AP Statistics Curriculum 2007 Prob Count - Revision history

Tdlee: Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/"

2020-03-03T17:08:36Z

Text replacement - "{{translate|pageName=http://wiki.stat.ucla.edu/socr/" to ""{{translate|pageName=http://wiki.socr.umich.edu/"

IvoDinov: /* Counting Principles in Probability Theory */

2013-04-17T01:52:08Z

‎Counting Principles in Probability Theory

IvoDinov: added the section on Poker Game Calculations

2011-04-25T15:45:49Z

added the section on Poker Game Calculations

IvoDinov: added a subsection See also (linking to multinomial arrangements and distribution)

2011-01-21T18:21:55Z

added a subsection See also (linking to multinomial arrangements and distribution)

Jenny: /* Number of integer solutions to linear equations */

2010-06-29T18:01:18Z

‎Number of integer solutions to linear equations

Jenny: /* Hands-on combination activity */

2010-06-29T18:00:22Z

‎Hands-on combination activity

Jenny: /* Combinations with repetitions */

2010-06-29T17:59:24Z

‎Combinations with repetitions

Jenny at 16:56, 28 June 2010

2010-06-28T16:56:12Z

IvoDinov: /* Combinations with repetitions */

2010-04-21T18:22:19Z

‎Combinations with repetitions

IvoDinov: /* Combinations with repetitions */

2010-04-21T16:01:15Z

‎Combinations with repetitions

← Older revision		Revision as of 17:08, 3 March 2020
Line 111:		Line 111:
	* SOCR Home page: http://www.socr.ucla.edu		* SOCR Home page: http://www.socr.ucla.edu

−	{{translate\|pageName=http://wiki.~~stat~~.~~ucla~~.edu~~/socr~~/index.php?title=AP_Statistics_Curriculum_2007_Prob_Count}}	+	"{{translate\|pageName=http://wiki.socr.umich.edu/index.php?title=AP_Statistics_Curriculum_2007_Prob_Count}}

@@ Line 2: / Line 2: @@
 === Counting Principles in Probability Theory===
-There are many useful counting principles to compute the number of ways that certain arrangements of objects can be formed. Examples of such problems are counting combinations or permutations of objects. For a given collection of objects {<math>a_i</math>}, counting methods help us describe protocols for computing numbers of arrangements of these objects when taken n at a time. For instance the number of different possible orderings of a deck of n cards is <math>1\times 2\times 3 \times 4 \times \cdots \times n = n!</math>
+There are many useful counting principles to compute the number of ways that certain arrangements of objects can be formed. Examples of such problems are counting combinations or permutations of objects. For a given collection of objects {<math>a_i</math>}, counting methods help us describe protocols for computing numbers of arrangements of these objects when taken n at a time. For instance the number of different possible orderings of a deck of n cards is <math>1\times 2\times 3 \times 4 \times \cdots \times n = n!</math> (i.e., as \(n \longrightarrow \infty\)).
 * [http://en.wikipedia.org/wiki/Stirling%27s_approximation Sterling Formula] for asymptotic behavior of n!

← Older revision		Revision as of 15:45, 25 April 2011
Line 60:		Line 60:

	* [http://www.math.hawaii.edu/~ramsey/Probability/PokerHands.html Additional information about other interesting 5-card arrangements may be found here].		* [http://www.math.hawaii.edu/~ramsey/Probability/PokerHands.html Additional information about other interesting 5-card arrangements may be found here].
		+
		+	====[[SOCR_EduMaterials_Activities_PokerExperiment \| Poker Game]] Calculations====
		+	There is a variety of events that may be of interest for 5-card (poker) games (see also [[AP_Statistics_Curriculum_2007_Prob_Count#Hands-on_combination_activity \| this section on card games]]. Some of these are:
		+	* '''A single Pair Match''': For instance, the hand with the pattern AABCD, where A, B, C and D are from the distinct kinds (denominations) of cards: aces, twos, threes, tens, jacks, queens, and kings (there are 13 denominations, and four suits, in the standard 52 card deck). The number of such hands is <math>{13 \choose 1}{4\choose 2}{12\choose 3}{4\choose 1}^3</math>. If all hands are equally likely, the probability of a single pair is obtained by dividing this number by the total number of 5-card hands possible (<math>{52\choose 5}=2,598,960</math>). Thus, P(1 pair only) = 0.422569.
		+
		+	* '''Two pairs''': For instance, the pattern AABBC where A, B, and C are from distinct kinds. The number of such hands is <math>{13\choose 2}{4\choose 2}{4\choose 2}{11\choose 1}{4\choose 1}</math>. And therefore, dividing by <math>{52\choose 5}=2,598,960</math>, the P(2 pairs) = 0.047539.
		+
		+	* '''One triple''': For example, the pattern AAABC where A, B, and C are from distinct kinds. The number of such hands is <math>{13\choose 1}{4\choose 3}{12\choose 2}{4\choose 1}^2</math>. Thus, the P(1 triple) = 0.021128.
		+
		+	* '''A Full House''': This includes patterns like AAABB where A and B are from distinct kinds. The number of such hands is <math>{13\choose 1}{4\choose 3}{12\choose 1}{4\choose 2}</math>. Thus, P(Full House)=0.001441.
		+
		+	* '''Four of a kind''': For instance, the pattern AAAAB where A and B are from distinct kinds. The number of such hands is <math>{13\choose 1}{4\choose 4}{12\choose 1}{4\choose 1}</math>. The P(4-of-a-kind)=0.000240.
		+
		+	* '''A straight''': Five cards in a sequence (e.g., 4,5,6,7,8), with aces allowed to be either 1 or 13 (low or high) and with the cards allowed to be of the same suit (e.g., all hearts) or from some different suits. The number of such hands is <math>10{4\choose 1}^5</math>. Thus, P(A Straight)=0.003940. But if you exclude Straight-Flushes AND Royal Flushes, the number of such hands is <math>10{4\choose 1}^5 - 36 - 4 = 10200</math>, the corresponding probability P(A Straight, but not a Straight or Royal Flush)=0.00392465.
		+
		+	* '''A Straight Flush''': All 5 cards are from the same suit and they form a straight. The number of such hands is 4*10, and P(Straight Flush)=0.0000153908.
		+
		+	* '''A Royal Flush''': This consists of the ten, jack, queen, king, and ace all in one suit. There are only 4 such hands. Thus, P(Royal Flush)=0.00000153908.
		+
		+	<center>[[Image:SOCR_EBook_Dinov_Probability_013008_Fig2.jpg\|400px]]</center>

	===Applications===		===Applications===

@@ Line 77: / Line 77: @@
 : If not all the money needs to be invested, let ''x_5'' be the left over money, then <math>x_1+x_2+x_3+x_4+x_5=20</math>:
 :: <math>{{24} \choose {4}} = 10,626.</math>
 ===[[EBook_Problems_Prob_Count|Problems]]===

@@ Line 64: / Line 64: @@
 ====Number of integer solutions to linear equations====
 * Preview: Suppose we have ''n'' balls that we want to randomly position in ''r'' distinguishable urns, assume <math>n\ge r</math>. What is the number of possible arrangements of these balls?
-** If the balls are distinguishable (labeled): <math>n^r</math> possible outcomes, where empty urns are permitted. Since each of the ''n'' balls can be placed in any of the ''r'' urns.
+** If the balls are distinguishable (labeled): <math>n^r</math> possible outcomes, where empty urns are permitted. Each of the ''n'' balls can be placed in any of the ''r'' urns.
 ** If the balls are indistinguishable: no empty urns are allowed – select ''r-1'' of all possible ''n-1'' dividing points between the ''n'' balls. Empty urns are allowed.

@@ Line 55: / Line 55: @@
 <center>[[Image:SOCR_EBook_Dinov_Probability_012908_Fig4.jpg|400px]]</center>
-* '''Alternative Approximation''': And we can also use the [http://socr.ucla.edu/htmls/exp/Poker_Experiment.html Poker Experiment] to estimate the probability of a flush (V=5). Choose a stopping criterion of ''V=5'' (this is flush). Then run the experiment. It will stop the first time you observe a ''flush''. Therefore, 1/(number of experiments) is an '''estimate''' of the desired probability of observing a flush (see image below) - in this case '''1/1,234.'''
+* '''Alternative Approximation''': We can also use the [http://socr.ucla.edu/htmls/exp/Poker_Experiment.html Poker Experiment] to estimate the probability of a flush (V=5). Choose a stopping criterion of ''V=5'' (this is flush). Then run the experiment. It will stop the first time you observe a ''flush''. Therefore, 1/(number of experiments) is an '''estimate''' of the desired probability of observing a flush (see image below) - in this case '''1/1,234.'''
 <center>[[Image:SOCR_EBook_Dinov_Probability_101708_Fig5.jpg|400px]]</center>

@@ Line 41: / Line 41: @@
 : Example selection: <math>\{1 1, 0 ,  1, 0, 0, 0, 0, 0, 0, 0\}</math> (one repetition).
 : Another selection: <math>\{0 , 0 ,  1, 0, 0, 1, 0, 0, 0, 1\}</math> (no repetitions).
-Now let's list these these arrangements as strings of 1's and 0's, where ones indicate doughnuts of the specified type, and zeros indicate spaces between doughnut categories (types):
+Now let's list these arrangements as strings of 1's and 0's, where ones indicate doughnuts of the specified type, and zeros indicate spaces between doughnut categories (types):
 :    <math>\{1 1 0 0 1 0 0 0 0 0 0 0\}</math>, first arrangement of 3 doughnuts containing two doughnuts of Type 1, one of Type 3, where the nine zeros represent the nine boundaries between the 10 doughnut types.
 :    <math>\{0 0 1 0 1 0 1 0 0 0 0 0\}</math>, second arrangement of 3 doughnuts containing one doughnut of Type 3, one of Type 4, and one of Type 5.

@@ Line 48: / Line 48: @@
 ====Hands-on combination activity====
-The [[SOCR_EduMaterials_Activities_CardExperiment | SOCR Card Experiment]] allow us to see how the number of combinations may be used to estimate probabilities of various events (in terms of card hand outcomes). Suppose we are interested in the probability that all 5 cards in a randomly drawn hand are of the same suit (''flush'').
+The [[SOCR_EduMaterials_Activities_CardExperiment | SOCR Card Experiment]] allows us to see how the number of combinations may be used to estimate probabilities of various events (in terms of card hand outcomes). Suppose we are interested in the probability that all 5 cards in a randomly drawn hand are of the same suit (''flush'').
 * '''Theoretical calculation''': The exact probability of a flush (all 5 cards are from the same suit) is computed as follows: The number of ''flush hands'' is <math>{{4} \choose {1}} \times {{13} \choose {5}}= 4 \times 1,287 = 5,148</math>, which reflects the number of ways to select one of the 4 suits [[Image:CardSuits_Aces.jpg|50px]] and the number of ways to select 5 from the 13 available card denominations [[Image:CardDenominations_Clubs.jpg|100px]]. The total number of 5-card poker hands is <math>{{52}\choose{5}}= 2,598,960</math>. Therefore, the probability <math>P(flush) ={5,148 \over 2,598,960} =0.00198079</math>. See also [[AP_Statistics_Curriculum_2007_Prob_Simul#Poker_Game |this section on poker game odds calculations]].

@@ Line 38: / Line 38: @@
 To understand this formula, consider the n=10 doughnuts as 10 categories <math>\{D_1, D_2, \cdots, D_{10}\}</math>, and the combination of 3 doughnuts , with repetitions, as arrangements. So several examples include:
-:    <math>\{1, 1, 0, 0, 1, 0, 0, 0, 1, 0, 0, 0\}</math>, first arrangement of 3 doughnuts containing two doughnuts of Type 1, one of Type 4.
+: Doughnut types:  <math>\{D_1 , D_2 ,  D_3 , D_4 , D_5 , D_6 , D_7 , D_8 , D_9 ,  D_{10}\}</math>
-:    <math>\{0, 0, 1,0, 1, 0, 1, 0, 0, 0, 0, 0\}</math>, second arrangement of 3 doughnuts containing one doughnut of Type 4, one of Type 6, and one of Type 8.
+: Example selection: <math>\{1 1, 0 ,  1, 0, 0, 0, 0, 0, 0, 0\}</math> (one repetition).
 : etc.
-We set an order of the categories and count how many from each category are chosen. Each combination will contain three 1’s, and nine 0's (between doughnuts-group category division). We use 0's to track the transitions. Thus, associated with each event is a binary string of 1’s (things to be chosen) and 0’s (transitions between categories).
+We set an order of the categories and count how many from each category are chosen. Each combination will contain three 1’s, and nine 0's (between doughnut-group category divisions). We use 0's to track the transitions between doughnut types. Thus, associated with each event is a binary string of 1’s (doughnut types to be chosen) and 0’s (9 transitions between the 10 doughnut categories). Hence we have <math>{{(n + k - 1)!} \over {k!(n - 1)!}} = {{10 + 3 - 1} \choose {3}} = {{12} \choose {3}} = 220</math> combinations (with possible doughnut type repetitions).
 ====Hands-on combination activity====

@@ Line 36: / Line 36: @@
 ====Combinations with repetitions====
 When the order does not matter and an object can be chosen more than once, then the number of combinations is <math>{{(n + k - 1)!} \over {k!(n - 1)!}} = {{n + k - 1} \choose {k}} = {{n + k - 1} \choose {n - 1}}</math>, where n is the number of objects from which you can choose and k is the number to be chosen. For example, if you have ten types of doughnuts (n) on a menu to choose from and you want three doughnuts (k) there are (10 + 3 − 1)! / 3!(10 − 1)! = 220 ways to choose.
 ====Hands-on combination activity====