Math/Science Initiative
 Professor Shai
Basic Probability
Introduction:
I am thinking of a
number. It is either 1, 10 or 100. If I asked you to guess
the number, you would have one out of three chances of getting it
right, or simply 1/3. That is all there is to basic
probability. You count how many possibilities there can be (3 in
this case); you count how many of those make you successful (1 in
this case); and you divide the number of successful outcomes by
the total number of outcomes.
If I said I was thinking of two numbers from the set 1, 10, 100 and you
had to guess either one to win, then your chance to win would be
2/3.
This is a very simple idea
but it can become very difficult as the things you are working with
become harder to count. Here are some simple examples to make
sure you get the idea:
What is the probability of
throwing heads when you flip a coin? There are two outcomes
(don't be a smartypants and suggest that the coin can land on it side)
and you win with one of them. So the chances are 1/2.
What is the chance of
throwing a 2 or a 5 when you roll one die? There are six
outcomes, and you win with two of them, so the chances are 2/6 = 1/3.
Now for a harder
example. You may recognize this as the start of a game of Craps.
What is the chance of
throwing a 7 or 11 when you roll two dice? Let's first do it wrong, just
to so you can see how it easy it is to make mistakes in
probability. There are 11 possible sums 2, 3, 5, 6, 7, 8, 9, 10,
11, 12. We win with two of this so the probability is
2/11... which is completely wrong! The reason it is wrong
is subtle. When you count the number if possible outcomes, you
must be sure you are counting the right thing. The possible
outcomes need to be things that equally likely to occur. We
counted 2 and 7 each once in the list of outcomes, but there are many
ways to get a 7 and only one way to get a 2. Really we
should have counted 7 six times: 16, 25, 34, 43, 52, 61.
Here is the correct way to
solve the problem. There are really 36 outcomes, as you can see
in the table below:

1

2

3

4

5

6

1

2

3

4

5

6

7

2

3

4

5

6

7

8

3

4

5

6

7

8

9

4

5

6

7

8

9

10

5

6

7

8

9

10

11

6

7

8

9

10

11

12

One die is listed vertically
and the other horizontally. Their sums appear in the table.
Out of these 36 there are 8 that add up to either 7 or 11, so the
chance of rolling a 7 or 11 is 8/36 = 2/9. You might have noticed
that the chance of not rolling a 7 or 11 is 7/9, or simply 1 
2/9.
This is general principle I in
probability  the complement
principle: The chance of not having something is equal to
1 minus the chance that it does happen.
Applications
 Casinos:
Why do we care about
probability? Because it helps up make predictions about the
future. For a gambler, meteorologist, and investor, probability
is their bread and butter. For example, let's say I want to bet
on a game where I roll two dice and I win if I get a 7 or 11. We already calculated that
the chance I win this game is 2/9. Does that mean that every nine
times I play, I will win twice and lose seven times? Absolutely
not! I might get lucky and win all nine times. It does
mean, however, that in the long run my winning percentage is expected
to be 2/9. So
if I play 9 million times, I expect to win around 2 million
times. It is
much more unlikely to win 9 million times in a row, than it is to win 9
times in a row. This
is called the law of large
numbers, which is quite hard to rigorously state but much easier to
understand inuitively. Let's get back to our game.
Say the casino offers to pay me
3:1 if I win. That is, if I bet $100 and I win, then they give me
my bet back plus a gain of $300. What will happen if I play this
game a lot? In the long
run I will win 2/9 and lose 7/9. Assuming I bet $100 every time
(big spender) then after nine times I gain $300 twice, and lose $100
seven times. This is a net loss of $100 for every $900 I spend on
the game. I might win in the short run by luck, but in the long
run I am playing a game that is hugely in favor of the casino.
That is how casinos make money. It is a nobrainer
business. Just sit back let probability and the law of large
numbers do its work.
In real life, casinos make their
games more favorable to make it more fun and exciting for the
player. For example, in the real game of Craps the chance of you
winning is about 493/1000, and the odds are 1:1. Hence if you
spend $1000 on Craps then you expect to win around $493 and lose around
$507 for a net loss of only about $14. For lots of people, this
might be a fair price just for the entertainment value of sitting
around a craps table and making $1000 worth of bets. Of course,
you might get lucky and win money, or get unlucky and lose all your
money, but in the long run, you will lose $14 per every $1000 you bet
playing Craps. Of course, the casino lives for the long run, they
don't care who gets lucky and unlucky as long as they take in lots of
players and they get their consistent $14 payoff. If Foxwoods has
100 Craps tables, and each one takes in $1,000,000 worth of bets a
day. Then Foxwoods expects to earn $1,400,000 a day from its
Craps tables alone. This leaves a lot of profit even after you
pay off the 500 or so employees needed to run the tables for the day.
In class we will analyze just why the odds of winning a game of Craps
is about 493/1000.
Applications  Basketball and
Freethrows
Here is a hard one because it is
not clear what to count. Let's say I am a 60% freethrow
shooter. That means that the number of freethrows I make divided
by the number I attempt equals .60 or 3/5. What is the chance
that I will make two freethrows in a row? Here too we can make a
table where the first freethrow attempt is shown horizontally and the
second vertically. An X marks that I made the freethrow, and an O
means I missed. I put three X's and two O's randomly for each of
the two throws. Inside the table, I put in a two letter
combination showing what happened on the two throws.

X

O

X

X

O

O

X,
O

O,
O

X,
O

X,
O

O,
O

X

X,
X

O,
X

X,
X

X,
X

O, X

O

X,
O

O,
O

X,
O

X, O

O,
O

X

X,
X

O,
X

X, X

X,
X

O,
X

X

X,
X

O, X

X,
X

X,
X

O,
X

After two throws there are 9 times
when both shots are made, 4 times when neither is made, and 12 times
when one out of two is made. So the chance you make both
freethrows is 9/25. You might have noticed that 9/25 is
just 3/5 x 3/5.
This is general principle II of
probability  the multiplication
principle: The chance of two independent events both
happening is the product of their individual probabilities.
Review Problem: You are sitting at home watching the NBA
finals. Shaq is on the freethrow line with no time left on the
clock. His freethrow shooting percentage is a dismal 68%.
He needs to make both shots to win the game. Your Uncle says that
he bets you $20 they don't win the game, and he will give you 2:1
odds. Do you take his bet?
Solution: The chance Shaq
will make both shots is .68 x .68 =
.4624. If you made this $20 bet 10,000 times then you would
expect to win 4624 times, and lose 5376 times. You win $40 4624
times, and you lose $20 5376. This is a net gain of $77,440. If
you make the bet once, you have an expected gain of about $7.74.
Take the bet!
Putting all the Principles to Use
I can throw a crumpled piece of paper across the class into a garbage
can 1 in 5 times. If I try five times, what's the chance I get it
in at least once?
Let's do it wrong. If I have 1/5 chance and I do it five times,
then I have 5 x 1/5 = 100% chance to get at least one. That means
if I throw it five times, I will certainly get at least one in.
Nonsense! Just watch me try  I miss five in a row fairly
often. Another way to see how silly this thinking is, is to
extend my attempts to ten throws. Is my chance to get at least
one 10 x 1/5 = 200% Double nonsense. Probabilities only
exist between 0% and 100% inclusive.
This leads to a nonprinciple of
probability  the addition nonprinciple. It is not true that: The chance of
at least one of two or more events occuring equals the sum of the
chances that each event occurs. Please never use this hideous
illogical but tempting nonprinciple.
Now let's do it right. This hard problem requires some careful
thinking, a good plan, and the use of our two main
principles. The chance of missing a throw is 4/5 (general principle I). The chance of missing all five
throws is 4/5 x 4/5 x 4/5 x 4/5 x 4/5 = .32768 (general principle II). The
chance of not missing all five, or equivalently the chance of making at
least one, is 1  .32768 = .67232 or about 67%.
Review Problem: Assuming that I am a 60% freethrow shooter, what
is the chance I get at least one freethrow out of 5 shots?
Solution: The chance
you miss a freethrow is 40% or 2/5
(general principle I). The chance you miss all 5 freethrows is
2/5 x 2/5 x 2/5 x 2/5 x 2/5 = .03125 (general principle II). The
chance you don't miss all 5 freethrows is equal to 1  .03125 = .96875
or about 97% (general principle I). Getting at least one is
the same as not missing all five, so the answer is about 97%.
Review Problem: I play the lottery 10,000 times, and my chance of
winning each time is 1 in 1,000,000, what's the chance I win at least
once?
Solution: The chance you
don't win each time is
999,999/1,000,000. The chance you don't win 10,000 times is
(999,999/1,000,000)^{10,000}. This is what calculators
are for! It comes to about .990498. The chance
of winning at least once is .009502, less than 1 in a 100.
Conclusion:
Basic probability is a matter of
counting. The chance that something occurs is simply the number
of times it occurs divided by the total number of possible
outcomes. There are two main principles in basic probability, the
complement principle and the multiplication principle. Make sure
not to use the erroneous addtion principle.
There are lots of harder
problems in probability and the difficulty lies in counting, which is
the next review topic in out list. To experiment with probability
go here
and click on probability.
Problems:
1. What's the chance to get an 8 or higher when you roll
two dice?
2. What's the chance that a couple with four children have
exactly 2 boys and 2 girls?
3. What's the chance that if I flip a coin 5 times, I will get
all heads?
4. If I shoot 65% freethrows, and I am on the line to take three
shots, what's the chance I will hit at least two out of three?
5. You are playing Scrabble. There are 20 tiles left that
you cannot see, and two of them are "e"s. If your opponent has an
"e" it is important for you to block a particular high scoring
potential spot on the board even at the expense of making your highest
scoring play. If your opponent does not have an "e", then you are
free to make your highest scoring play without worrying about opening
up a good spot for them.
a. Your opponent has 1 tile. What's the chance
that it is an "e"? What's the chance that it is not an "e"?
b. Your opponent has 2 tiles. What's the
chance that both tiles are not "e"? What's the chance that at
least one is?
c. Your opponent has 7 tiles. What's the chance that all his/her
tiles
are not "e"? What's the chance that at least one is?
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