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Introduction to Discrete Probability

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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 paircannot include two pair, three of a kind, four of a kind, or full house
- Two paircannot include three of a kind, four of a kind, or full house
- Three of a kindcannot include four of a kind or full house
- Straightcannot include straight flush or royal flush
- Flushcannot include straight flush or royal flush
- Full house
- Four of a kind
- Straight flushcannot 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):
- Nothing1,302,5400.5012
- One pair1,098,2400.4226
- Two pair123,5520.0475
- Three of a kind54,9120.0211
- Straight10,2000.00392
- Flush5,1080.00197
- Full house3,7440.00144
- Four of a kind6240.000240
- Straight flush360.0000139
- Royal flush40.00000154

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

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

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

- First card is an A, second card is a 10, J, Q, or K
- 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 Ace: C(4,1)
- 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)

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

- 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

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

- 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?

- 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?

- Learning the wheel’s biases

- 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!
- That’s 1quadrillion

- See http://en.wikipedia.org/wiki/Martingale_(roulette_system) for more info

- Martingale betting strategy

Monty Hall Paradox

- 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?
- Does that chance change if I tell you more info about what’s in the deck?

- 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

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

- 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
- Just because you know more info about the other doors doesn’t change your chances
- You didn’t know this info beforehand!

- Just because you know more info about the other doors doesn’t change your chances
- 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…

- 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 investment is $10
- 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!

- We are assuming that your nut hand will win

- 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

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

- 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

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

- Called a “flush draw”
- 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”

- Both on how they bet in general
- 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)