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# Introduction to Discrete Probability - PowerPoint PPT Presentation

Introduction to Discrete Probability. Epp, section 6. x CS 202 Aaron Bloomfield. Terminology. Experiment A repeatable procedure that yields one of a given set of outcomes Rolling a die, for example Sample space The range of outcomes possible For a die, that would be values 1 to 6 Event

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### Introduction to Discrete Probability

Epp, section 6.x

CS 202

Aaron Bloomfield

• Experiment

• A repeatable procedure that yields one of a given set of outcomes

• Rolling a die, for example

• Sample space

• The range of outcomes possible

• For a die, that would be values 1 to 6

• Event

• One of the sample outcomes that occurred

• If you rolled a 4 on the die, the event is the 4

• The probability of an event occurring is:

• Where E is the set of desired events (outcomes)

• Where S is the set of all possible events (outcomes)

• Note that 0 ≤ |E| ≤ |S|

• Thus, the probability will always between 0 and 1

• An event that will never happen has probability 0

• An event that will always happen has probability 1

• Something with a probability of 0 will never occur

• Something with a probability of 1 will always occur

• You cannot have a probability outside this range!

• Note that when somebody says it has a “100% probability)

• That means it has a probability of 1

• What is the probability of getting “snake-eyes” (two 1’s) on two six-sided dice?

• Probability of getting a 1 on a 6-sided die is 1/6

• Via product rule, probability of getting two 1’s is the probability of getting a 1 AND the probability of getting a second 1

• Thus, it’s 1/6 * 1/6 = 1/36

• What is the probability of getting a 7 by rolling two dice?

• There are six combinations that can yield 7: (1,6), (2,5), (3,4), (4,3), (5,2), (6,1)

• Thus, |E| = 6, |S| = 36, P(E) = 6/36 = 1/6

### Poker

• You are given 5 cards (this is 5-card stud poker)

• The goal is to obtain the best hand you can

• The possible poker hands are (in increasing order):

• No pair

• One pair (two cards of the same face)

• Two pair (two sets of two cards of the same face)

• Three of a kind (three cards of the same face)

• Straight (all five cards sequentially – ace is either high or low)

• Flush (all five cards of the same suit)

• Full house (a three of a kind of one face and a pair of another face)

• Four of a kind (four cards of the same face)

• Straight flush (both a straight and a flush)

• Royal flush (a straight flush that is 10, J, K, Q, A)

• What is the chance ofgetting a royal flush?

• That’s the cards 10, J, Q, K, and A of the same suit

• There are only 4 possible royal flushes

• Possibilities for 5 cards: C(52,5) = 2,598,960

• Probability = 4/2,598,960 = 0.0000015

• Or about 1 in 650,000

• What is the chance of getting 4 of a kind when dealt 5 cards?

• Possibilities for 5 cards: C(52,5) = 2,598,960

• Possible hands that have four of a kind:

• There are 13 possible four of a kind hands

• The fifth card can be any of the remaining 48 cards

• Thus, total possibilities is 13*48 = 624

• Probability = 624/2,598,960 = 0.00024

• Or 1 in 4165

• What is the chance of getting a flush?

• That’s all 5 cards of the same suit

• We must do ALL of the following:

• Pick the suit for the flush: C(4,1)

• Pick the 5 cards in that suit: C(13,5)

• As we must do all of these, we multiply the values out (via the product rule)

• This yields

• Possibilities for 5 cards: C(52,5) = 2,598,960

• Probability = 5148/2,598,960 = 0.00198

• Or about 1 in 505

• Note that if you don’t count straight flushes (and thus royal flushes) as a “flush”, then the number is really 5108

• What is the chance of getting a full house?

• That’s three cards of one face and two of another face

• We must do ALL of the following:

• Pick the face for the three of a kind: C(13,1)

• Pick the 3 of the 4 cards to be used: C(4,3)

• Pick the face for the pair: C(12,1)

• Pick the 2 of the 4 cards of the pair: C(4,2)

• As we must do all of these, we multiply the values out (via the product rule)

• This yields

• Possibilities for 5 cards: C(52,5) = 2,598,960

• Probability = 3744/2,598,960 = 0.00144

• Or about 1 in 694

• The possible poker hands are (in increasing order):

• Nothing

• One pair cannot include two pair, three of a kind, four of a kind, or full house

• Two pair cannot include three of a kind, four of a kind, or full house

• Three of a kind cannot include four of a kind or full house

• Straight cannot include straight flush or royal flush

• Flush cannot include straight flush or royal flush

• Full house

• Four of a kind

• Straight flush cannot include royal flush

• Royal flush

• What is the chance of getting a three of a kind?

• That’s three cards of one face

• Can’t include a full house or four of a kind

• We must do ALL of the following:

• Pick the face for the three of a kind: C(13,1)

• Pick the 3 of the 4 cards to be used: C(4,3)

• Pick the two other cards’ face values: C(12,2)

• We can’t pick two cards of the same face!

• Pick the suits for the two other cards: C(4,1)*C(4,1)

• As we must do all of these, we multiply the values out (via the product rule)

• This yields

• Possibilities for 5 cards: C(52,5) = 2,598,960

• Probability = 54,912/2,598,960 = 0.0211

• Or about 1 in 47

• The possible poker hands are (in increasing order):

• Nothing 1,302,540 0.5012

• One pair 1,098,240 0.4226

• Two pair 123,552 0.0475

• Three of a kind 54,912 0.0211

• Straight 10,200 0.00392

• Flush 5,108 0.00197

• Full house 3,744 0.00144

• Four of a kind 624 0.000240

• Straight flush 36 0.0000139

• Royal flush 4 0.00000154

### Back to theory again

• Let E be an event in a sample space S. The probability of the complement of E is:

• Recall the probability for getting a royal flush is 0.0000015

• The probability of not getting a royal flush is 1-0.0000015 or 0.9999985

• Recall the probability for getting a four of a kind is 0.00024

• The probability of not getting a four of a kind is 1-0.00024 or 0.99976

• Let E1 and E2 be events in sample space S

• Then p(E1 U E2) = p(E1) + p(E2) – p(E1 ∩ E2)

• Consider a Venn diagram dart-board

p(E1 U E2)

S

E1

E2

• If you choose a number between 1 and 100, what is the probability that it is divisible by 2 or 5 or both?

• Let n be the number chosen

• p(2|n) = 50/100 (all the even numbers)

• p(5|n) = 20/100

• p(2|n) and p(5|n) = p(10|n) = 10/100

• p(2|n) or p(5|n) = p(2|n) + p(5|n) - p(10|n)

= 50/100 + 20/100 – 10/100

= 3/5

• This is a statistical analysis, not a moral/ethical discussion

• What if you gamble \$1, and have a ½ probability to win \$10?

• If you play 100 times, you will win (on average) 50 of those times

• Each play costs \$1, each win yields \$10

• For \$100 spent, you win (on average) \$500

• Average win is \$5 (or \$10 * ½) per play for every \$1 spent

• What if you gamble \$1 and have a 1/100 probability to win \$10?

• If you play 100 times, you will win (on average) 1 of those times

• Each play costs \$1, each win yields \$10

• For \$100 spent, you win (on average) \$10

• Average win is \$0.10 (or \$10 * 1/100) for every \$1 spent

• One way to determine if gambling is worth it:

• probability of winning * payout ≥ amount spent

• Or p(winning) * payout ≥ investment

• Of course, this is a statistical measure

• Many older lotto games you have to choose 6 numbers from 1 to 48

• Total possible choices is C(48,6) = 12,271,512

• Total possible winning numbers is C(6,6) = 1

• Probability of winning is 0.0000000814

• Or 1 in 12.3 million

• If you invest \$1 per ticket, it is only statistically worth it if the payout is > \$12.3 million

• As, on the “average” you will only make money that way

• Of course, “average” will require trillions of lotto plays…

• Modern powerball lottery is a bit different

• Source: http://en.wikipedia.org/wiki/Powerball

• You pick 5 numbers from 1-55

• Total possibilities: C(55,5) = 3,478,761

• You then pick one number from 1-42 (the powerball)

• Total possibilities: C(42,1) = 42

• By the product rule, you need to do both

• So the total possibilities is 3,478,761* 42 = 146,107,962

• While there are many “sub” prizes, the probability for the jackpot is about 1 in 146 million

• You will “break even” if the jackpot is \$146M

• Thus, one should only play if the jackpot is greater than \$146M

• If you count in the other prizes, then you will “break even” if the jackpot is \$121M

### Blackjack

• You are initially dealt two cards

• 10, J, Q and K all count as 10

• Ace is EITHER 1 or 11 (player’s choice)

• You can opt to receive more cards (a “hit”)

• You want to get as close to 21 as you can

• If you go over, you lose (a “bust”)

• You play against the house

• If the house has a higher score than you, then you lose

• Getting 21 on the first two cards is called a blackjack

• Or a “natural 21”

• Assume there is only1 deck of cards

• Possible blackjack blackjack hands:

• First card is an A, second card is a 10, J, Q, or K

• 4/52 for Ace, 16/51 for the ten card

• = (4*16)/(52*51) = 0.0241 (or about 1 in 41)

• First card is a 10, J, Q, or K; second card is an A

• 16/52 for the ten card, 4/51 for Ace

• = (16*4)/(52*51) = 0.0241 (or about 1 in 41)

• Total chance of getting a blackjack is the sum of the two:

• p = 0.0483, or about 1 in 21

• How appropriate!

• More specifically, it’s 1 in 20.72 (0.048)

• Another way to get 20.72

• There are C(52,2) = 1,326 possible initial blackjack hands

• Possible blackjack blackjack hands:

• Pick your 10 card: C(16,1)

• Total possibilities is the product of the two (64)

• Probability is 64/1,326 = 1 in 20.72 (0.048)

• Getting 21 on the first two cards is called a blackjack

• Assume there is an infinite deck of cards

• So many that the probably of getting a given card is not affected by any cards on the table

• Possible blackjack blackjack hands:

• First card is an A, second card is a 10, J, Q, or K

• 4/52 for Ace, 16/52 for second part

• = (4*16)/(52*52) = 0.0236 (or about 1 in 42)

• First card is a 10, J, Q, or K; second card is an A

• 16/52 for first part, 4/52 for Ace

• = (16*4)/(52*52) = 0.0236 (or about 1 in 42)

• Total chance of getting a blackjack is the sum:

• p = 0.0473, or about 1 in 21

• More specifically, it’s 1 in 21.13 (vs. 20.72)

• In reality, most casinos use “shoes” of 6-8 decks for this reason

• It slightly lowers the player’s chances of getting a blackjack

• And prevents people from counting the cards…

• Counting cards means keeping track of which cards have been dealt, and how that modifies the chances

• There are “easy” ways to do this – count all aces and 10-cards instead of all cards

• Yet another way for casinos to get the upper hand

• It prevents people from counting the “shoes” of 6-8 decks of cards

• After cards are discarded, they are added to the continuous shuffling machine

• Many blackjack players refuse to play at a casino with one

• So they aren’t used as much as casinos would like

• Most people think that a single-deck blackjack table is better, as the player’s odds increase

• And you can try to count the cards

• But it’s usually not the case!

• Normal rules have a 3:2 payout for a blackjack

• If you bet \$100, you get your \$100 back plus 3/2 * \$100, or \$150 additional

• Most single-deck tables have a 6:5 payout

• You get your \$100 back plus 6/5 * \$100 or \$120 additional

• This lowered benefit of being able to count the cards OUTWEIGHSthe benefit of the single deck!

• And thus the benefit of counting the cards

• Even with counting cards

• You cannot win money on a 6:5 blackjack table that uses 1 deck

• Remember, the house always wins

Blackjack probabilities: when to hold

• House usually holds on a 17

• What is the chance of a bust if you draw on a 17? 16? 15?

• Assume all cards have equal probability

• Bust on a draw on a 18

• 4 or above will bust: that’s 10 (of 13) cards that will bust

• 10/13 = 0.769 probability to bust

• Bust on a draw on a 17

• 5 or above will bust: 9/13 = 0.692 probability to bust

• Bust on a draw on a 16

• 6 or above will bust: 8/13 = 0.615 probability to bust

• Bust on a draw on a 15

• 7 or above will bust: 7/13 = 0.538 probability to bust

• Bust on a draw on a 14

• 8 or above will bust: 6/13 = 0.462 probability to bust

• If the dealer’s visible card is an Ace, the player can buy insurance against the dealer having a blackjack

• There are then two bets going: the original bet and the insurance bet

• If the dealer has blackjack, you lose your original bet, but your insurance bet pays 2-to-1

• So you get twice what you paid in insurance back

• Note that if the player also has a blackjack, it’s a “push”

• If the dealer does not have blackjack, you lose your insurance bet, but your original bet proceeds normal

• Is this insurance worth it?

• If the dealer shows an Ace, there is a 4/13 = 0.308 probability that they have a blackjack

• Assuming an infinite deck of cards

• Any one of the “10” cards will cause a blackjack

• If you bought insurance 1,000 times, it would be used 308 (on average) of those times

• Let’s say you paid \$1 each time for the insurance

• The payout on each is 2-to-1, thus you get \$2 back when you use your insurance

• Thus, you get 2*308 = \$616 back for your \$1,000 spent

• Or, using the formula p(winning) * payout ≥ investment

• 0.308 * \$2 ≥ \$1

• 0.616 ≥ \$1

• Thus, it’s not worth it

• Buying insurance is considered a very poor option for the player

• Hence, almost every casino offers it

• These tables tell you the best move to do on each hand

• The odds are still (slightly) in the house’s favor

• The house always wins…

• If you make two or three mistakes an hour, you lose any advantage

• And, in fact, cause a disadvantage!

• You lose lots of money learning to count cards

• Then, once you can do so, you are banned from the casinos

• Although the casino has the upper hand, the odds are much closer to 50-50 than with other games

• Notable exceptions are games that you are not playing against the house – i.e., poker

• You pay a fixed amount per hand

As seen ina casino

• This wheel is spun if:

• You place \$1 on the “spin the wheel” square

• You get a natural blackjack

• You lose the dollar either way

• You win the amount shown on the wheel

• The amounts on the wheel are:

• 30, 1000, 11, 20, 16, 40, 15, 10, 50, 12, 25, 14

• Average is \$103.58

• Chance of a natural blackjack:

• p = 0.0473, or 1 in 21.13

• So use the formula:

• p(winning) * payout ≥ investment

• 0.0473 * \$103.58 ≥ \$1

• \$4.90 ≥ \$1

• But the house always wins! So what happened?

As seen ina casino

• Note that not all amounts have an equal chance of winning

• There are 2 spots to win \$15

• There is ONE spot to win \$1,000

• Etc.

• If you weight each “spot” by the amount it can win, you get \$1609 for 30 “spots”

• That’s an average of \$53.63 per spot

• So use the formula:

• p(winning) * payout ≥ investment

• 0.0473 * \$53.63 ≥ \$1

• \$2.54 ≥ \$1

• Still not there yet…

• I think the wheel is weighted so the \$1,000 side of the wheel is heavy and thus won’t be chosen

• As the “chooser” is at the top

• But I never saw it spin, so I can’t say for sure

• Take the \$1,000 out of the 30 spot discussion of the last slide

• That leaves \$609 for 29 spots

• Or \$21.00 per spot

• So use the formula:

• p(winning) * payout ≥ investment

• 0.0473 * \$21 ≥ \$1

• \$0.9933 ≥ \$1

• And I’m probably still missing something here…

• Remember that the house always wins!

### Roulette

• A wheel with 38 spots is spun

• Spots are numbered 1-36, 0, and 00

• European casinos don’t have the 00

• A ball drops into one of the 38 spots

• A bet is placed as to which spot or spots the ball will fall into

• Money is then paid out if the ball lands in the spot(s) you bet upon

• Bets can be placed on:

• A single number

• Two numbers

• Four numbers

• All even numbers

• All odd numbers

• The first 18 nums

• Red numbers

Probability:

1/38

2/38

4/38

18/38

18/38

18/38

18/38

• Bets can be placed on:

• A single number

• Two numbers

• Four numbers

• All even numbers

• All odd numbers

• The first 18 nums

• Red numbers

Probability:

1/38

2/38

4/38

18/38

18/38

18/38

18/38

Payout:

36x

18x

9x

2x

2x

2x

2x

• It has been proven that proven that no advantageous strategies exist

• Including:

• Learning the wheel’s biases

• Casino’s regularly balance their Roulette wheels

• Using lasers (yes, lasers) to check the wheel’s spin

• What casino will let you set up a laser inside to beat the house?

• It has been proven that proven that no advantageous strategies exist

• Including:

• Martingale betting strategy

• Where you double your bet each time (thus making up for all previous losses)

• It still won’t work!

• You can’t double your money forever

• It could easily take 50 times to achieve finally win

• If you start with \$1, then you must put in \$1*250 = \$1,125,899,906,842,624 to win this way!

• The Monty Hall problem paradox

• Consider a game show where a prize (a car) is behind one of three doors

• The other two doors do not have prizes (goats instead)

• After picking one of the doors, the host (Monty Hall) opens a different door to show you that the door he opened is not the prize

• Do you change your decision?

• Your initial probability to win (i.e. pick the right door) is 1/3

• What is your chance of winning if you change your choice after Monty opens a wrong door?

• After Monty opens a wrong door, if you change your choice, your chance of winning is 2/3

• Thus, your chance of winning doubles if you change

• Huh?

• Consider a dealt hand of cards

• Assume they have not been seen yet

• What is the chance of drawing a flush?

• Does that chance change if I speak words after the experiment has completed?

• No!

• Words spoken after an experiment has completed do not change the chance of an event happening by that experiment

• No matter what is said

• Consider 100 doors

• You choose one

• Monty opens 98 wrong doors

• Do you switch?

• Your initial chance of being right is 1/100

• Right before your switch, your chance of being right is still 1/100

• You didn’t know this info beforehand!

• Your final chance of being right is 99/100 if you switch

• You have two choices: your original door and the new door

• The original door still has 1/100 chance of being right

• Thus, the new door has 99/100 chance of being right

• The 98 doors that were opened were not chosen at random!

• Monty Hall knows which door the car is behind

• Reference: http://en.wikipedia.org/wiki/Monty_Hall_problem

### A bit more theory

• Assume you have a 5/6 chance for an event to happen

• Rolling a 1-5 on a die, for example

• What’s the chance of that event happening twice in a row?

• Cases:

• Event happening neither time: 1/6 * 1/6 = 1/36

• Event happening first time: 5/6 * 1/6 = 5/36

• Event happening second time: 1/6 * 5/6 = 5/36

• Event happening both times: 5/6 * 5/6 = 25/36

• For an event to happen twice, the probabilityis the product of the individual probabilities

• Assume you have a 5/6 chance for an event to happen

• Rolling a 1-5 on a die, for example

• What’s the chance of that event happening at least once?

• Cases:

• Event happening neither time: 1/6 * 1/6 = 1/36

• Event happening first time: 5/6 * 1/6 = 5/36

• Event happening second time: 1/6 * 5/6 = 5/36

• Event happening both times: 5/6 * 5/6 = 25/36

• It’s 35/36!

• For an event to happen at least once, it’s 1 minus the probability of it never happening

• Or 1 minus the compliment of it never happening

• Consider an event that has a 1 in 3 chance of happening

• Probability is 0.333

• Which is a 1 in 3 chance

• Or 2:1 odds

• Meaning if you play it 3 (2+1) times, you will lose 2 times for every 1 time you win

• This, if you have x:y odds, you probability is y/(x+y)

• The y is usually 1, and the x is scaled appropriately

• For example 2.2:1

• That probability is 1/(1+2.2) = 1/3.2 = 0.313

• 1:1 odds means that you will lose as many times as you win

### Texas Hold’em

Reference:

http://teamfu.freeshell.org/poker_odds.html

• The most popular poker variant today

• Every player starts with two face down cards

• Called “hole” or “pocket” cards

• Hence the term “ace in the hole”

• Five cards are placed in the center of the table

• These are common cards, shared by every player

• Initially they are placed face down

• The first 3 cards are then turned face up, then the fourth card, then the fifth card

• You can bet between the card turns

• You try to make the best 5-card hand of the seven cards available to you

• Your two hole cards and the 5 common cards

• Hand progression

• Note that anybody can fold at any time

• Cards are dealt: 2 “hole” cards per player

• 5 community cards are dealt face down (how this is done varies)

• Bets are placed based on your pocket cards

• The first three community cards are turned over (or dealt)

• Called the “flop”

• Bets are placed

• The next community card is turned over (or dealt)

• Called the “turn”

• Bets are placed

• The last community card is turned over (or dealt)

• Called the “river”

• Bets are placed

• Hands are then shown to determine who wins the pot

• Pocket: your two face-down cards

• Pocket pair: when you have a pair in your pocket

• Flop: when the initial 3 community cards are shown

• Turn: when the 4thcommunity card is shown

• River: when the 5th community card is shown

• Nuts (or nut hand): the best possible hand that you can hope for with the cards you have and the not-yet-shown cards

• Outs: the number of cards you need to achieve your nut hand

• Pot: the money in the center that is being bet upon

• Fold: when you stop betting on the current hand

• Call: when you match the current bet

• Pick any poker hand

• We’ll choose a royal flush

• There are only 4 possibilities (1 of each suit)

• There are 7 cards dealt

• Total of C(52,7) = 133,784,560 possibilities

• Chance of getting that in a Texas Hold’em game:

• Choose the 5 cards of your royal flush: C(4,1)

• Choose the remaining two cards: C(47,2)

• Product rule: multiply them together

• Result is 4324 (of 133,784,560) possibilities

• Or 1 in 30,940

• Or probability of 0.000,032

• This is much more common than 1 in 649,740 for stud poker!

• But nobody does Texas Hold’em probability that way, though…

An example of a hand usingTexas Hold’em terminology

• Your pocket hand is J♥, 4♥

• The flop shows 2♥, 7♥, K♣

• There are two cards still to be revealed (the turn and the river)

• Your nut hand is going to be a flush

• As that’s the best hand you can (realistically) hope for with the cards you have

• There are 9 cards that will allow you to achieve your flush

• Any other heart

• Thus, you have 9 outs

• There are 47 unknown cards

• The two unturned cards, the other player’s cards, and the rest of the deck

• There are 9 outs (the other 9 hearts)

• What’s the chance you will get your flush?

• Rephrased: what’s the chance that you will get an out on at least one of the remaining cards?

• For an event to happen at least once, it’s 1 minus the probability of it never happening

• Chances:

• Out on neither turn nor river 38/47 * 37/46 = 0.65

• Out on turn only 9/47 * 38/46 = 0.16

• Out on river only 38/47 * 9/46 = 0.16

• Out on both turn and river 9/47 * 8/46 = 0.03

• All the chances add up to 1, as expected

• Chance of getting at least 1 out is 1 minus the chance of not getting any outs

• Or 1-0.65 = 0.35

• Or 1 in 2.9

• Or 1.9:1

• What if you miss your out on the turn

• Then what is the chance you will hit the out on the river?

• There are 46 unknown cards

• The two unturned cards, the other player’s cards, and the rest of the deck

• There are still 9 outs (the other 9 hearts)

• What’s the chance you will get your flush?

• 9/46 = 0.20

• Or 1 in 5.1

• Or 4.1:1

• The odds have significantly decreased!

• These odds are called the hand odds

• I.e. the chance that you will get your nut hand

• So far we’ve seen the odds of getting a given hand

• Assume that you are playing with only one other person

• If you win the pot, you get a payout of two times what you invested

• As you each put in half the pot

• This is called the pot odds

• Well, almost – we’ll see more about pot odds in a bit

• After the flop, assume that the pot has \$20, the bet is \$10, and thus the call is \$10

• Payout (if you match the bet and then win) is \$40

• Your pot odds are 30:10 (not 40:10, as your call is not considered as part of the odds)

• Or 3:1

• When is it worth it to continue?

• What if you have 3:1 hand odds (0.25 probability)?

• What if you have 2:1 hand odds (0.33 probability)?

• What if you have 1:1 hand odds (0.50 probability)?

• Note that we did not consider the probabilities before the flop

• Pot payout is \$40, investment is \$10

• Use the formula: p(winning) * payout ≥ investment

• When is it worth it to continue?

• We are assuming that your nut hand will win

• A safe assumption for a flush, but not a tautology!

• What if you have 3:1 hand odds (0.25 probability)?

• 0.25 * \$40 ≥ \$10

• \$10 =\$10

• If you pursue this hand, you will make as much as you lose

• What if you have 2:1 hand odds (0.33 probability)?

• 0.33 * \$40 ≥ \$10

• \$13.33 > \$10

• Definitely worth it to continue!

• What if you have 1:1 hand odds (0.50 probability)?

• 0.5 * \$40 ≥ \$10

• \$20 >\$10

• Definitely worth it to continue!

• Pot odds is the ratio of the amount in the pot to the amount you have to call

• In other words, we don’t consider any previously invested money

• Only the current amount in the pot and the current amount of the call

• The reason is that you are considering each bet as it is placed, not considering all of your (past and present) bets together

• If you considered all the amounts invested, you must then consider the probabilities at each point that you invested money

• Instead, we just take a look at each investment individually

• Technically, these are mathematically equal, but the latter is much easier (and thus more realistic to do in a game)

• In the last example, the pot odds were 3:1

• As there was \$30 in the pot, and the call was \$10

• Even though you invested some money previously

• Assume the pot is \$100, and the call is \$10

• Thus, the pot odds are 100:10 or 10:1

• You invest \$10, and get \$110 if you win

• Thus, you have to win 1 out of 11 times to break even

• Or have odds of 10:1

• If you have better odds, you’ll make money in the long run

• If you have worse odds, you’ll lose money in the long run

• Pot is now \$20, investment is \$10

• Pot odds are thus 2:1

• Use the formula: p(winning) * payout ≥ investment

• When is it worth it to continue?

• What if you have 3:1 hand odds (0.25 probability)?

• 0.25 * \$30 ≥ \$10

• \$7.50 <\$10

• What if you have 2:1 hand odds (0.33 probability)?

• 0.33 * \$30 ≥ \$10

• \$10 = \$10

• If you pursue this hand, you will make as much as you lose

• What if you have 1:1 hand odds (0.50 probability)?

• 0.5 * \$30 ≥ \$10

• \$15 >\$10

• The only time it is worth it to continue is when the pot odds outweigh the hand odds

• Meaning the first part of the pot odds is greater than the first part of the hand odds

• If you do not follow this rule, you will lose money in the long run

• Consider the following hand progression:

• Your hand: almost a flush (4 out of 5 cards of one suit)

• Called a “flush draw”

• Perhaps because one more draw can make it a flush

• On the flop: \$5 pot, \$10 bet and a \$10 call

• Your call: match the bet or fold?

• Pot odds: 1.5:1

• Hand odds: 1.9:1 (or 0.35)

• The pot odds do not outweigh the hand odds, so do not continue

• Consider the following hand progression:

• Your hand: almost a flush (4 out of 5 cards of one suit)

• Called a flush draw

• On the flop: now a \$30 pot, \$10 bet and a \$10 call

• Your call: match the bet or fold?

• Pot odds: 4:1

• Hand odds: 1.9:1 (or 0.35)

• The pot odds dooutweigh the hand odds, so do continue

• There are other odds to consider:

• Expected odds (what you expect other players in the game to bet on)

• Your knowledge of the players

• Both on how they bet in general

• How often do they bluff, etc.

• And any “things” that give away their hand

• I.e. not keeping a “poker face”

• Etc.

• What is the probably the worst pocket to be dealt in Texas Hold’em?

• Alternatively, what is the worst initial two cards to be dealt in any poker game?

• 2 and 7 of different suits

• They are low cards, different suits, and you can’t do anything with them (they are just out of straight range)