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1. Sample Space and Probability Part IV: Pascal Triangle and Bernoulli Trials ECE 302 Spring 2012 Purdue University, School of ECE Prof. Ilya Pollak ConnecGon between Pascal triangle and
probability theory: Number of successes in a
sequence of independent Bernoulli trials •  A Bernoulli trial is any probabilisGc experiment with two
possible outcomes Ilya Pollak
ConnecGon between Pascal triangle and
probability theory: Number of successes in a
sequence of independent Bernoulli trials •  A Bernoulli trial is any probabilisGc experiment with two
possible outcomes, e.g., –  Will CiGgroup become insolvent during next 12 months? –  Democrats or Republicans in the next elecGon? –  Will Dow Jones go up tomorrow? –  Will a new drug cure at least 80% of the paGents? Ilya Pollak
ConnecGon between Pascal triangle and
probability theory: Number of successes in a
sequence of independent Bernoulli trials •  A Bernoulli trial is any probabilisGc experiment with two
possible outcomes, e.g., –  Will CiGgroup become insolvent during next 12 months? –  Democrats or Republicans in the next elecGon? –  Will Dow Jones go up tomorrow? –  Will a new drug cure at least 80% of the paGents? •  Terminology: someGmes the two outcomes are called
“success” and “failure.” •  Suppose the probability of success is p. What is the
probability of k successes in n independent trials? Ilya Pollak
Probability of k successes in n
independent Bernoulli trials •  n independent coin tosses, P(H) = p Ilya Pollak
Probability of k successes in n
independent Bernoulli trials •  n independent coin tosses, P(H) = p •  E.g., P(HTTHHH) = p(1-­‐p)(1-­‐p)p3 = p4(1-­‐p)2 Ilya Pollak
Probability of k successes in n
independent Bernoulli trials •  n independent coin tosses, P(H) = p •  E.g., P(HTTHHH) = p(1-­‐p)(1-­‐p)p3 = p4(1-­‐p)2 •  P(specific sequence with k H’s and (n-­‐k) T’s) = pk (1-­‐p)n-­‐k Ilya Pollak
Probability of k successes in n
independent Bernoulli trials • 
• 
• 
• 
n independent coin tosses, P(H) = p E.g., P(HTTHHH) = p(1-­‐p)(1-­‐p)p3 = p4(1-­‐p)2 P(specific sequence with k H’s and (n-­‐k) T’s) = pk (1-­‐p)n-­‐k P(k heads) = (number of k-­‐head sequences) ·∙ pk (1-­‐p)n-­‐k Ilya Pollak
Probability of k successes in n
independent Bernoulli trials • 
• 
• 
• 
n independent coin tosses, P(H) = p E.g., P(HTTHHH) = p(1-­‐p)(1-­‐p)p3 = p4(1-­‐p)2 P(specific sequence with k H’s and (n-­‐k) T’s) = pk (1-­‐p)n-­‐k P(k heads) = (number of k-­‐head sequences) ·∙ pk (1-­‐p)n-­‐k Ilya Pollak
An interesGng property of binomial
coefficients Since P(zero H's) + P(one H) + P(two H's) + … + P(n H's) = 1,
n ⎛ ⎞
n k
it follows that ∑⎜ ⎟ p (1− p) n−k = 1.
k
k= 0 ⎝ ⎠
Another way to show the same thing is to realize that
n ⎛ ⎞
n k
n−k
n
n
p
(1−
p)
=
(
p
+
(1−
p))
=
1
= 1.
∑⎜⎝ k ⎟⎠
k= 0
Ilya Pollak
Binomial probabiliGes: illustraGon Ilya Pollak
Binomial probabiliGes: illustraGon Ilya Pollak
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Comments on binomial
probabiliGes and the bell curve •  Summing many independent random
contribuGons usually leads to the bell-­‐shaped
distribuGon. Ilya Pollak
Comments on binomial
probabiliGes and the bell curve •  Summing many independent random
contribuGons usually leads to the bell-­‐shaped
distribuGon. •  This is called the central limit theorem (CLT). •  We have not yet covered the tools to precisely
state the CLT, but we will later in the course. Ilya Pollak
Comments on binomial
probabiliGes and the bell curve •  Summing many independent random
contribuGons usually leads to the bell-­‐shaped
distribuGon. •  This is called the central limit theorem (CLT). •  We have not yet covered the tools to precisely
state the CLT, but we will later in the course. •  The behavior of the binomial distribuGon for
large n shown above is a manifestaGon of the
CLT. Ilya Pollak
InteresGngly, we get the bell curve even for
asymmetric binomial probabiliGes Ilya Pollak
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This tells us how to empirically esGmate the
probability of an event! •  To esGmate the probability p based on n flips,
divide the observed number of H’s by the total
number of experiments: k/n. •  To see the distribuGon of k/n for any n, simply
rescale the x-­‐axis in the distribuGon of k. •  This distribuGon will tell us –  What we should expect our esGmate to be, on
average, and –  What error we should expect to make, on average Ilya Pollak
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Note:
o  for 50 flips, the most likely outcome is the correct one, 0.8
o  it’s also close to the “average” outcome
o  it’s very unlikely to make a mistake of more than 0.2
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If p=0.8, when estimating based on 1000 flips,
it’s extremely unlikely to make a mistake of
more than 0.05.
Ilya Pollak
If p=0.8, when estimating based on 1000 flips,
it’s extremely unlikely to make a mistake of
more than 0.05.
•  Hence, when the goal is to forecast a two-way
election, and the actual p is reasonably far from
1/2, polling a few hundred people is very likely
to give accurate results.
Ilya Pollak
If p=0.8, when estimating based on 1000 flips,
it’s extremely unlikely to make a mistake of
more than 0.05.
•  Hence, when the goal is to forecast a two-way
election, and the actual p is reasonably far from
1/2, polling a few hundred people is very likely
to give accurate results.
•  However,
o  independence is important;
o  getting a representative sample is important
(for a country with 300M population, this is
tricky!)
o  when the actual p is extremely close to 1/2
(e.g., the 2000 presidential election in Florida or
the 2008 senatorial election in Minnesota),
pollsters’ forecasts are about as accurate as a
random guess.
Ilya Pollak
The 2008 Franken-­‐Coleman elecGon •  Franken 1,212,629 votes •  Coleman 1,212,317 votes •  In our analysis, we will disregard third party
candidate who got 437,505 votes (he actually
makes pre-­‐elecGon polling even more
complicated) •  EffecGvely, p ≈ 0.500064 Ilya Pollak
ProbabiliGes for fracGons of Franken vote in pre
-­‐elecGon polling based on n=2.5M (more than all
Franken and Coleman votes combined) •  Even though we are unlikely to make
an error of more than 0.001, this is not
enough because p-0.5=0.000064!
•  Note: 42% of the area under the bell
curve is to the left of 1/2.
•  When the election is this close, no poll
can accurately predict the outcome.
•  In fact, the noise in the voting process
itself (voting machine malfunctions,
human errors, etc) becomes very
important in determining the outcome.
Ilya Pollak
EsGmaGng the probability of success in
a Bernoulli trial: summary •  As the number n of independent experiments
increases, the empirical fracGon of
occurrences of success becomes close to the
actual probability of success, p. •  The error goes down proporGonately to n1/2.
I.e., error aler 400 trials is twice as small as
aler 100 trials. •  This is called the law of large numbers. •  This result will be precisely described later in
the course. Ilya Pollak