SOCR EduMaterials Activities BirthdayExperiment - Revision history

IvoDinov at 15:55, 4 April 2008

2008-04-04T15:55:34Z

IvoDinov: /* Exact Calculations */

2008-04-02T17:35:13Z

‎Exact Calculations

IvoDinov at 17:31, 2 April 2008

2008-04-02T17:31:52Z

IvoDinov: /* The Birthday Paradox */

2008-04-02T17:29:36Z

‎The Birthday Paradox

IvoDinov: /* The Birthday Paradox */

2008-04-02T17:24:11Z

‎The Birthday Paradox

IvoDinov: /* The Birthday Paradox */

2008-04-02T15:46:30Z

‎The Birthday Paradox

IvoDinov at 15:45, 2 April 2008

2008-04-02T15:45:13Z

IvoDinov: /* Experiment */

2007-06-17T00:16:07Z

‎Experiment

PriscillaChui: /* Applications */

2007-05-16T01:11:25Z

‎Applications

PriscillaChui: /* Goal */

2007-05-13T04:01:04Z

‎Goal

@@ Line 9: / Line 9: @@
 == Goal ==
-The purpose of this experiment is to draw attention toward the behaviors of random sampling with replacement.
+The purpose of this experiment is to draw attention toward the behaviors of random sampling with replacement. The [[SOCR_EduMaterials_Activities_Birthday |second part of the Birthday materials provide hands-on activities]].
 == Experiment ==

@@ Line 48: / Line 48: @@
 Denote by ''P(E)'' the probability of the [[AP_Statistics_Curriculum_2007_Prob_Basics#Event_Manipulations|complementary event]], E = {All ''N'' birthdays are ''different''}. Then,
-:<math>P(E) = 1 \times \left(1-\frac{1}{365}\right) \times \left(1-\frac{2}{365}\right)  \cdots \left(1-\frac{N-1}{365}\right) = { 365 \times 364 \cdots (365-N+1) \over 365^N } = { 365! \over 365^N (365-N)!}</math>
+:<math>P(E) = 1 \times \left(1-\frac{1}{365}\right) \times \left(1-\frac{2}{365}\right)  \times\cdots \times\left(1-\frac{N-1}{365}\right)</math>
 because the second person cannot have the same birthday as the first one, <math>\left(1-\frac{1}{365}\right)=\frac{364}{365}</math>, the third cannot have the same birthday as the first two, <math>\left(1-\frac{2}{365}\right)=\frac{363}{365}</math>, the forth cannot have the same birthday as the first three, <math>\left(1-\frac{3}{365}\right)=\frac{362}{365}</math>, etc.

@@ Line 13: / Line 13: @@
 == Experiment ==
-Go to the [http://www.socr.ucla.edu/htmls/SOCR_Experiments.html SOCR Experiments] and select the Birthday Experiment from the drop-down list of experiments on the top left. The image below shows the initial view of this experiment:
+Go to the [http://www.socr.ucla.edu/htmls/SOCR_Experiments.html SOCR Experiments] and select the '''Birthday Experiment''' from the drop-down list of experiments on the top left. The image below shows the initial view of this experiment:
 <center>[[Image:SOCR_Activities_BirthdayExperiment_Chui_050407_Fig1.jpg|300px]]</center>
@@ Line 56: / Line 56: @@
 :<math> P(E^c) = 1 - P(E).</math>
-For ''N'' = 23,  <math>P(E^c) \approx 0.507</math>, and for ''N'' = 100,  <math>P(E^c) \approx 0.9999996</math>.
+For ''N'' = 23,  <math>P(E^c) \approx 0.507</math>, and for ''N'' = 100,  <math>P(E^c) \approx 0.9999996</math>. You can explicitely compute all these probabilities for any ''N'' using the '''Birthday Experiment''' in [http://www.socr.ucla.edu/htmls/SOCR_Experiments.html SOCR Experiments].
 {{translate|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_Activities_BirthdayExperiment}}

@@ Line 41: / Line 41: @@
 The reason for such a high probability is that any of the 20 people can compare their birthday with any other one, not just you comparing your birthday to anybody else’s.
-* '''Approximate Argument''': There are <math>{N \choose 2} = 20*19/2=190</math> ways to select a pair or people from a pool of 20 people. Assume there are 365 days in a year, P(one-particular-pair-same-B-day)=1/365, and
+====Approximate Argument====
 P(one-particular-pair-failure)=1-1/365 ~ 0.99726. For N=20, let the event E={No 2 people have the same birthday}. Then E is the event {all 190 pairs fail (i.e., have different birthdays)}, then (''assuming independence of each of these individual events!'') <math>P(E) = P(failure)^{190} = 0.99726^{190} = 0.59</math>. Hence,  P(at-least-one-success)=1-0.59=0.41, quite high. Note: for ''N=42, P > 0.9''.
-* '''Exact Calculations''': Denote the probability of the [[AP_Statistics_Curriculum_2007_Prob_Basics#Event_Manipulations|complementary event]], E = {All ''N'' birthdays are ''different''}, by ''P(E)''. Then,
+====Exact Calculations====
-:<math>P(E) = 1 \cdot \left(1-\frac{1}{365}\right) \cdot \left(1-\frac{2}{365}\right)  \cdots \left(1-\frac{N-1}{365}\right) = { 365 \cdot 364 \cdots (365-N+1) \over 365^N } = { 365! \over 365^N (365-N)!}</math>
+:<math>P(E) = 1 \times \left(1-\frac{1}{365}\right) \times \left(1-\frac{2}{365}\right)  \cdots \left(1-\frac{N-1}{365}\right) = { 365 \times 364 \cdots (365-N+1) \over 365^N } = { 365! \over 365^N (365-N)!}</math>
-because the second person cannot have the same birthday as the first one (364/365), the third cannot have the same birthday as the first two (363/365), etc.
+because the second person cannot have the same birthday as the first one, <math>\left(1-\frac{1}{365}\right)=\frac{364}{365}</math>, the third cannot have the same birthday as the first two, <math>\left(1-\frac{2}{365}\right)=\frac{363}{365}</math>, the forth cannot have the same birthday as the first three, <math>\left(1-\frac{3}{365}\right)=\frac{362}{365}</math>, etc.
 The [[AP_Statistics_Curriculum_2007_Prob_Basics#Types_of_probabilities|event]] of at least two of the ''N'' people having the same birthday is [[AP_Statistics_Curriculum_2007_Prob_Basics#Event_Manipulations|complementary]] to E, i.e., we are interested in the probability <math>P(E^c)</math>, which is equal to

@@ Line 37: / Line 37: @@
 ===The Birthday Paradox===
-The ''Birthday Paradox'' is not a real paradox, despite the fact that its statement may sound a little counter-intuitive, initially. Suppose we have a random group of N people. What is the change that at least two people have the same birthday? For example, if N=20, P(one-or-more-Birthday-matches) > 0.4. Main confusion arises from the fact that in real life we rarely meet people having the same birthday as us, and we meet more than 20 people.
+The ''Birthday Paradox'' is not a real paradox, despite the fact that its statement may initially sound a little counter-intuitive. Suppose we have a random group of N people. What is the change that at least two people have the same birthday? For example, if N=20, P(one-or-more-Birthday-matches) > 0.4. Main confusion arises from the fact that in real life we rarely meet people having the same birthday as us, and we meet more than 20 people.
 The reason for such a high probability is that any of the 20 people can compare their birthday with any other one, not just you comparing your birthday to anybody else’s.
-There are <math>{N \choose 2} = 20*19/2=190</math> ways to select a pair or people from a pool of 20 people. Assume there are 365 days in a year, P(one-particular-pair-same-B-day)=1/365, and
+* '''Approximate Argument''': There are <math>{N \choose 2} = 20*19/2=190</math> ways to select a pair or people from a pool of 20 people. Assume there are 365 days in a year, P(one-particular-pair-same-B-day)=1/365, and
-P(one-particular-pair-failure)=1-1/365 ~ 0.99726.
+P(one-particular-pair-failure)=1-1/365 ~ 0.99726. For N=20, let the event E={No 2 people have the same birthday}. Then E is the event {all 190 pairs fail (i.e., have different birthdays)}, then (''assuming independence of each of these individual events!'') <math>P(E) = P(failure)^{190} = 0.99726^{190} = 0.59</math>. Hence,  P(at-least-one-success)=1-0.59=0.41, quite high. Note: for ''N=42, P > 0.9''.
-For N=20, let the event E={No 2 people have the same birthday}. Then E is the event {all 190 pairs fail (i.e., have different birthdays)}, then <math>P(E) = P(failure)^{190} = 0.99726^{190} = 0.59</math>. Hence,  P(at-least-one-success)=1-0.59=0.41, quite high.
+* '''Exact Calculations''': Denote the probability of the [[AP_Statistics_Curriculum_2007_Prob_Basics#Event_Manipulations|complementary event]], E = {All ''N'' birthdays are ''different''}, by ''P(E)''. Then,
-Note: for ''N=42, P > 0.9''.
+:<math>P(E) = 1 \cdot \left(1-\frac{1}{365}\right) \cdot \left(1-\frac{2}{365}\right)  \cdots \left(1-\frac{N-1}{365}\right) = { 365 \cdot 364 \cdots (365-N+1) \over 365^N } = { 365! \over 365^N (365-N)!}</math>
 {{translate|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_Activities_BirthdayExperiment}}

← Older revision		Revision as of 15:45, 2 April 2008
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	For example, researchers are interested to know the probability of selecting a male who is born on May 15, 1986.		For example, researchers are interested to know the probability of selecting a male who is born on May 15, 1986.
		+
		+	===The Birthday Paradox===
		+	The ''Birthday Paradox'' is not a real paradox, despite the fact that its statement may sound a little counter-intuitive, initially. Suppose we have a random group of N people. What is the change that at least two people have the same birthday? For example, if N=23, P(one-or-more-Birthday-matches) > 0.5. Main confusion arises from the fact that in real life we rarely meet people having the same birthday as us, and we meet more than 23 people.
		+
		+	The reason for such a high probability is that any of the 23 people can compare their birthday with any other one, not just you comparing your birthday to anybody else’s.
		+
		+	There are <math>{N \choose 2} = 20*19/2=190</math> ways to select a pair or people from a pool of 20 people. Assume there are 365 days in a year, P(one-particular-pair-same-B-day)=1/365, and
		+	P(one-particular-pair-failure)=1-1/365 ~ 0.99726.
		+
		+	For N=20, let the event E={No 2 people have the same birthday}. Then E is the event {all 190 pairs fail (i.e., have different birthdays)}, then <math>P(E) = P(failure)^{190} = 0.99726^{190} = 0.59</math>. Hence, P(at-least-one-success)=1-0.59=0.41, quite high.
		+
		+	Note: for ''N=42, P > 0.9''.
		+

	{{translate\|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_Activities_BirthdayExperiment}}		{{translate\|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_Activities_BirthdayExperiment}}

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 == Applications ==
-The Birthday Experiment may be used for many different types of events that involve selecting individual elements from a large population. For example, setting variable V as the desired event in the Birthday Experiment may represent a quality (e.g. birth date, age, height, etc.) for every person in a city. Similarly, variable I may represent two distinct variables that are being represented (e.g. gender, left/right-handed, married/single, etc.). Note that the probability density graph could be symbolized as a hypothesis in this experiment.
+The Birthday Experiment may be used for many different types of events that involve selecting individual elements from a large population. Setting variable V as the desired event in the Birthday Experiment may represent a quality (e.g. birth date, age, height, etc.) for every person in a city and variable I as two distinct variables that are being represented (e.g. gender, left/right-handed, married/single, etc.). Note that the probability density graph could be symbolized as a hypothesis in this experiment.
 {{translate|pageName=http://wiki.stat.ucla.edu/socr/index.php?title=SOCR_EduMaterials_Activities_BirthdayExperiment}}

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 The purpose of this experiment is to draw attention toward the behaviors of random sampling with replacement.
-'''*Experiment:'''
+== Experiment ==
 Go to the SOCR Experiments [[http://www.socr.ucla.edu/htmls/SOCR_Experiments.html]] and select the Birthday Experiment from the drop-down list of experiments on the top left. The image below shows the initial view of this experiment:
@@ Line 28: / Line 28: @@
 <center>[[Image:SOCR_Activities_BirthdayExperiment_Chui_050407_Fig3.jpg|300px]]</center>
 As the number of trials increase, the empirical density function graph in red begins to look more similar to the probability density graph in blue.
 == Applications ==