Economics 2500.03

 

Introductory Statistics

Section A

 

Professor Barry Smith

 

 

 

 

 

 

 

 

Puzzle #1: Matching Birthdays (A)

 

Suppose you have a group of N people (say, a class of 95 students). What is the chance that 2 or more people in the group will have birthdays on the same day?

 

 

 

 

Size of Group (N)

Probability of at least 1 match (P)

5

0.027

10

0.117

20

0.411

23

0.507

30

0.706

40

0.891

60

0.994

 

Answer to Birthday Problem (A)

 

In a group of N people, what is the chance that two or more will have a birthday on the same day?

 

Suppose there are 365 equally likely birthdays (i.e. forget about February 29)

 

Note that the following is true:

 

Probability that two or more people in a group N share the same birthday (P)

+ probability that fewer than two in a group of N share the same birthday (Q)

=1

 

That is P + Q = 1 (one of the two events will happen)

 

We are after P but we will get Q (it is easier to figure out) and then get P by: P = 1-Q.

 

So, we want to figure out the chance of nobody matching.

 

First, we want to figure out how many patterns of birthdays we could have with N people. The first person could have his/her birthday in one of 365 ways (on any of 365 days). The same is true for the second person.

 

The same is true for the N’th person. Thus, there are different possible birthday patterns.

 

Now, of the 365N patterns, some will contain matching numbers and others won’t. If there are no matching birthdays then:

There are 365 ways to choose the first person’s birthday there are 364 ways to choose the 2nd person’s birthday (note: the –1 is to make sure that there is no match.

There are 365 – N +1 ways to choose the Nth person’s birthday.

 

Thus, of the 365N possible patterns,

 

365 · 364 · 363 ¼ (365-N+1)

 

arise with no matching birthdays

 

Thus:

 

This is approximately equal to:

Table I (shown in class) was more carefully calculated.

 

Here is a simpler version of the version of the problem:

Suppose there are only four birthdays: 1, 2, 3, or 4.

(That is, some world where there are 4 days in a year)

 

In a class of 3, how many ‘patterns’ could you get? (43=64)

 

 

1,1,1

2,1,1

3,1,1

4,1,1

1,1,2

2,1,2

*3,1,2

*4,1,2

1,1,3

*2,1,3

3,1,3

*4,1,3

1,1,4

*2,1,4

*3,1,4

4,1,4

1,2,1

2,2,1

*3,2,1

*4,2,1

1,2,2

2,2,2

3,2,2

4,2,2

*1,2,3

2,2,3

3,2,3

*4,2,3

*1,2,4

2,2,4

*3,2,4

4,2,4

1,3,1

*2,3,1

3,3,1

*4,3,1

*1,3,2

2,3,2

3,3,2

*4,3,2

1,3,3

2,3,3

3,3,3

4,3,3

*1,3,4

*2,3,4

3,3,4

4,3,4

1,4,1

*2,4,1

*3,4,1

4,4,1

*1,4,2

2,4,2

*3,4,2

4,4,2

*1,4,3

*2,4,3

3,4,3

4,4,3

1,4,4

2,4,4

3,4,4

4,4,4

 

I put a * beside all the cases where there are no matches.

 

So,

 

Puzzle #2: Matching Birthdays (B)

 

Suppose you want to find someone whose birthday is the same as yours. How many strangers (say, in a shopping mall) will you have to approach before having a 50% chance of getting a match?

 

NOTE: If you carried out the experiment, you could get lucky and find a match on the first try. Similarly, you could be unlucky and get no match after 365 tries. Suppose you visited many many malls and kept records of how many people you asked before getting a match. Make a list of the number of tries ranked from smallest to largest. Choose the number (of tries) half way along the list … this is the number we want.

 

Theoretical Solution:

 

no. of tries 253

 

Answer to Birthday Problem (B)

 

This one is easier than problem (A).

 

How many strangers will you have to approach before having a 50% chance of getting a match on birthdays?

 

Let R be the numbers of people you approach.

 

First, there are 364 ways that a person you approach won’t have a matching birthday.

 

Thus, the chance that any person you approach won’t match is .

 

The chance that R people won’t match is:

 

So: Prob. of a match after R tries is

You have to solve for R such that P=.5

 

Thus

 

Discrimination?

 

University of California at Berkeley (1973)

Graduate Admissions (Stylized Version)

 

 

 

Men

Women

 

applied

admitted

admit rate

applied

admitted

admit rate

Total

8300

3700

.45

4100

1500

.37

 

Evidence of discrimination against women?

 

 

ARTS

2300

700

.304

3000

939

.313

SCIENCE

6000

3000

.50

1100

561

.51

 

Looks like discrimination against men!

 

Explanation:

 

Overall admit rate to ARTS

Overall admit rate to SCIENCE

 

 

 

QUESTION: … does this hold for wages too?

 

(SIMPSON’S PARADOX)

 

Swindles

 

  1. For 7 weeks in a row a stock analyst correctly tells you whether a given stock will go up or down. On the 8th week s/he demands a large fee to tell you how the stock will move. Should you pay it? (Suppose that your ‘portfolio’ is large enough so that if the analyst was correct and you bought the stock you would earn more than the demanded fee.)