Download Algebra 1 Summer Institute 2014 The Poker Manipulation Summary

Algebra 1 Summer Institute 2014
The Poker Manipulation
Summary
Goals
During this activity,
participants will have the
opportunity to study the
applications of probabilities
in the poker game. Different
examples involving the
counting principle,
permutations, and
combinations will be
studied but the focus will be
on investigating and
interpreting probabilities for
different types of 5-card
poker hands: a pair, two
pairs, full house, four cards
of a kind, royal flush, and
straight flush.
Participant Handouts
 Practice with different
counting methods like
permutations and
combinations
 Practice finding
probabilities of
dependents and
independent events
1. The Poker Manipulation
Materials
Technology
Source
Estimated Time
Paper
Poker Cards
LCD Projector
Facilitator Laptop
NCTM
Journal of Statistics
Education Volume
90 minutes
20, Number 2
(2012)
Mathematics Standards
Common Core State Standards for Mathematics
MAFS. 7.SP.3: Investigate chance processes and develop, use, and evaluate probability
models
3.5: Understand that the probability of a chance event is a number between 0 and 1
that expresses the likelihood of the event occurring. Larger numbers indicate
greater likelihood. A probability near 0 indicates an unlikely event, a probability
around ½ indicates an event that is neither unlikely nor likely, and a probability
near 1 indicates a likely event.
3.6: Approximate the probability of a chance event by collecting data on the chance
process that produces it and observing its long-run relative frequency, and predict
the approximate relative frequency given the probability. For example, when
rolling a number cube 600 times, predict that a 3 or 6 would be rolled roughly 200
times, but probably not exactly 200 times.
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Algebra 1 Summer Institute 2014
3.7: Develop a probability model and use it to find probabilities of events. Compare
probabilities from a model to observed frequencies; if the agreement is not good,
explain possible sources of the discrepancy.
a. Develop a uniform probability model by assigning equal probability to all
outcomes, and use the model to determine probabilities of events. For
example, if a student is selected a random from a class, find the probability
that Jane will be selected and the probability that a girl will be selected.
b. Develop a probability model (which may not be uniform) by observing
frequencies in data generated from a chance process. For example, find the
approximate probability that a spinning penny will land heads up or that a
tossed paper clip will land open end down. Do the outcomes for the spinning
penny appear to be equally likely based on the observed frequencies.
3.8: Find the probabilities of compound events using organized lists, tables, trees, and
simulation.
a. Understand that, just as with simple events, the probability of a compound
event is the fraction of outcomes in the sample space for which the compound
event occurs.
b. Represent sample spaces for the compound events using methods such as
organized lists, tables and tree diagrams. For an event described in everyday
language (e.g.,”rolling double sixes”), identify the outcomes in the sample
space which compose the event.
c. Design and use a simulation to generate frequencies for compound events. For
example, use random digits as a simulation tool to approximate the answer to
the question: If 40% of donors have type A blood, what is the probability that
it will take at least 4 donors to find one with type A blood?
MAFS.912.S-IC.1: Understand and evaluate random processes underlying statistical
experiments
1.1: Understand statistics as a process for making inferences about population
parameters based on a random sample from that population.
1.2: Decide if a specified model is consistent with results from a given datagenerating process, e.g., using simulation. For example, a model says that a
spinning coin falls heads up with probability 0.5. Would a result of 5 tails in a row
cause you to question the model?
Standards for Mathematical Practice
1. Make sense of problems and persevere in solving them
2. Reason abstractly and quantitatively
3. Construct viable arguments and critique the reasoning of others
4. Model with mathematics
5. Use tools appropriately
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Algebra 1 Summer Institute 2014
Instructional Plan
During this activity, participants will have the opportunity to study the applications of
probabilities in the poker game. Different examples involving the counting principle,
permutations, and combinations will be studied but the focus will be on investigating and
interpreting probabilities for different types of 5-card poker hands: a pair, two pairs, full
house, four cards of a kind, royal flush, and straight flush.
The counting Principle (Slide 2)
If there are two events E1 and E2 where the first can happen in n1 different ways and the
second in n2 different ways, then together the events can occur in n1 x n2 different ways,
assuming that the events are not influencing each other.
This generalizes to k events E1, E2, ..., Ek with the number of possibilities for the
corresponding events n1, n2, ..., nk. The total number of possibilities is n1 x n2 x ...x nk.
1. Let’s suppose we select new uniforms for a team. The pants are coming in 2
styles, shirts in 3 styles, and hats in 4 styles. In how many different ways can a 3piece uniform be selected? (Slide 3)
To determine how many choices there are in total, make a tree diagram, which
will show 24 branches. Thus, there are 2 x 3 x 4 choices in all.
2. How many combinations exist for a lock that opens with a sequence a 3 numbers
from 1 to 40? (Slide 4)
The total number is 40 x 40 x 40 = 64,000.
Permutations (Slide 5)
A permutation of some or all of the elements of a set is any arrangement of the elements
in definite order. For example, for the set {a, b, c} there are 6 permutations: abc, acb, bac,
bca, cab, cba.
To find the number of permutations without listing them, the fundamental counting
principle can be used:



There are 3 ways to fill the first position.
For each way to fill the first position there are 2 ways to fill the second position.
For each way to fill the first and the second positions there is only 1 way to fill the
third position. In this way, the number of permutations of the elements of {a, b, c}
is 3 x 2 x 1 = 3! = 6.
3. Suppose we have to arrange 8 students in a row. In how many ways can you do
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Algebra 1 Summer Institute 2014
this? (Slide 5)
The answer is 8 x 7 x 6 x 5 x 4 x 3 x 2 x 1 = 8! = 40,320
Combinations (Slide 7)
A combination is a way of choosing k objects out of a collection of n, or is a subset of k
objects from a set of n. The number of ways of choosing k objects out of n is designated
C(n, k) = nCk and is usually read as “n choose k”.
𝑛𝐶𝑘 = 𝐶(𝑛, 𝑘) =
𝑃(𝑛, 𝑘)
𝑛!
=
(𝑛 − 𝑘)! 𝑘!
𝑘!
When you count combinations of elements of a set, the order in which they are listed is
disregarded.
4. In the game of poker a hand consists of five cards dealt from a deck of 52. How
many different poker hands are there? (Slide 8)
We start out by considering permutations of 5 out of 52.
P(52, 5) = 52 x 51 x 50 x 49 x 48
Each hand will be counted more than once. How many times will each hand be
counted? A given hand of 5 cards can be arranged in 5! = 5 x 4 x 3 x 2 x 1 = 120
different ways, so the total number of hands is
52 x 51 x 50 x 49 x 48 / 120 = 2,598, 960.
Using the Combination Formula: (Slide 9)
52𝐶5 = 𝐶(52, 5) =
52!
52!
52 ∙ 51 ∙ 50 ∙ 49 ∙ 48
=
=
= 2,598,960
(52 − 5)! 5! 47! ∙ 5!
5∙4∙3∙2∙1
The Poker Hand (Slide 10)
Poker is a game that is played all over the world. There are many forms of poker, but they
differ only in minor details and all follow the same basic principles. In general, the
players use a standard deck of fifty-two playing cards. Certain five card combinations are
recognized in all forms of poker and the ranking of these combinations gives a poker
hand its strength.
The 52 cards are divided into 4 suits - clubs, diamonds, hearts and spades – and each suit
is divided into 13 ranks: 2, 3, 4, 5, 6, 7, 8, 9, 10, Jack, Queen, King, and Ace. In the game
of poker a hand usually consists of 5 cards drawn from a standard deck of 52 cards.
Participants should set up the cards and discuss all possible hands of five cards they could
get during a poker game. Facilitators will observe the interaction of participants during
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Algebra 1 Summer Institute 2014
the project, check for understanding and consider all different strategies they used for
making and defend/change predictions. The information recorded by participants will be
shared in the final discussion.
The types of winning hand from highest to lowest are: (Slide 11)
Royal Flush – consists of a ten, jack, queen, king and ace from the same suit.
Straight Flush – five cards that are both straight and a flush
Four of a kind – four cards of the same rank
Full House – 3 cards of the same rank and two others of the same rank
Flush – five cards in the same suit
Straight – five cards that can be arranged so that each card has rank 1 greater than
previous card
7. 3 of a kind – 3 cards of the same rank and two others not of the same rank
8. Two pairs – 2 cards of one rank and two cards of another rank
9. Single pair – two cards of same rank and none of the other 3 cards of same rank.
1.
2.
3.
4.
5.
6.
Below is an example of how the demonstration might go for two-pair.
If a hand contains two 7‟s, two 8‟s, and one queen, the hand is “two-pair, 8‟s and 7‟s”. It
has the same hand-value whether it is dealt in the order 8877Q or 877Q8 or any other
order. To specify the hand, we will need to specify the denominations for each pair (i.e.
the 8‟s and the 7‟s) as well as the single; and then specify suits for each card (since for
example holding 7♠ and 7♦ is different from holding 7♥ and 7♣).
a. Specify Denominations. We must ultimately select three different denominations
(two for pairs and one more for the single). Participants will likely need guidance
in assessing whether the order in which these denominations are selected is
important. In fact, it is only partially important. There are C(13,2) choices for
denominations of our pairs (it doesn’t matter whether the 7’s or 8’s come first),
but then conditionally 11 choices for the denomination of the single (it does
matter that the Q is alone, and not one of the pairs). Observant participants
as we could just as easily decide
the denomination of the single first.
b. Specify Suits. For each pair we must take two of four suits, while for the
unmatched card we must pick only one. Note that in no case does the order of
selection for suits make a difference. However, if we select hearts-clubs and
diamonds-clubs for the pairs, it does make a difference which pair is associated to
which group of suits. Thus the multiplication rule will apply:
choosing the suits for the highest pair, choosing the suits for the lowest pair, and
then choosing the suit for the single.
c. Put it together. For every way there is to specify the denominations, there are
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Algebra 1 Summer Institute 2014
applying the multiplication rule, there are:
Therefore, again
123,552
different hands at the rank of two-pair.
Solutions: (Slides 12 to 21)
1. Royal Flush – consists of a ten, jack, queen, king and ace from the same suit.
There are only 4 hands of this kind. The probability to get a royal flush is
4/2,598,960 = 0.00000153908.
2. Straight Flush – five cards that are both straight and a flush. The cards are
arranged so that each one has a rank 1 higher than the previous card. The Ace can
be used as the lowest and as the highest-ranking card, so there are 10 possible
straights in a suit (ex: 6, 7, 8, 9, 10). For 4 suits, there are 4 x 10 = 40 straights
flushes, but if we exclude the royal flushes, the number of hands will be 40 – 4 =
36. Because the total number of hands of 5 cards is 52C5 = 2,598,960, the
probability to get a straight flush is 36/2,598,960 = 0.0000138517.
3. Four of a Kind – this hand has the pattern AAAAB, where A and B are two cards
of different kind. There are 13 possibilities or 13C1 for card A rank and 12C1 =
12 possibilities for card B rank. Because we need 4 cards of one specific rank
(card A), we can find out that there are 4C4 =1.To fill in the last card, there are
4C1 =4 ways to choose it. In total, there are13x12x4x1 = 624 possible ways of
having hands with 5 cards, four of which have the same rank. The probability to
get a hand of this kind is 624/2,598,960 = 0.000240.
4. Full House – this hand has the pattern AAABB, where A and B are two cards of
different kind. For 3 of a kind (AAA), there are 13C1 = 13 ways of choosing the
rank. To choose 3 cards of that rank, there are 4C3 = 4 ways to do that. For 2 of a
kind (BB), there are 12C1 = 12 ways of choosing the rank and 4C2 = 6 ways to
have 2 cards of that rank. The number of such hands is 13 x 4 x 12 x 6 = 3744, so
the probability to get one of these is 3744/2,598,960 = 0.001441.
5. Flush – this hand includes 5 cards from the same suit, excluding the straight
flushes. The number of hands will be (13C5)( 4C1) – 40 = 5108, with probability
approximately 0.0019654.
6. Straight – this hand includes 5 cards in a sequence where each card has rank 1
greater than previous card, and the cards allowed to be from same suit or different
suits. The number of hands of this kind is 10 x ( 4C1) x ( 4C1) x ( 4C1) x ( 4C1) x
( 4C1) = 10,240, with probability 0.003940. If straight flushes and royal flushes
are excluded, the number of hands will be 10,200, with probability 0.00392465.
7. Triple (AAABC) – in this case the hand should contain 3 different ranks, for
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Algebra 1 Summer Institute 2014
cards A, B, respectively C. Looking at one suit, the card rank for the triple AAA
can be found as 13C1 = 13 and the ranks for the other two cards BC is 12C2 (the
order of the ranks is not important here). For a rank of card A, there are 4C3 = 4
triples, and for chosen ranks of cards B and C, there is the same number of ways
of choosing card B, respectively card C : 4C1= 4 ways. The number of hands will
be (13C1) (4C3) (12C2) (4C1) (4C1) = 54,912. The probability to get this kind of
hand is 0.021128.
8. Two pairs (AABBC) – since for the group AABB the order of ranks is not
important, the number of ways of choosing 2 ranks is 13C2. The card C has 11C1
= 11 options for the rank. The groups AA and BB can be each chosen in 4C2 = 6
different ways for a given rank, and card C can be chosen in 4C1 = 4 ways. The
total number of hands is ( 13C2) ( 4C1) ( 4C2) ( 11C1)( 4C1) = 123,552 and the
probability for having a hand of this kind is 0.047539.
9. One pair (AABCD) – there are 4 different ranks in this hand. To choose the rank
for the pair, we have 13C1 = 13 ways to do it. To find the ranks for cards B, C,
and D, we have to consider that the order for their ranks is not important, so there
are 12C3 ways. For given ranks, pair AA should come in 4C2 = 6 ways and cards
B, C or D in 4C1 = 4 ways each. There are ( 13C1) ( 4C2) ( 12C3) ( 4C1)( 4C1)(
4C1) = 1,098,240. The chances of getting a one-pair hand are
1,098,240/2,598,960 =0.422569 or about 42%.
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