Download Probability: Bernoulli Trials, Expected Value, and More About

Probability: Bernoulli Trials,
Expected Value, and More
About Human Beings!
CSCI 2824, Fall 2012!
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Assignments
• For this week: Read Chapter 6, sec. 4; Chapter 7,
sections 1-2
• Problem Set 5 due Thursday Dec. 13
• FCQs on Thursday
• FINAL EXAM: Saturday, Dec. 15 4:30-7 PM
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Flipping a weighted coin
•  Suppose we’re going to flip a coin 4 times,
but the probability of the coin coming up
“heads” is 0.6 (not, as usual, 0.5). (Each of
these independent repeated trials of a
probabilistic event is called a Bernoulli
trial.)
•  What’s the probability of getting exactly
two heads and two tails? Flipping a weighted coin
•  Let’s take a look at the probability of a
particular sequence: say, HTTT
•  The probability of getting this exact
sequence (by our “tree” method) is 0.6*0.43
•  By similar reasoning, the probability of
getting, say, HTHT is 0.62*0.42
Flipping a weighted coin
•  Now, there are C(4, 2) ways getting exactly
two heads out of four flips; and each of
these sequences has probability 0.62*0.42
•  So, the overall probability of getting two
heads is 6 * (0.62 * 0.42) = 0.346
•  As a check: this is a little bit less than the
probability of getting two heads with four
flips of a fair coin (0.375). A general theorem about
Bernoulli trials
•  Suppose each of our trials has a probability
p of “success”. •  Then the probability of getting exactly k
successes out of n trials is just:
C(n, k) * pk * (1-p)(n-k)
A Problem in Human Intuition
Which of these sequences of coin tosses is
more likely?:
HHHHHHHH!
! !HTTHTHTH!
Expectation (or Expected Value)
•  Suppose we try the following Yahtzee-like
gamble:
•  We roll three dice. If three of a kind show
up, we get $10. If only two of a kind show
up, we get nothing. If all three dice have
distinct values, we pay $0.75. Is this a good
game to play over time?
Here’s how to solve this problem:
•  Probability of getting three of a kind: 1/36
•  Probability of two of a kind: 90/216
•  Probability of all three distinct: 120/216
•  So our expected value for the game is:
(1/36 * 10) + (90/216 * 0) + (120/216 * -0.75)
= -0.14
In other words, on average, we lose 14 cents per
play of this game.
What do we mean by “expected
value”, anyway?
•  Suppose we have a “reward” Ri associated
with each possible event i in our sample
space. Then the expected value of our
sample space is just the sum of Ri times the
probability of getting that Ri over the whole
space:
Σi Ri*pi
Expected Value of N Bernoulli
Trials…
•  Suppose we flip a fair coin eight times.
What is the expected number of heads?
0 * C(8,0) * (0.58) +
1 * C(8,1) * (0.57 * 0.5) +
2 * C(8,2) * (0.56 * 0.52) + …
8 * C(8,8) * (0.58)
Expected Value of N Bernoulli
Trials…
•  Suppose we flip a “0.6-heads” coin eight
times. What is the expected number of
heads?
0 * C(8,0) * (0.48) +
1 * C(8,1) * (0.47 * 0.6) +
2 * C(8,2) * (0.46 * 0.62) + …
8 * C(8,8) * (0.68)
A Theorem We Can Prove By
Induction…
•  In n Bernoulli trials with success probability
p, the expected number of successes is just
n*p.
•  To take an example: the expected number of
heads after flipping a fair coin 100 times is
50.
•  The expected number of heads after flipping
a 0.6-heads coin 100 times is 60.
How do we show this?
•  After 1 trial, the expected number of
successes is just 1*p = p. (That’s what it
means to have a probability p of success!)
•  So now, let’s say that if we perform m
trials, up to a given value of m (where m is
at least 1), we expect to have p*m
successes.
What we’ve proved so far (probability
p of winning in each trial)
0 * Prob[0 out of m]
+
1 * Prob[1 out of m]
+ 2 * Prob[2 out of m]
…
+ m * Prob[m out of m]
= p*m After m+1 trials
0 * Prob[0 out of m] * (1-p)
+ 1 * Prob[1 out of m] * (1-p)
+ 2 * Prob[2 out of m] * (1-p)
…
+ m * Prob[m out of m] * (1-p)
+ 1 * Prob[0 out of m] * p
+ 2 * Prob[1 out of m] * p
+ … (m+1) * Prob[m out of m] * p
= (1-p)* mp + [green part] + 1 * Prob[0 out of m] * p
+ 2 * Prob[1 out of m] * p
+ … (m+1) * Prob[m out of m] * p
= (1+ 0) Prob[0 out of m] * p
+ (1+ 1) Prob[1 out of m] * p
+ (1 + 2) Prob[2 out of m] * p
+ … (1 + m) Prob [m out of m] * p
= p*(Prob[0 out of m] + Prob[1 out of m] + … Prob[m out of m])
+ p*mp
= p + p*mp
After m+1 trials
0 * Prob[0 out of m] * (1-p)
+ 1 * Prob[1 out of m] * (1-p)
+ 2 * Prob[2 out of m] * (1-p)
…
+ m * Prob[m out of m] * (1-p)
+ 1 * Prob[0 out of m] * p
+ 2 * Prob[1 out of m] * p
+ … (m+1) * Prob[m out of m] * p
= (1-p)* mp + [green part]
= (1-p)*mp + p + p*mp
= (m+1) p A Famous Experiment on
“Guessing” by Gerd Gigerenzer
We create a test on American cities
(populations) with lots of questions of the
form:
“Which is bigger: SAN JOSE or SAN
ANTONIO?”
We then administer this test to a classroom of
American students and a classroom of German
students; the German students do better.
A Famous Experiment on
“Guessing” by Gerd Gigerenzer
Now, we create a test on German cities
(populations) with lots of questions of the
form:
“Which is bigger: DORTMUND or BREMEN?”
We then administer this test to a classroom of
American students and a classroom of German
students; now the American students do better.
"
"
Suppose we have to choose between pairs drawn from a
list of 100. Further suppose:"
"
a. When both objects are recognized, we have a 60
percent chance of getting the right answer. (E.g., is
Munich a bigger city than Berlin?)"
"
b. When both objects are unrecognized, we have a 50
percent chance. (Essentially, we’re just “flipping a coin”:
is Dortmund bigger than Duisberg?)"
"
c. When one object is unrecognized, we have an 80
percent chance of getting the right answer. (Is Munich
bigger than Dortmund?)"
"
Three people take the test, which has 100 * 99 /2 = 4950 questions.
One (person A) recognizes each and every object in the list. His
score is:!
.6 * (100 * 99 / 2) = 2970!
Person B doesn’t know a thing about the objects in the list. His
score is:!
0.5 * (100 * 99/2) = 2475!
Person C knows half the list. His score is:!
0.5 * (50 * 49 / 2) + 0.6 * (50 * 49 / 2) + 0.8 * (50 * 50) = !
612.5 + 735 + 2000 = 3347.5!
Moral: A little ignorance can sometimes help.!
Another “probability effect”
•  Estimate the proportion of English words
that begin with the letter “K” versus those
that have “K” in the third position.
The Conjunction Effect!
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!Bill is 34 years old. He is intelligent, but
unimaginative, compulsive, and
generally lifeless. In school, he was
strong in mathematics but weak in
social studies and humanities.!
Bill is a doctor, and his hobby is playing poker.!
Bill is an architect.!
Bill is an accountant.!
Bill plays jazz for a hobby.!
Bill surfs for a hobby.!
Bill is a reporter.!
Bill is an accountant who plays jazz for a hobby.!
Bill climbs mountains for a hobby.!