# A primer on probabilities

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```					A primer on probabilities

Faigman & Park
Scientific Method for Lawyers
Feb. 8, 2012
First topic: Probabilities and odds
The equally likely outcomes of flipping two fair
coins are:

HH HT TH TT

The probability that both coins will be heads is 1
out of 4. The odds in favor of them both being
heads is 1 to 3. The odds against them both
being heads is 3 to 1.
The odds are 4-1 against a horse winning.
What is the probability that it will win?

1.   80%
2.   75%
50%
3.   25%
33%
4.   20%
17%

0%
%

%

%

%
80

75

25

20
A horse has a 10% chance of winning.
What are to odds against it winning?
1. 10-1
92%
2. 9-1
3. 1-10
4. 1-9

8%
0%           0%
-1

1

9
10
9-

1-
10

1-
You have one fair six-sided die. What is the
probability that on the first roll you will roll either a
2 or a 3?

100%
1.   1/6
2.   2/6
3.   4/9
4.   other

0%          0%    0%
6

6

9

r
he
1/

2/

4/

ot
When two events are mutually exclusive,
the probability that either one or the other
will occur is equal to the probability that
the first will occur plus the probability that
the second will occur.
If you draw a card at random from standard deck of 52
cards, what is the probability that it will be either an ace or
a diamond?
Note: there are 13 diamonds and 4 aces in the deck.
You draw a card at random from standard deck of 52
cards. What is the probability that it will be either an ace or
a diamond?
[Note: there are 13 diamonds and 4 aces in the deck.]

82%
1.   16/52
2.   17/52
3.   18/52
4.   13/52 * 4/52
9%                     9%
0%

2
2

2

2

/5
/5

/5

/5

*4
16

17

18

2
/5
13

A         A         A        A

2        2         2        2

3        3         3        3

4        4         4        4

5        5         5        5

6        6         6        6

7        7         7        7

8        8         8        8

9        9         9        9

10        10        10       10

J        J         J        J

Q         Q         Q        Q

K         K         K        K

A         A         A        A

2        2         2        2

3        3         3        3

4        4         4        4

5        5         5        5

6        6         6        6

7        7         7        7

8        8         8        8

9        9         9        9

10        10        10       10

J        J         J        J

Q         Q         Q        Q

K         K         K        K

For any events E, F, the probability of either E or F (or
both) occurring is equal to the probability of E occurring
plus the probability of F occurring minus the probability
of E and F occurring together. That is,

P(E OR F) = P(E) + P(F) - P(E AND F)
You have two fair six-sided dice. You are offered a
proposition bet that you will not roll a 6 on either of the dice.
If you roll a six on either die (or both), the bettor will pay
you off at \$2.02 for every \$1 you bet. In other words, you
will get back your \$1 plus \$2.02 from him. When you lose,
you lose only your \$1. He says the true odds are 2-1
against rolling a six on either die, so you are getting a slight
edge on every bet.
Here is his reasoning:

1/6 + 1/6 = 2/6 or 1/3

1/3 probability = 2-1 odds against
The bettor is offering you 2.02-1 odds.
Should you take the bet?
(Assume that you’re risk-neutral and will take any bet that
would put you ahead in the long run)

1. Yes                                 64%
2. No
36%

s

No
Ye
probabilities together. Why didn’t it apply to
the prior example?
1. The events were                           82%
not mutually
exclusive
2. The question was
not whether either                                      18%
event 1 or event 2                                                0%
would happen.

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3. Other
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Outcomes – rolling two dice
P(E OR F) = P(E) + P(F) - P(E AND F)

1.1       1.2      1.3     1.4     1.5       1.6

2.1       2.2      2.3     2.4     2.5       2.6

3.1       3.2      3.3     3.4     3.5       3.6

4.1       4.2      4.3     4.4     4.5       4.6

5.1       5.2      5.3     5.4     5.5       5.6

6.1       6.2      6.3     6.4     6.5       6.6
36 equally likely outcomes.
You win your bet in 11 of them.
11/36= .31 probability
.31 probability = 69-31 odds against
69-31 odds = 2.23-1 odds against
Bettor was offering 2.02-1 odds against

1.1   1.2          1.3          1.4           1.5   1.6

2.1   2.2          2.3          2.4           2.5   2.6

3.1   3.2          3.3          3.4           3.5   3.6

4.1   4.2          4.3          4.4           4.5   4.6

5.1   5.2          5.3          5.4           5.5   5.6

6.1   6.2          6.3          6.4           6.5   6.6
Multiplication rule method.
First, a side question. The probability
that you will roll a six on both dice is --
91%
1.   1/6 + 1/6 = 2/6
2.   1/6 * 1/6 = 1/36
3.   5/6 * 5/6 = 25/36
4.   Other
9%
0%                             0%

6
6

6

er
2/

3

/3

th
1/

25
=

O
=
6

=
/6
1/

6
*1
+

5/
6

6
1/

*
1/

6
5/
The two events are independent. The dice
are not in communication with each other.
So you can use the special multiplication
rule.
P(E AND F) = P(E) * P(F)
P(6 and 6) = 1/6 * 1/6 = 1/36
But that’s not what you’re interested in. You
don’t want to know what the probability of
rolling a six on both dice is.
You want to know the probability that you
will get a six on either the first die or the
second die (or both).
The probability that the event of not getting a
six on the first die will occur jointly with the
event of not getting a six on the second die
is--
92%
1.   1/6 + 1/6 = 2/6
2.   5/6 + 5/6 = 10/6
3.   1/6 * 1/6 = 1/36
4.   5/6 * 5/6 = 25/36
5.   Other                             0%
8%
0%                   0%

...

...
..
..

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1.
1.

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=
=

=
=

6

O
6

/6

5/
6
1/

5/

*1
+

*
+

6
6

6

5/
6
1/

1/
5/
The probability of not getting a six on either
die is 25/36. Applying the subtraction rule,
the probability of yes getting a six on either
die is
88%
1.   11/36
2.   12/36
3.   25/36
4.   other
13%
0%    0%

r
6

6

6

he
/3

/3

/3
12

25
11

ot
11/36 is the same answer that we got by the
simple counting method.
1.1    1.2     1.3    1.4    1.5     1.6

2.1    2.2     2.3    2.4    2.5     2.6

3.1    3.2     3.3    3.4    3.5     3.6

4.1    4.2     4.3    4.4    4.5     4.6

5.1    5.2     5.3    5.4    5.5     5.6

6.1    6.2     6.3    6.4    6.5     6.6
Question 2, p. 3
You are going to draw a card from a deck, then
replace the card, shuffle the deck, and draw
another card. What it the chance of drawing a

1.    ½
2.   ¼
3.   1/16
4.   clueless                          10%
0%                   0%

ss
½

16
¼

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Since the two events are independent (you
replaced the card), you can use the
special multiplication rule to solve the
previous problem.
Chance of spade first time: ¼
Chance of spade second time: ¼
Chance of spade both times: 1/16
Special multiplication rule
a/k/a product rule
• When two events are independent, the
probability that both events will occur is
equal to the probability that the first event
will occur times the probability that the
second event will occur. That is,

•    P(E AND F) = P(E) * P(F)
You are playing draw poker and have three spades in your
cards. What is the probability that you will draw two
(There are 47 unknown cards, of which 10 are spades)
100%

1.   ¼ * ¼ = 6.2%
2.   10/47 * 10/47 = 4.5%
3.   10/47 * 9/46 = 4.2%
4.   Clueless

0%            0%                              0%

...

..
2%

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=.
47

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6.

6

e
/4
0/

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=

*1

*9

C
*¼

7

7
/4

/4
¼

10

10
General Multiplication Rule
(applicable when events are not independent)
The probability that both event E and event F will
occur is equal to the probability that event E will
occur times the probability that event F will occur
given that Event E has occurred. That is,

P(E AND F) = P(E) * P(F|E)
Q-4, p. 3.
You are offered a proposition bet. You put up \$1 and get to
roll three dice. If any of the three dice comes up as a six,
you win \$1. If none of them come up as a six, you lose \$1.
Should you take the bet?
100%

1. Yes
2. No
3. Indifferent

0%               0%
s

t
o

en
Ye

N

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ffe
di
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5/6 * 5/6 * 5/6 = 125/216

125/216 = 58% chance of not getting a six
Transposition error
P(E|G) ≠ P (G|E)
Example: Thinking that the probability of
finding the blood match, given that
someone other than the defendant was
the source, is equal to the probability that
defendant was the not the source, given
that there is a blood match.
P(E|NS) ≠ P(NS|E)
Bayes’ rule could potentially solve the
problem of getting from P(E|NS) to
P(NS|E), that is, it is a way of using
evidence of the probability that you’d find
the evidence if defendant was not the
source, combining it with other evidence,
and coming up with a probability that
defendant is not the source, given that you
found the evidence.
Bayes’ Rule
Posterior odds = Likelihood ratio * prior odds
The likelihood ratio is the probability of
finding the new information, given that the
fact is true, divided by the probability of
finding the new information, given that the
fact is not true.
Posterior odds = likelihood ratio * prior odds
Hypo. Prior odds that def is the father are 1-3. His blood
matches the profile that the father would have. You’re 100
times more likely to find that match if he’s the father than if
he’s not. The posterior odds that he is the father are

1.   Unknowable
67%
2.   1-300
3.   100-3
4.   100-1
17%                             17%
5.   300-1
0%              0%                0%

6.   clueless

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30

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Example of Bayesian table
(based on a likelihood ratio of 25)

Prior        Posterior
0%    ➔           0%
5%    ➔          57%
10%    ➔          74%
15%    ➔          82%
20%    ➔          86%
25%    ➔          89%
30%    ➔          91%
35%    ➔          93%
Problems with using a Bayes
chart
-- Hard for jury to adjust for possible error in
estimates
--”dwarfing soft variables”
Bayes applied to statistical
significance
A study indicates that rubbing sticks
together is associated with rain, P = .05

Here’s a slide from a prior lesson . . .
To say that a result is statistically significant
at P = .05 is to state

1. The probability of                            82%
getting the result
given that the null
hypothesis is true.
2. The probability that
the null hypothesis is                                       9%                     9%
0%
true given the result.

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3. Both of the above.

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To state that there is a statistically significant
difference (P=.05) between the stick-rubbing group
and the control group is to state that
1. The probability of getting
the result, given that the
null hypothesis is true, is
5%                                           64%
2. Given the experimental
results, there is a 5%
chance that stick-rubbing                                    27%
makes no difference
9%
3. Both of the above

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Suppose that, given your experience, the
information you get from meteorologists,
and everything else you know about the
world, your subjective estimate is that the
odds are a thousand to one against stick-
rubbing being associated with rain.
Odds 1000 to 1 against =
1 to 1000 in favor
After you learn of the experiment showing
the association of stick-rubbing and rain, P
= .05, you should change your estimate of
the probability.
Assuming that the experiment is completely
valid, the probability of getting that result,
given that it is due to chance, is .05
Likelihood ratio =

Probability of evidence, given proposition is
true, divided by probability of evidence,
given falsity
Assume that the probability you’d get the
result if stick-rubbing is associated with
rain is 100%

The probability of getting the result if stick-
rubbing is not associated with rain = 5%
The likelihood ratio is 100% divided by 5%
=20.
ratio is 20. Your posterior odds that stick rubbing
is associated with rain are --
1.   20 to 1 in favor                                                                100%

2.   1 to 1000 in favor
3.   1000 to 1 in favor
4.   20 to 1000 in favor

0%         0%               0%

.. .

...
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10
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20

10

20
The end.

```
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