or K) = 52 = 128 639 1798 = 70: 546 to (20) ) 1.