An introduction to Bayes theorem and the idea of updating your state of knowledge with examples. Also, time travel and robot assassins.

1. Bayes
DATA SCIENCE BOOTCAMP

2. Thug
Life.
Gangsta.

3. http://oscarbonilla.com/2009/05/visualizing-bayes-theorem/!

4. People
with
cancer
People
without
cancer
Everybody

5. P(A) =
| A |
|U |
If
I
picked
a
random
person
from
the
universe,
what
is
the
probability
of
him/her
having
cancer?

7. People
that
test
posiCve
for
cancer
People
that
test
negaCve
for
cancer
Everybody

8. P(B) =
| B |
|U |
If
I
picked
a
random
person
from
the
universe,
what
is
the
probability
of
him/her
tesCng
posiCve
for
cancer?

10. People
without
cancer
that
test
posiCve
People
without
cancer
that
test
negaCve
People
with
cancer
that
test
posiCve
People
with
cancer
that
test
negaCve

11. People
without
cancer
that
test
posiCve
(B
–
AB)
People
with
cancer
that
test
posiCve
People
with
cancer
that
test
negaCve
(A
–
AB)
People
without
cancer
that
test
negaCve

12. P(A, B) =
| AB |
|U |
If
I
picked
a
random
person
from
the
universe,
what
is
the
prob.
of
them
having
cancer
AND
tesCng
posiCve
for
cancer?
Joint
probability

13. P(A | B) =
| AB |
| B |
If
I
picked
a
random
person
that
tested
posi.ve,
what
is
the
prob.
of
him/her
actually
having
cancer?
CondiConal
probability

14. If
I
picked
a
random
person
that
tested
posi.ve,
what
is
the
prob.
of
him/her
actually
having
cancer?
CondiConal
probability
P(A | B) =
| AB |
| B |

15. If
I
picked
a
random
person
that
tested
posi.ve,
what
is
the
prob.
of
him/her
actually
having
cancer?
P(A | B) =
| AB |
| B |
=
| AB | |U |
| B | |U |
=
P(A, B)
P(B)

16. P(B | A) =
| AB |
| A |
If
I
picked
a
random
person
that
has
cancer,
what
is
the
prob.
of
him/her
tesCng
posiCve?
CondiConal
probability

17. P(B | A) =
| AB |
| A |
If
I
picked
a
random
person
that
has
cancer,
what
is
the
prob.
of
him/her
tesCng
posiCve?
CondiConal
probability

18. If
I
picked
a
random
person
that
has
cancer,
what
is
the
prob.
of
him/her
tesCng
posiCve?
P(B | A) =
| AB |
| A |
=
| AB | |U |
| A | |U |
=
P(A, B)
P(A)

19. P(A | B) =
P(A, B)
P(B)
P(test
posiCve
|
has
cancer)
P(B | A) =
P(A, B)
P(A)
P(has
cancer
|
test
posiCve)

20. P(A, B) = P(A | B)P(B)P(B | A)P(A) = P(A, B)
P(test
posiCve
|
has
cancer)
P(has
cancer
|
test
posiCve)

21. = P(A | B)P(B)P(B | A)P(A) = P(A, B)
P(test
posiCve
|
has
cancer)
P(has
cancer
|
test
posiCve)

22. P(B | A)P(A) = P(A | B)P(B)
P(test
posiCve
|
has
cancer)
P(has
cancer
|
test
posiCve)

23. P(test
posiCve
|
has
cancer)
P(B | A)P(A) = P(A | B)P(B)
P(has
cancer
|
test
posiCve)
P(test
posiCve)
P(has
cancer)

24. P(A | B) =
P(B | A)P(A)
P(B)
P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)

25. P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?

26. P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01

27. P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01
P(test
posiCve
|
has
cancer)
=
0.80

28. P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01
P(test
posiCve
|
has
cancer)
=
0.80
P(test
posiCve)
=
P(test
posiCve|has
cancer)
P(has
cancer)
+
P(test
posiCve|not
cancer)
P(not
cancer)

29. P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01
P(test
posiCve
|
has
cancer)
=
0.80
P(test
posiCve)
=
P(test
posiCve|has
cancer)
P(has
cancer)
+
P(test
posiCve|not
cancer)
P(not
cancer)
=
0.80
P(has
cancer)
+
0.096
P(not
cancer)

30. P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01
P(test
posiCve
|
has
cancer)
=
0.80
P(test
posiCve)
=
P(test
posiCve|has
cancer)
P(has
cancer)
+
P(test
posiCve|not
cancer)
P(not
cancer)
=
0.80
P(has
cancer)
+
0.096
P(not
cancer)
=
0.80
*
0.01
+
0.096*(
1
–
0.80
)

31. P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01
P(test
posiCve
|
has
cancer)
=
0.80
P(test
posiCve)
=
P(test
posiCve|has
cancer)
P(has
cancer)
+
P(test
posiCve|not
cancer)
P(not
cancer)
=
0.80
P(has
cancer)
+
0.096
P(not
cancer)
=
0.80
*
0.01
+
0.096*(
1
–
0.80
)
P(test
posiCve)
=
0.103

32. P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01
P(test
posiCve
|
has
cancer)
=
0.80
P(test
posiCve)
=
0.103
P(has
cancer|
test
posiCve)
=
?

33. P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01
P(test
posiCve
|
has
cancer)
=
0.80
P(test
posiCve)
=
0.103
P(has
cancer|
test
posiCve)
=
=
0.078
0.80
*
0.01
0.103

34. 1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01
P(test
posiCve
|
has
cancer)
=
0.80
P(test
posiCve)
=
0.103
P(has
cancer|
test
posiCve)
=
=
0.078
0.80
*
0.01
0.103
ONLY
7.8%

35. 1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
P(has
cancer)
=
0.01
P(test
posiCve
|
has
cancer)
=
0.80
P(test
posiCve)
=
0.103
P(has
cancer|
test
posiCve)
=
=
0.078
0.80
*
0.01
0.103
ONLY
7.8%
Most
doctors
guessed
~80%

36. 1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?

37. 1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?

38. 1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?

39. 1%
of
women
at
age
forty
who
parCcipate
in
rouCne
screening
have
breast
cancer.
80%
of
women
with
breast
cancer
will
get
posiCve
mammograms.
9.6%
of
women
without
breast
cancer
will
also
get
posiCve
mammograms.
A
woman
in
this
age
group
had
a
posiCve
mammography
in
a
rouCne
screening.
What
is
the
probability
that
she
actually
has
breast
cancer?
|A|/|U|=
1%
|AB|/|A|
=
80%
(|B|-­‐|AB|)/|U|
=
9.6%
|AB|/|B|
=
7.8%

40. P(A | B) =
P(B | A)P(A)
P(B)
P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)

41. P(A | B) =
P(B | A)P(A)
P(B)
P(test
posiCve
|
has
cancer)
P(has
cancer|
test
posiCve)
P(test
posiCve)
P(has
cancer)
posterior
likelihood
prior
evidence

42. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(A | B) =
P(B | A)P(A)
P(B)

43. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
=
#
nominated
/
#
released
P(A | B) =
P(B | A)P(A)
P(B)

46. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
#
released
P(A | B) =
P(B | A)P(A)
P(B)

49. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
=
10
/
600
P(A | B) =
P(B | A)P(A)
P(B)

50. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(A | B) =
P(B | A)P(A)
P(B)

51. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
=
#
MS
movies
/
#
all
P(A | B) =
P(B | A)P(A)
P(B)

52. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
=
#
MS
movies
/
600
P(A | B) =
P(B | A)P(A)
P(B)

55. #
MS
movies
74
movies
In
40
years
1.85
Meryl
Movies
per
year
on
average

56. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
≈
1.85
/
600
=
0.3%
P(A | B) =
P(B | A)P(A)
P(B)

57. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
≈
1.85
/
600
=
0.3%
P(Meryl
Streep
in
the
movie|oscar
nominaCon)
=
#
nominated
MS
movies
/
all
nominated
movies
P(A | B) =
P(B | A)P(A)
P(B)

58. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
≈
1.85
/
600
=
0.3%
P(Meryl
Streep
in
the
movie|oscar
nominaCon)
=
#
nominated
MS
movies
/
10
P(A | B) =
P(B | A)P(A)
P(B)

59. 19
nomina.ons
40
years
à
0.475
nomina.ons
per
year!

60. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
≈
1.85
/
600
=
0.3%
P(Meryl
Streep
in
the
movie|oscar
nominaCon)
=
#
nominated
MS
movies
/
10
P(A | B) =
P(B | A)P(A)
P(B)

61. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
≈
1.85
/
600
=
0.3%
P(Meryl
Streep
in
the
movie|oscar
nominaCon)
=
0.475
/
10
P(A | B) =
P(B | A)P(A)
P(B)

62. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
≈
1.85
/
600
=
0.3%
P(Meryl
Streep
in
the
movie|oscar
nominaCon)=4.8%
0.475
/
10
P(A | B) =
P(B | A)P(A)
P(B)

63. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
≈
1.85
/
600
=
0.3%
P(Meryl
Streep
in
the
movie|oscar
nominaCon)=4.8%
0.475
/
10
P(nom|MS)
=
P(MS|nom)P(nom)
/
P(MS)
P(A | B) =
P(B | A)P(A)
P(B)

64. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
≈
1.85
/
600
=
0.3%
P(Meryl
Streep
in
the
movie|oscar
nominaCon)=4.8%
0.475
/
10
P(nom|MS)
=
P(MS|nom)P(nom)
/
P(MS)
=
4.8%
*
1.67%
/
0.3%
P(A | B) =
P(B | A)P(A)
P(B)

65. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(
oscar
nominaCon
)
≈
10
/
600
=
1.67%
P(Meryl
Streep
in
the
movie)
≈
1.85
/
600
=
0.3%
P(Meryl
Streep
in
the
movie|oscar
nominaCon)=4.8%
0.475
/
10
P(nom|MS)
=
P(MS|nom)P(nom)
/
P(MS)
=
4.8%
*
1.67%
/
0.3%
=
25.7%
P(A | B) =
P(B | A)P(A)
P(B)

66. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
P(nom|MS)
=
P(MS|nom)P(nom)
/
P(MS)
=
4.8%
*
1.67%
/
0.3%
=
25.7%
P(A | B) =
P(B | A)P(A)
P(B)

67. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
=
#
nominated
MS
movies
/
#
MS
movies
P(nom|MS)
=
P(MS|nom)P(nom)
/
P(MS)
=
4.8%
*
1.67%
/
0.3%
=
25.7%
P(A | B) =
P(B | A)P(A)
P(B)

68. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
=
#
nominated
MS
movies
/
#
MS
movies
=
(
0.475
nom
per
year)
/
(1.85
MS
movies
per
year)
P(nom|MS)
=
P(MS|nom)P(nom)
/
P(MS)
=
4.8%
*
1.67%
/
0.3%
=
25.7%
P(A | B) =
P(B | A)P(A)
P(B)

69. P(
oscar
nominaCon
|
Meryl
Streep
in
the
movie)
=
?
=
#
nominated
MS
movies
/
#
MS
movies
=
(
0.475
nom
per
year)
/
(1.85
MS
movies
per
year)
=
0.475
/
1.85
=
25.7%
P(nom|MS)
=
P(MS|nom)P(nom)
/
P(MS)
=
4.8%
*
1.67%
/
0.3%
=
25.7%
P(A | B) =
P(B | A)P(A)
P(B)

70. Meryl
movies
that
are
not
nominated
Non-­‐Meryl
movies
that
don’t
get
nominated
Nominated
Meryl
movies
0.475
per
year
Nominated
movies
10
per
year
Meryl
movies
1.85
per
year
Nominated
non-­‐Meryl
movies

71. UpdaCng
the
state
of
knowledge
step
by
step
with
new
informaCon

72. UpdaCng
the
state
of
knowledge
step
by
step
with
new
informaCon
or
An Introduction to
Time Travel

75. Date Jan. 23, 2015.
Time Unknown.

76. Date Jan. 23, 2015.
Time Unknown.
I arrived from 2055,
in Union Square subway station,
naked.

77. Date Jan. 23, 2015.
Time Unknown.
I arrived from 2055,
in Union Square subway station,
naked.

78. Date Jan. 23, 2015.
Time Unknown.
I arrived from 2055,
in Union Square subway station,
naked.
Between 9 AM and 10 AM today, a scientist will
perform an experiment, starting a chain of
events resulting in a black hole in NYC.

80. Date Jan. 23, 2015.
Time Unknown.
I arrived from 2055,
in Union Square subway station,
naked.
Between 9 AM and 10 AM today, a scientist will
perform an experiment, starting a chain of
events resulting in a black hole in NYC.
I’ve been sent back to stop her.

82. What time is it? Is it 9 AM?

83. What time is it? Is it 9 AM?
Time travel technology is only precise enough to
target a 24 hour window.

84. What time is it? Is it 9 AM?
Time travel technology is only precise enough to
target a 24 hour window.
What is my current certainty that it is 9 AM?

85. What time is it? Is it 9 AM?
Time travel technology is only precise enough to
target a 24 hour window.
What is my current certainty that it is 9 AM?
Any hour in that 24-hour window is equally likely.

86. What time is it? Is it 9 AM?
Time travel technology is only precise enough to
target a 24 hour window.
What is my current certainty that it is 9 AM?
Any hour in that 24-hour window is equally likely.
P(9AM) = 1/24 = 0.0416

87. What time is it? Is it 9 AM?
Time travel technology is only precise enough to
target a 24 hour window.
What is my current certainty that it is 9 AM?
Any hour in that 24-hour window is equally likely.
P(9AM) = 1/24 = 0.0416
4.16%
Right
Time

88. What time is it? Is it 9 AM?
Time travel technology is only precise enough to
target a 24 hour window.
What is my current certainty that it is 9 AM?
Any hour in that 24-hour window is equally likely.
P(9AM) = 1/24 = 0.0416
I need to gather more information.
4.16%
Right
Time

89. 4.16%
Right
Time

90. What time is it? Is it 9 AM?
Time travel technology is only precise enough to
target a 24 hour window.
What is my current certainty that it is 9 AM?
Any hour in that 24-hour window is equally likely.
P(9AM) = 1/24 = 0.0416
I need to gather more information.
5-Train is approaching. I can see that it is full.
Checking my databanks…..
4.16%
Right
Time

91. Checking my databanks…..Records found:
4.16%
Right
Time

92. Checking my databanks…..Records found:
Between 9 and 10 AM, 5-train is full 70% of the time
4.16%
Right
Time

93. Checking my databanks…..Records found:
Between 9 and 10 AM, 5-train is full 70% of the time
Other times, it is full 20% of the time on average
4.16%
Right
Time

94. Checking my databanks…..Records found:
Between 9 and 10 AM, 5-train is full 70% of the time
Other times, it is full 20% of the time on average
P(5-train full|9AM) = 0.7
4.16%
Right
Time

95. Checking my databanks…..Records found:
Between 9 and 10 AM, 5-train is full 70% of the time
Other times, it is full 20% of the time on average
P(5-train full|9AM) = 0.7
My current certainty, prior to the new information, is
P(9AM) = 0.0416
4.16%
Right
Time

96. Checking my databanks…..Records found:
Between 9 and 10 AM, 5-train is full 70% of the time
Other times, it is full 20% of the time on average
P(5-train full|9AM) = 0.7
My current certainty, prior to the new information, is
P(9AM) = 0.0416
P(5-train full) = P(5-train full|9AM) P(9AM)
+ P(5-train full|not 9AM) P(not 9AM)
4.16%
Right
Time

97. Checking my databanks…..Records found:
Between 9 and 10 AM, 5-train is full 70% of the time
Other times, it is full 20% of the time on average
P(5-train full|9AM) = 0.7
My current certainty, prior to the new information, is
P(9AM) = 0.0416
P(5-train full) = P(5-train full|9AM) P(9AM)
+ P(5-train full|not 9AM) P(not 9AM)
= 0.7 * 0.0416 + 0.2 * (1 – 0.0416)
= 0.2208
4.16%
Right
Time

98. P(5-train full|9AM) = 0.7
P(5-train full) = 0.2208
P(9AM) = 0.0416
4.16%
Right
Time

99. P(5-train full|9AM) = 0.7
P(5-train full) = 0.2208
P(9AM) = 0.0416
P(5-train full|9AM) P(9AM)
P(9AM|5-train full)=
P(5-train full)
4.16%
Right
Time

100. P(5-train full|9AM) = 0.7
P(5-train full) = 0.2208
P(9AM) = 0.0416
current
updated P(5-train full|9AM) P(9AM)
P(9AM|5-train full)=
P(5-train full)
4.16%
Right
Time

101. P(5-train full|9AM) = 0.7
P(5-train full) = 0.2208
P(9AM) = 0.0416
0.7 * 0.0416
P(9AM|5-train full)=
0.2208
4.16%
Right
Time

102. P(5-train full|9AM) = 0.7
P(5-train full) = 0.2208
P(9AM) = 0.0416
P(9AM|5-train full)= 0.1319
13.19%
Right
Time

103. My current certainty is 13.19%.
I need to gather more information.
13.19%
Right
Time

104. 13.19%
Right
Time

105. My current certainty is 13.19%.
I need to gather more information.
Train doors open. I can hear someone’s radio.
It’s NPR News, live.
13.19%
Right
Time

106. My current certainty is 13.19%.
I need to gather more information.
Train doors open. I can hear someone’s radio.
It’s NPR News, live.
Checking database……
13.19%
Right
Time

107. My current certainty is 13.19%.
I need to gather more information.
Train doors open. I can hear someone’s radio.
It’s NPR News.
Checking database……Records found:
NPR News is broadcast 3 times during the day:
9AM-10AM, 1PM-2PM, 7PM-8PM.
13.19%
Right
Time

108. My current certainty is 13.19%.
I need to gather more information.
Train doors open. I can hear someone’s radio.
It’s NPR News.
Checking database……Records found:
NPR News is broadcast 3 times during the day:
9AM-10AM, 1PM-2PM, 7PM-8PM.
P(News |9AM) = 1.0
P(News|not 9AM) = 2 / 23 = 0.087
13.19%
Right
Time

109. My current certainty is 13.19%.
I need to gather more information.
Train doors open. I can hear someone’s radio.
It’s NPR News.
Checking database……Records found:
NPR News is broadcast 3 times during the day:
9AM-10AM, 1PM-2PM, 7PM-8PM.
P(News |9AM) = 1.0
P(News|not 9AM) = 2 / 23 = 0.087
P(9AM|5tf)= 0.1319
13.19%
Right
Time

110. My current certainty is 13.19%.
I need to gather more information.
Train doors open. I can hear someone’s radio.
It’s NPR News.
Checking database……Records found:
NPR News is broadcast 3 times during the day:
9AM-10AM, 1PM-2PM, 7PM-8PM.
P(News |9AM) = 1.0
P(News|not 9AM) = 2 / 23 = 0.087
P(9AM|5tf)= 0.1319
P(News|5tf) = P(News|9AM)P(9AM|5tf)
+ P(News|not 9AM)P(not 9AM|5tf)
13.19%
Right
Time

111. My current certainty is 13.19%.
I need to gather more information.
Train doors open. I can hear someone’s radio.
It’s NPR News.
Checking database……Records found:
NPR News is broadcast 3 times during the day:
9AM-10AM, 1PM-2PM, 7PM-8PM.
P(News |9AM) = 1.0
P(News|not 9AM) = 2 / 23 = 0.087
P(9AM|5tf)= 0.1319
P(News|5tf) = P(News|9AM)P(9AM|5tf)
+ P(News|not 9AM)P(not 9AM|5tf)
= 1.0* 0.1319 + 0.087 * (1 - 0.1319)
= 0.2074
13.19%
Right
Time

112. P(News|5tf) = 0.2074
P(News |9AM,5tf) = P(News |9AM) = 1.0
P(9AM|5tf)= 0.1319
13.19%
Right
Time

113. P(News|5tf) = 0.2074
P(News |9AM,5tf) = P(News |9AM) = 1.0
P(9AM|5tf)= 0.1319
P(News|9AM,5tf) P(9AM|5tf)
P(9AM|News,5tf) =
P(News|5tf)
13.19%
Right
Time

114. P(News|5tf) = 0.2074
P(News |9AM,5tf) = P(News |9AM) = 1.0
P(9AM|5tf)= 0.1319
1.0 * 0.1319
P(9AM|News,5tf) =
0.2074
13.19%
Right
Time

115. P(News|5tf) = 0.2074
P(News |9AM,5tf) = P(News |9AM) = 1.0
P(9AM|5tf)= 0.1319
P(9AM|News,5tf) = 0.6359
63.59%
Right
Time

116. My current certainty is 63.59%.
I need to gather more information.
63.59%
Right
Time

117. My current certainty is 63.59%.
I need to gather more information.
I’ll ask this person. “What time is it?”
63.59%
Right
Time

118. 63.59%
Right
Time

119. My current certainty is 63.59%.
I need to gather more information.
I’ll ask this person. “What time is it?”
He says “It’s nine.”
63.59%
Right
Time

120. My current certainty is 63.59%.
I need to gather more information.
I’ll ask this person. “What time is it?”
He says “It’s nine.”
Loading human interactions model……..
63.59%
Right
Time

121. My current certainty is 63.59%.
I need to gather more information.
I’ll ask this person. “What time is it?”
He says “It’s nine.”
Loading human interactions model……..
P(“Nine” | 9AM) = 0.96 (watch may be broken, etc.)
63.59%
Right
Time

122. My current certainty is 63.59%.
I need to gather more information.
I’ll ask this person. “What time is it?”
He says “It’s nine.”
Loading human interactions model……..
P(“Nine” | 9AM) = 0.96 (watch may be broken, etc.)
P(“Nine” |not 9AM) = 1/24 = 0.042 (9 PM -> “Nine”)
63.59%
Right
Time

123. My current certainty is 63.59%.
I need to gather more information.
I’ll ask this person. “What time is it?”
He says “It’s nine.”
Loading human interactions model……..
P(“Nine” | 9AM) = 0.96 (watch may be broken, etc.)
P(“Nine” |not 9AM) = 1/24 = 0.042 (9 PM -> “Nine”)
P(9AM|N,5tf)= 0.6359
63.59%
Right
Time

124. My current certainty is 63.59%.
I need to gather more information.
I’ll ask this person. “What time is it?”
He says “It’s nine.”
Loading human interactions model……..
P(“Nine” | 9AM) = 0.96 (watch may be broken, etc.)
P(“Nine” |not 9AM) = 1/24 = 0.042 (9 PM -> “Nine”)
P(9AM|N,5tf)= 0.6359
P(“Nine”|N,5tf) = P(“Nine”|9AM)P(9AM|N,5tf)
+ P(“Nine”|not 9AM)P(not 9AM|N,5tf)
63.59%
Right
Time

125. My current certainty is 63.59%.
I need to gather more information.
I’ll ask this person. “What time is it?”
He says “It’s nine.”
Loading human interactions model……..
P(“Nine” | 9AM) = 0.96 (watch may be broken, etc.)
P(“Nine” |not 9AM) = 1/24 = 0.042 (9 PM -> “Nine”)
P(9AM|N,5tf)= 0.6359
P(“Nine”|N,5tf) = P(“Nine”|9AM)P(9AM|N,5tf)
+ P(“Nine”|not 9AM)P(not 9AM|N,5tf)
= 0.96 * 0.6359 + 0.042 * (1 – 0.6359)
= 0.6258
63.59%
Right
Time

126. P(“Nine”|N,5tf) = 0.6258
P(“Nine” |9AM,N,5tf) = P(“Nine” | 9AM) = 0.96
P(9AM|N,5tf)= 0.6359
63.59%
Right
Time

127. P(“Nine”|N,5tf) = 0.6258
P(“Nine” |9AM,N,5tf) = P(“Nine” | 9AM) = 0.96
P(9AM|N,5tf)= 0.6359
P(“Nine”|9AM,N,5tf) P(9AM|N,5tf)
P(9AM|“Nine” ,N,5tf) =
P(“Nine”|N,5tf)
63.59%
Right
Time

128. P(“Nine”|N,5tf) = 0.6258
P(“Nine” |9AM,N,5tf) = P(“Nine” | 9AM) = 0.96
P(9AM|N,5tf)= 0.6359
0.96 * 0.6359
P(9AM|“Nine” ,N,5tf) =
0.6258
63.59%
Right
Time

129. P(“Nine”|N,5tf) = 0.6258
P(“Nine” |9AM,N,5tf) = P(“Nine” | 9AM) = 0.96
P(9AM|N,5tf)= 0.6359
P(9AM|“Nine” ,N,5tf) = 0.9756
97.56%
Right
Time

130. 97.56%
Right
TimeMy latest posterior is 97.56%.

131. My latest posterior is 97.56%.
I am now sufficiently certain that I arrived at the right
time.
97.56%
Right
Time

132. 97.56%
Right
Time

133. My latest posterior is 97.56%.
I am now sufficiently certain that I arrived at the right
time.
Time to kill a scientist.
97.56%
Right
Time

134. The
Cookie
Problem
Dude,
you
have
a
problem

135. Bowl
1
Bowl
2

138. Which
Bowl?

139. Our
Cme
traveler
had
two
hypotheses
H1
:
It
is
9
AM
(9
to
10)
H2
:
It
is
not
9
AM

140. Our
Cme
traveler
had
two
hypotheses
H1
:
It
is
9
AM
(9
to
10)
H2
:
It
is
not
9
AM
At
first,
P(H1)
=
P(9AM)
=
4.16%
P(H2)
=
P(not
9AM)
=
95.84%

141. Our
Cme
traveler
had
two
hypotheses
H1
:
It
is
9
AM
(9
to
10)
H2
:
It
is
not
9
AM
At
first,
P(H1)
=
P(9AM)
=
4.16%
P(H2)
=
P(not
9AM)
=
95.84%
Updates
with
new
informaCon

142. Our
Cme
traveler
had
two
hypotheses
H1
:
It
is
9
AM
(9
to
10)
H2
:
It
is
not
9
AM
At
first,
P(H1)
=
P(9AM)
=
4.16%
P(H2)
=
P(not
9AM)
=
95.84%
Updates
with
new
informaCon
In
the
end,
P(H1)
=
P(
9AM
|“Nine”,N,5tf)
=
97.56%
P(H2)
=
P(not
9AM|“Nine”,N,5tf)
=
2.44%

143. Cookie
problem
also
has
two
hypotheses
H1
:
It
is
Bowl
1
H2
:
It
is
Bowl
2

144. Cookie
problem
also
has
two
hypotheses
H1
:
It
is
Bowl
1
H2
:
It
is
Bowl
2
At
first,
P(H1)
=
P(Bowl
1)
=
50%
P(H2)
=
P(Bowl
2)
=
50%

145. Cookie
problem
also
has
two
hypotheses
H1
:
It
is
Bowl
1
H2
:
It
is
Bowl
2
At
first,
P(H1)
=
P(Bowl
1)
=
50%
P(H2)
=
P(Bowl
2)
=
50%
Update
with
one
cookie
informaCon:
Vanilla
P(Bowl
1
|
Vanilla)
=
?

146. Cookie
problem
also
has
two
hypotheses
H1
:
It
is
Bowl
1
H2
:
It
is
Bowl
2
At
first,
P(H1)
=
P(Bowl
1)
=
50%
P(H2)
=
P(Bowl
2)
=
50%
Update
with
one
cookie
informaCon:
Vanilla
P(Bowl
1
|
Vanilla)
=
?
P(Bowl
2
|
Vanilla)
=
1
-­‐
P(Bowl
1
|
Vanilla)
=
?

147. Prior:
P(Bowl
1)
=
50%
50.00%
Bowl 1

148. Prior:
P(Bowl
1)
=
50%
Likelihood:
P(Vanilla
|
Bowl
1)
=
#
vanilla
/
#
all
50.00%
Bowl 1

149. Prior:
P(Bowl
1)
=
50%
Likelihood:
P(Vanilla
|
Bowl
1)
=
30
/
40
=
0.75
50.00%
Bowl 1

150. Prior:
P(Bowl
1)
=
50%
P(Bowl
2)
=
50%
Likelihood:
P(Vanilla
|
Bowl
1)
=
30
/
40
=
0.75
P(Vanilla
|
Bowl
2)
=
20
/
40
=
0.50
50.00%
Bowl 1

151. Prior:
P(Bowl
1)
=
50%
P(Bowl
2)
=
50%
Likelihood:
P(Vanilla
|
Bowl
1)
=
30
/
40
=
0.75
P(Vanilla
|
Bowl
2)
=
20
/
40
=
0.50
Evidence:
P(Vanilla)
=
P(Vanilla|Bowl
1)P(Bowl
1)
+
P(Vanilla|Bowl
2)P(Bowl
2)
50.00%
Bowl 1

152. Prior:
P(Bowl
1)
=
50%
P(Bowl
2)
=
50%
Likelihood:
P(Vanilla
|
Bowl
1)
=
30
/
40
=
0.75
P(Vanilla
|
Bowl
2)
=
20
/
40
=
0.50
Evidence:
P(Vanilla)
=
P(Vanilla|Bowl
1)P(Bowl
1)
+
P(Vanilla|Bowl
2)P(Bowl
2)
=
0.75
*
0.50
+
0.50
*
0.50
=
0.625
50.00%
Bowl 1

153. P(Bowl
1)
=
0.50
P(Vanilla
|
Bowl
1)
=
0.75
P(Vanilla)
=
0.625
50.00%
Bowl 1

154. P(Bowl
1)
=
0.50
P(Vanilla
|
Bowl
1)
=
0.75
P(Vanilla)
=
0.625
P(Vanilla|Bowl 1) P(Bowl 1)
P(Bowl 1|Vanilla) =
P(Vanilla)
50.00%
Bowl 1

155. P(Bowl
1)
=
0.50
P(Vanilla
|
Bowl
1)
=
0.75
P(Vanilla)
=
0.625
0.75 * 0.50
P(Bowl 1|Vanilla) =
0.625
50.00%
Bowl 1

156. P(Bowl
1)
=
0.50
P(Vanilla
|
Bowl
1)
=
0.75
P(Vanilla)
=
0.625
P(Bowl 1|Vanilla) = 0.6
60.00%
Bowl 1

157. P(Bowl
1)
=
0.50
P(Vanilla
|
Bowl
1)
=
0.75
P(Vanilla)
=
0.625
P(Bowl 1|Vanilla) = 0.6
P(Bowl 2|Vanilla) = 1 – 0.6 = 0.4
60.00%
Bowl 1

158. P(Bowl
1)
=
0.50
P(Vanilla
|
Bowl
1)
=
0.75
P(Vanilla)
=
0.625
P(Bowl 1|Vanilla) = 0.6
P(Bowl 2|Vanilla) = 0.4
I
can
get
more
informaCon
by
eaCng
another
cookie!
60.00%
Bowl 1

159.
Update
2
(An
excuse
to
eat
more
cookies)
60.00%
Bowl 1
Seriously,
seek
help

160. Prior:
P(Bowl
1)
=
0.6
60.00%
Bowl 1

161. Prior:
P(Bowl
1)
=
0.6
Likelihood:
P(Vanilla
|
Bowl
1)
=
29
/
39
=
0.7436
60.00%
Bowl 1

162. Prior:
P(Bowl
1)
=
0.6
P(Bowl
2)
=
0.4
Likelihood:
P(Vanilla
|
Bowl
1)
=
29
/
39
=
0.7436
P(Vanilla
|
Bowl
2)
=
19
/
39
=
0.4872
60.00%
Bowl 1

163. Prior:
P(Bowl
1)
=
0.6
P(Bowl
2)
=
0.4
Likelihood:
P(Vanilla
|
Bowl
1)
=
29
/
39
=
0.7436
P(Vanilla
|
Bowl
2)
=
19
/
39
=
0.4872
Evidence:
P(Vanilla)
=
P(Vanilla|Bowl
1)P(Bowl
1)
+
P(Vanilla|Bowl
2)P(Bowl
2)
=
0.7436
*
0.6
+
0.4872
*
0.4
=
0.641
60.00%
Bowl 1

164. P(Bowl
1)
=
0.6
P(Vanilla
|
Bowl
1)
=
0.7436
P(Vanilla)
=
0.641
60.00%
Bowl 1

165. P(Bowl
1)
=
0.6
P(Vanilla
|
Bowl
1)
=
0.7436
P(Vanilla)
=
0.641
P(Vanilla|Bowl 1) P(Bowl 1)
P(Bowl 1|Vanilla) =
P(Vanilla)
60.00%
Bowl 1

166. P(Bowl
1)
=
0.6
P(Vanilla
|
Bowl
1)
=
0.7436
P(Vanilla)
=
0.641
0.7436 * 0.6
P(Bowl 1|Vanilla) =
0.641
60.00%
Bowl 1

167. P(Bowl
1)
=
0.6
P(Vanilla
|
Bowl
1)
=
0.7436
P(Vanilla)
=
0.641
P(Bowl 1|Vanilla) = 0.696
69.60%
Bowl 1

168. P(Bowl
1)
=
0.6
P(Vanilla
|
Bowl
1)
=
0.7436
P(Vanilla)
=
0.641
P(Bowl 1|Vanilla) = 0.696
P(Bowl 2|Vanilla) = 0.304
69.60%
Bowl 1

169.
Update
3
(MOAR
COOKIEEEEES)
69.60%
Bowl 1
Ummm….

170. Prior:
P(Bowl
1)
=
0.696
P(Bowl
2)
=
0.304
Likelihood:
P(Chocolate
|
Bowl
1)
=
10
/
38
=
0.2632
P(Chocolate
|
Bowl
2)
=
20
/
38
=
0.5263
Evidence:
P(Chocolate)
=
P(Chocolate|Bowl
1)P(Bowl
1)
+
P(Chocolate|Bowl
2)P(Bowl
2)
=
0.2632
*
0.696
+
0.5263
*
0.304
=
0.343
69.60%
Bowl 1

171. P(Bowl
1)
=
0.696
P(Chocolate
|
Bowl
1)
=
0.2632
P(Chocolate
)
=
0.343
P(Chocolate
|Bowl 1) P(Bowl 1)
P(Bowl 1|Chocolate
) =
P(Chocolate
)
69.60%
Bowl 1

172. P(Bowl
1)
=
0.696
P(Chocolate
|
Bowl
1)
=
0.2632
P(Chocolate
)
=
0.343
0.2632 * 0.696
P(Bowl 1|Chocolate
) =
0.343
69.60%
Bowl 1

173. P(Bowl
1)
=
0.696
P(Chocolate
|
Bowl
1)
=
0.2632
P(Chocolate
)
=
0.343
P(Bowl 1|Chocolate
) = 0.5338
53.38%
Bowl 1

174. P(Bowl
1)
=
0.696
P(Chocolate
|
Bowl
1)
=
0.2632
P(Chocolate
)
=
0.343
P(Bowl 1|Chocolate
) = 0.5338
P(Bowl 2|Chocolate
) = 0.4662
53.38%
Bowl 1

175. P(Bowl
1)
=
0.696
P(Chocolate
|
Bowl
1)
=
0.2632
P(Chocolate
)
=
0.343
P(Bowl 1|Chocolate
) = 0.5338
P(Bowl 2|Chocolate
) = 0.4662
After eating Vanilla, Vanilla, Chocolate,
I am 53.38% certain that this is Bowl 1.
(So basically I have no idea, I’ll have to eat all cookies)
53.38%
Bowl 1

176. The
LocomoCve
Problem
Choo
choo,
baby

177. A
railroad
numbers
its
locomoCves
in
order
1…...N.
One
day
you
see
a
locomoCve
with
the
number
60.
EsCmate
how
many
locomoCves
the
railroad
has.
60

178. H1
:
There
is
1
train
H2
:
There
are
2
trains
H3
:
There
are
3
trains
H4
:
There
are
4
trains
.
.
.
H1000
:
There
are
1000
trains
60

179. H1
:
There
is
1
train
0.1%
H2
:
There
are
2
trains
0.1%
H3
:
There
are
3
trains
0.1%
…
H200
:
There
are
200
trains
0.1%
…
H1000
:
There
are
1000
trains
0.1%
60

180. H1
:
There
is
1
train
0.1%
H2
:
There
are
2
trains
0.1%
H3
:
There
are
3
trains
0.1%
…
H200
:
There
are
200
trains
0.1%
…
H1000
:
There
are
1000
trains
0.1%
60
Update

181. H1
:
There
is
1
train
0.1%
0%
H2
:
There
are
2
trains
0.1%
0%
H3
:
There
are
3
trains
0.1%
0%
…
H200
:
There
are
200
trains
0.1%
0.4%
…
H1000
:
There
are
1000
trains
0.1%
0.3%
60
Update

182. Our
Cme
traveler
had
two
hypotheses
H1
:
It
is
9
AM
(9
to
10)
H2
:
It
is
not
9
AM
H1
H2
At
first,
prior
dist.
P(H1)
=
P(9AM)
=
4.16%
P(H2)
=
P(not
9AM)
=
95.84%

183. Our
Cme
traveler
had
two
hypotheses
H1
:
It
is
9
AM
(9
to
10)
H2
:
It
is
not
9
AM
H1
H2
posterior
dist.

184. Our
Cme
traveler
had
two
hypotheses
H1
:
It
is
9
AM
(9
to
10)
H2
:
It
is
not
9
AM
H1
H2
posterior
dist.

185. Our
Cme
traveler
had
two
hypotheses
H1
:
It
is
9
AM
(9
to
10)
H2
:
It
is
not
9
AM
H1
H2
posterior
dist.
Our
Cme
traveler
had
two
hypotheses
H1
:
It
is
9
AM
(9
to
10)
H2
:
It
is
not
9
AM
H1
H2
In
the
end,
P(H1)
=
P(
9AM
|“Nine”,N,5tf)
=
97.56%
P(H2)
=
P(not
9AM|“Nine”,N,5tf)
=
2.44%

186. H1
:
There
is
1
train
0.1%
0%
H2
:
There
are
2
trains
0.1%
0%
H3
:
There
are
3
trains
0.1%
0%
…
H200
:
There
are
200
trains
0.1%
0.4%
…
H1000
:
There
are
1000
trains
0.1%
0.3%
60
Update

187. H1
:
There
is
1
train
0.1%
H2
:
There
are
2
trains
0.1%
H3
:
There
are
3
trains
0.1%
…
H200
:
There
are
200
trains
0.1%
…
H1000
:
There
are
1000
trains
0.1%
… … … …
H1
H2
H3
H4
H5
…
H60
…
H200
...
H1000
…

188. H1
:
There
is
1
train
0.1%
0%
H2
:
There
are
2
trains
0.1%
0%
H3
:
There
are
3
trains
0.1%
0%
…
H200
:
There
are
200
trains
0.1%
0.4%
…
H1000
:
There
are
1000
trains
0.1%
0.3%
… … … …
H1
H2
H3
H4
H5
…
H60
…
H200
...
H1000
…
Update