A game show with goats, a girl named Florida, and
other curious questions in conditional probability
John Pike
Math 1600
Cornell University
Spring 2016
Monty Hall Problem
You are a contestant on a game show and are given the choice of
three doors: A car is behind one of the doors, and goats are behind
the other two.
You pick a door, and the host opens another door to reveal a goat.
(The host knows the location of the car. If you picked a goat door,
the host reveals the other goat door; if you picked the car door, the
host ips a coin to decide which of the others to open.)
The host then asks if you want to switch to the door he did not
open. Should you switch?
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3 / 18
If you don't switch doors, you win if and only if you initially picked
the door with the car. The chance that this happens is 1/3.
If you do switch, you win if and only if you initially picked a door
without the car, which happens with probability 2/3.
3 / 18
Suppose instead that after you chose a door, an unwitting audience
member is called up to open one of the other two doors.
If they open the door containing the car, clearly you should choose
that door. What if they open a door not containing the car?
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Suppose instead that after you chose a door, an unwitting audience
member is called up to open one of the other two doors.
If they open the door containing the car, clearly you should choose
that door. What if they open a door not containing the car?
Door
Revealed
r
Ca
Initial
Choice
Go
at
1
·0=0
3
II: P (pick car, goat revealed) =
C ar
Door
Revealed
I: P (pick car, car revealed) =
Goat
1
1
3
2
3
C ar
0
III: P (pick goat, car revealed) =
1
1
·1=
3
3
2 1
1
· =
3 2
3
1
2
Goat
1
2
IV: P (pick goat, goat revealed) =
2 1
1
· =
3 2
3
4 / 18
Suppose instead that after you chose a door, an unwitting audience
member is called up to open one of the other two doors.
If they open the door containing the car, clearly you should choose
that door. What if they open a door not containing the car?
Door
Revealed
r
Ca
Initial
Choice
Go
at
1
·0=0
3
II: P (pick car, goat revealed) =
C ar
Door
Revealed
I: P (pick car, car revealed) =
Goat
1
1
3
2
3
C ar
0
III: P (pick goat, car revealed) =
1
1
·1=
3
3
2 1
1
· =
3 2
3
1
2
Goat
1
2
IV: P (pick goat, goat revealed) =
With the switching strategy, you win with
With the staying strategy, you win with
2 1
1
· =
3 2
3
P(III) + P(IV) = 2/3.
P(III) + P(II) = 2/3.
4 / 18
Sibling Paradoxes
You visit a town in which every household has exactly two children,
each of which is equally likely to be a boy or a girl independently of
the other. You meet a parent at random.
What is the probability that they have two daughters?
5 / 18
Sibling Paradoxes
You visit a town in which every household has exactly two children,
each of which is equally likely to be a boy or a girl independently of
the other. You meet a parent at random.
What is the probability that they have two daughters?
(oldest sex, youngest sex), the
(G , G ), (G , B), (B, G ), (B, B), and each is equally
probability of two daughters is 1/4.
Writing the sibling combinations as
possibilities are
likely, so the
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What is the probability that they have two daughters if they answer
yes when asked if they have at least one daughter?
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What is the probability that they have two daughters if they answer
yes when asked if they have at least one daughter?
(G , G ), (G , B),
is 1/3.
In this case, the equiprobable possibilities are
(B, G ),
so the probability of two daughters
6 / 18
What is the probability that they have two daughters if they answer
yes when asked if they have at least one daughter?
(G , G ), (G , B),
is 1/3.
In this case, the equiprobable possibilities are
(B, G ),
so the probability of two daughters
What is the probability that they have two daughters if they answer
yes when asked if their eldest child is a girl?
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What is the probability that they have two daughters if they answer
yes when asked if they have at least one daughter?
(G , G ), (G , B),
is 1/3.
In this case, the equiprobable possibilities are
(B, G ),
so the probability of two daughters
What is the probability that they have two daughters if they answer
yes when asked if their eldest child is a girl?
(G , G ), (G , B),
daughters is 1/2.
Given that their eldest is a girl, the possibilities are
both equally likely, so the probability of two
6 / 18
What if they answer yes when asked if they have a daughter born
on Tuesday?
7 / 18
What if they answer yes when asked if they have a daughter born
on Tuesday?
Writing
(oldest
birth day,youngest birth day) and coloring green for
girl, blue for boy, the equally likely possibilities are
(Tue,Mon),(Tue,Tue),(Tue,Wed),(Tue,Thu),(Tue,Fri),(Tue,Sat),(Tue,Sun),
(Tue,Mon),(Tue,Tue),(Tue,Wed),(Tue,Thu),(Tue,Fri),(Tue,Sat),(Tue,Sun),
(Mon,Tue),(Tue,Tue),(Wed,Tue),(Thu,Tue),(Fri,Tue),(Sat,Tue),(Sun,Tue),
(Mon,Tue),(Wed,Tue),(Thu,Tue),(Fri,Tue),(Sat,Tue),(Sun,Tue)
7 / 18
What if they answer yes when asked if they have a daughter born
on Tuesday?
Writing
(oldest
birth day,youngest birth day) and coloring green for
girl, blue for boy, the equally likely possibilities are
(Tue,Mon),(Tue,Tue),(Tue,Wed),(Tue,Thu),(Tue,Fri),(Tue,Sat),(Tue,Sun),
(Tue,Mon),(Tue,Tue),(Tue,Wed),(Tue,Thu),(Tue,Fri),(Tue,Sat),(Tue,Sun),
(Mon,Tue),(Tue,Tue),(Wed,Tue),(Thu,Tue),(Fri,Tue),(Sat,Tue),(Sun,Tue),
(Mon,Tue),(Wed,Tue),(Thu,Tue),(Fri,Tue),(Sat,Tue),(Sun,Tue)
Of the 27 possible outcomes, the 13 in the second and fourth rows
correspond to two girls, so the probability in question is 13/27.
7 / 18
Recall that the probability of two girls when the parent answers yes
to the question Do you have at least one daughter? is 1/3.
Suppose instead that the way you learned that they have at least
one daughter is that a girl walked up to the parent you were talking
with and said Hi Dad! Does this change anything?
8 / 18
Recall that the probability of two girls when the parent answers yes
to the question Do you have at least one daughter? is 1/3.
Suppose instead that the way you learned that they have at least
one daughter is that a girl walked up to the parent you were talking
with and said Hi Dad! Does this change anything?
Presumably, it is twice as likely that you encounter a girl thusly in a
two girl family as in a one girl family.
The three possibilities are still
(G , G ), (G , B), (B, G ),
but now the
rst is twice as likely as the others.
Thus we now compute the probability of two girls as 1/2.
8 / 18
Probability
The standard way of treating such questions rigorously is to model
them in terms of a probability space
(Ω, F, P).
9 / 18
Probability
The standard way of treating such questions rigorously is to model
them in terms of a probability space
Ω
is the
sample space.
(Ω, F, P).
It consists of all possible outcomes of a
random experiment or all possible states of a random system.
9 / 18
Probability
The standard way of treating such questions rigorously is to model
them in terms of a probability space
Ω
is the
sample space.
(Ω, F, P).
It consists of all possible outcomes of a
random experiment or all possible states of a random system.
F
is the
σ -eld
of events. It consists of all subsets of outcomes
about which we can make meaningful statements:
For each
E ∈ F,
we can ask whether an outcome in
E
occurred.
9 / 18
Probability
The standard way of treating such questions rigorously is to model
them in terms of a probability space
Ω
is the
sample space.
(Ω, F, P).
It consists of all possible outcomes of a
random experiment or all possible states of a random system.
F
is the
σ -eld
of events. It consists of all subsets of outcomes
about which we can make meaningful statements:
For each
E ∈ F,
we can ask whether an outcome in
E
occurred.
P : F → [0, 1] is the probability measure. It assigns probabilities
to events in F in a consistent manner. For example, we have
P(Ω) = 1 and E ∩ F = ∅ implies P(E ∪ F ) = P(E ) + P(F ).
9 / 18
Probability
The standard way of treating such questions rigorously is to model
them in terms of a probability space
Ω
is the
sample space.
(Ω, F, P).
It consists of all possible outcomes of a
random experiment or all possible states of a random system.
F
is the
σ -eld
of events. It consists of all subsets of outcomes
about which we can make meaningful statements:
For each
E ∈ F,
we can ask whether an outcome in
E
occurred.
P : F → [0, 1] is the probability measure. It assigns probabilities
to events in F in a consistent manner. For example, we have
P(Ω) = 1 and E ∩ F = ∅ implies P(E ∪ F ) = P(E ) + P(F ).
If the experiment is rolling a fair six-sided die, then
Ω = {1, 2, 3, 4, 5, 6}, F = 2Ω ,
and
P(E ) = |E | /6.
9 / 18
Conditional probability addresses the problem of updating one's
assessment of probabilities when presented with additional
information.
10 / 18
Conditional probability addresses the problem of updating one's
assessment of probabilities when presented with additional
information.
For example, if I roll a die in another room, then you would evaluate
the odds of me rolling an even number as
P ({2, 4, 6}) =
1
2.
But if I told you that I rolled a number greater than three, you
would know that the possible values are 4, 5, 6.
Since these outcomes were originally equally likely, they should
remain so after you learned that the value exceeded 3.
10 / 18
Conditional probability addresses the problem of updating one's
assessment of probabilities when presented with additional
information.
For example, if I roll a die in another room, then you would evaluate
the odds of me rolling an even number as
P ({2, 4, 6}) =
1
2.
But if I told you that I rolled a number greater than three, you
would know that the possible values are 4, 5, 6.
Since these outcomes were originally equally likely, they should
remain so after you learned that the value exceeded 3.
Thus, when given this additional information, your evaluation of the
probability changes from
2
1
2 to 3 .
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P(E |F ), the probability
information that F has occurred.
In general, we are interested in computing
that the event
E
occurs given the
11 / 18
P(E |F ), the probability
information that F has occurred.
In general, we are interested in computing
that the event
Knowing that
E
F
occurs given the
has occurred means that we need to eectively
shrink our original sample space
Ω
to only include outcomes in
The possible events are of the form
A ∩ F:
For
A
to have occurred,
the experiment must have resulted in an outcome in
Since we must now assign
F
F.
A ∩ F.
probability 1 and our information
A, B ⊂ F has
probability by P(F ).
about the relative likelihoods of events
we need to divide our original
not changed,
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P(E |F ), the probability
information that F has occurred.
In general, we are interested in computing
that the event
Knowing that
E
F
occurs given the
has occurred means that we need to eectively
shrink our original sample space
Ω
to only include outcomes in
The possible events are of the form
A ∩ F:
For
A
to have occurred,
the experiment must have resulted in an outcome in
Since we must now assign
F
A ∩ F.
probability 1 and our information
A, B ⊂ F has
probability by P(F ).
about the relative likelihoods of events
we need to divide our original
Our updated probability is thus
the
F.
P (E |F ) =
conditional probability of E given F .
not changed,
P(E ∩ F )
,
P(F )
11 / 18
E = {(G , G )},
F = {(G , G ), (G , B), (B, G )}, H = {(G .G ), (G , B)}.
In the sibling paradox problem, consider
12 / 18
E = {(G , G )},
F = {(G , G ), (G , B), (B, G )}, H = {(G .G ), (G , B)}.
In the sibling paradox problem, consider
We compute
P (E |F ) =
P(E ∩ F )
P ({(G , G )})
=
=
P(F )
P ({(G , G ), (G , B), (B, G )})
1
4
3
4
=
1
3
and
P (E |H ) =
P(E ∩ H)
P ({(G , G )})
=
=
P(H)
P ({(G , G ), (G , B)})
1
4
1
2
1
= .
2
12 / 18
Bayes' Rule
If
E, F ∈ F
have
P(E ), P(F ) > 0,
then rearranging the conditional
probability formula gives
P(E |F )P(F ) = P(E ∩ F ) = P(F |E )P(E ).
In particular, we have
Bayes' rule
P (E |F ) =
P (E ∩ F )
P(F |E )P(E )
=
.
P(F )
P(F )
13 / 18
Bayes' Rule
If
E, F ∈ F
have
P(E ), P(F ) > 0,
then rearranging the conditional
probability formula gives
P(E |F )P(F ) = P(E ∩ F ) = P(F |E )P(E ).
In particular, we have
Bayes' rule
P (E |F ) =
The interpretation is that
P (E ∩ F )
P(F |E )P(E )
=
.
P(F )
P(F )
P(F |E )/P(F ) is
for or against E .
a measure of the
evidence that
F
To obtain the
posterior probability, P(E |F ),
provides
probability, P(E ),
we multiply the
prior
by this factor.
13 / 18
Often it is convenient to write
P (F ) = P (F ∩ E )+P F ∩ E
C
= P(F |E )P(E )+P(F E C )P(E C )
so that Bayes' rule reads
P (E |F ) =
P(F |E )P(E )
.
P(F |E )P(E ) + P(F |E C )P(E C )
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Often it is convenient to write
P (F ) = P (F ∩ E )+P F ∩ E
C
= P(F |E )P(E )+P(F E C )P(E C )
so that Bayes' rule reads
P (E |F ) =
P(F |E )P(E )
.
P(F |E )P(E ) + P(F |E C )P(E C )
More generally, we have
Theorem
Suppose that E1 , E2 , ..., En ∈ F are disjoint with P(Ej ) > 0 for
S
j = 1, . . . , n and nk=1 Ek = Ω. Then for any F ∈ F with
P(F ) > 0, and any j = 1, . . . , n we have
P(F |Ej )P(Ej )
.
k=1 P(F |Ek )P(Ek )
P (Ej |F ) = Pn
14 / 18
Bayesian Monty Hall
Without loss of generality, call the door you originally picked
the one revealed to be empty
A, B, C
let B∅ be
a,
and
b.
a, b, c ,
empty is b .
Write
for the events that the car is behind doors
and
the event that the door shown to be
15 / 18
Bayesian Monty Hall
Without loss of generality, call the door you originally picked
A, B, C
let B∅ be
a,
and
b.
the one revealed to be empty
a, b, c ,
empty is b .
Write
for the events that the car is behind doors
and
the event that the door shown to be
When Monty revealed, we get
P (C |B∅ ) =
=
P (B∅ |C ) P (C )
P (B∅ |C ) P (C ) + P (B∅ |B ) P (B) + P (B∅ |A ) P (A)
1·
1
3
1
1· 3
1
+0· 3
+
1
2
·
1
3
=
2
3
.
15 / 18
Bayesian Monty Hall
Without loss of generality, call the door you originally picked
A, B, C
let B∅ be
a,
and
b.
the one revealed to be empty
a, b, c ,
empty is b .
Write
for the events that the car is behind doors
and
the event that the door shown to be
When Monty revealed, we get
P (C |B∅ ) =
=
P (B∅ |C ) P (C )
P (B∅ |C ) P (C ) + P (B∅ |B ) P (B) + P (B∅ |A ) P (A)
1·
1
3
1
1· 3
1
+0· 3
+
1
2
·
1
3
=
2
3
.
When the audience member revealed, we get
P (C |B∅ ) =
=
P (B∅ |C ) P (C )
P (B∅ |C ) P (C ) + P (B∅ |B ) P (B) + P (B∅ |A ) P (A)
1
2
·
1
2 ·
1
3 +0·
1
3
1
3
+
1
2
·
1
3
=
1
2
.
15 / 18
Bayesian Siblings
Let
Y
be the event that parent answered yes when asked if they
had a daughter.
P ((G , G ) |Y ) =
P (Y |(G , G ) ) P ((G , G ))
P (Y |(G , G ) ) P ((G , G )) + P (Y |(G , B) ) P ((G , B)) + P (Y |(B, G ) ) P ((B, G )) + P (Y |(B, B
=
1 · 14
1
1
= 43 = .
3
1 · 14 + 1 · 14 + 1 · 41 + 0 · 41
4
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Bayesian Siblings
Let
Y
be the event that parent answered yes when asked if they
had a daughter.
P ((G , G ) |Y ) =
P (Y |(G , G ) ) P ((G , G ))
P (Y |(G , G ) ) P ((G , G )) + P (Y |(G , B) ) P ((G , B)) + P (Y |(B, G ) ) P ((B, G )) + P (Y |(B, B
=
Let
D
1 · 14
1
1
= 43 = .
3
1 · 14 + 1 · 14 + 1 · 41 + 0 · 41
4
be the event that the child who revealed themselves during
the conversation was a girl.
P ((G , G ) |D ) =
P (D |(G , G ) ) P ((G , G ))
P (D |(G , G ) ) P ((G , G )) + P (D |(G , B) ) P ((G , B)) + P (D |(B, G ) ) P ((B, G )) + P (D |(B, B
=
1
1 · 14
= 41 =
1
1
1
1
1
1
1· 4 + 2 · 4 + 2 · 4 +0· 4
2
1
.
2
16 / 18
A Girl Named Florida
Suppose we learn by means of straightforward questioning that a
parent of two children has a daughter named Florida.
We assume that every girl is named Florida with probability
p ∈ (0 , 1 )
regardless of whether the family already has a daughter
named Florida. Boys are never named Florida.
17 / 18
A Girl Named Florida
Suppose we learn by means of straightforward questioning that a
parent of two children has a daughter named Florida.
We assume that every girl is named Florida with probability
p ∈ (0 , 1 )
regardless of whether the family already has a daughter
named Florida. Boys are never named Florida.
If
F
is the event that the parent has a daughter named Florida,
P ((G , G ) |F ) =
P (F |(G , G ) ) P ((G , G ))
P (F |(G , G ) ) P ((G , G )) + P (F |(G , B) ) P ((G , B)) + P (F |(B, G ) ) P ((B, G )) + P (F |(B, B)
p(1 − p) + (1 − p)p + p 2 · 14
=
[p(1 − p) + (1 − p)p + p 2 ] · 14 + p · 14 + p · 14 + 0 · 41
1 2p − p 2 2−p
= 1 4
=
.
2) + 1 p
4−p
(
2
p
−
p
4
2
17 / 18
Note that
1
and
2 for
P ((G , G ) |F ) =
p = 0.
2−p
1
4−p takes values between 3 for
Thus the rarer the name (the smaller value of
p ),
p=1
the closer the
probability of the other child being a daughter is to
1
2.
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Note that
1
and
2 for
P ((G , G ) |F ) =
p = 0.
2−p
1
4−p takes values between 3 for
Thus the rarer the name (the smaller value of
p ),
p=1
the closer the
probability of the other child being a daughter is to
1
2.
This same analysis also solves the Thursday problem: Each girl is
born on Thursday independently with probability
P (2
girls |girl born on Thursday )
=
p=
−
4−
2
1
7
1
7
1
7 , so
=
13
27
.
18 / 18