Stay?
Switch?
Object: Maximize the probability
for getting the prize.
Object: Maximize the probability
for getting the prize.
With only two doors to choose
from, take the door with more
than 50% probability of
containing the prize.
Object: Maximize the probability
for getting the prize.
With only two doors to choose
from, take the door with more
than 50% probability of
containing the prize.
Problem: The probabilities are
not uniquely specified.
Want to handle this using Bayesian thinking. We
start with the situation before Monty Hall opened
one of the doors and want to find the probabilities
after he has opened them.
Nomenclature follows:
N
M
M1
M2
N
M
M1
M2
Want to handle this using Bayesian thinking. We
start with the situation before Monty Hall opened
one of the doors and want to find the probabilities
after he has opened them.
Nomenclature follows:
N: The right door has been chosen
M: The wrong door has been chosen. The prize is
behind door M1 or M2 (sub-models).
Data:
D1: Monty Hall opens a door
D2: The prize is not behind that
door
D3: The rightmost (or leftmost) of
the door you didn't chose was
opened.
Data:
D1: Monty Hall opens a door
D2: The prize is not behind that
door
D3: The rightmost (or leftmost) of
the door you didn't chose was
opened.
Data:
D1: Monty Hall opens a door
D2: The prize is not behind that
door
D2: The prize is not behind that
door
This has the form of the lesson in
clip 5. One sub-model of M has
been eliminated. Thus the
probability of M must decrease.
(Or at most stay the same, if the probability of the
prize being behind the opened door is zero).
Problem: By symmetry, each
door has a 1/3 probability of
being correct. So prior to the
data;
Pr(N)=1/3
Pr(M)=Pr(M1)+Pr(M2)=2/3
So, Pr(M | D2)<2/3
But, switching is best if
Pr(M)>1/2, which is still possible.
Pr(M | D2)
1/3
Staying maximizes
the probability of
getting the prize
1/2
2/3
Switch maximizes
the probability of
getting the prize
Pr(M | D2)
1/2
Do not switch!
2/3
Switch!
Complete specification, strategy 1:
Complete specification, strategy 1:
D1: Monty Hall chooses to open a
door without regard to whether
you've chosen the right door or
not, easiest choice: always. Thus
D1 is irrelevant.
Pr(D1|N)=Pr(D1)=1. Thus Pr(N|
D1)=Pr(N).
Complete specification, strategy 1:
D1: Monty Hall chooses to open a
door always. D1 is irrelevant. Pr(D1|
N)=Pr(D1)=1. Thus Pr(N|D1)=Pr(N).
D2: Monty Hall always choses a nonoccupied door, thus D2 is also
irrelevant. Pr(D2|N)=Pr(D2)=1. Thus:
Pr(N | D2, D1)=Pr(N|D1)=Pr(N)=1/3
Easiest way to look at
strategy 1:
You'll get the same data
whether you've chosen the
right door or not. Thus the
probability stays the same.
However, the alternate is now
one single door.
Your choice, N
Possibility 1:
Possibility 2:
Possibility 3:
M1
M2
$
$
$
Your choice, N
Possibility 1:
Possibility 2:
Possibility 3:
M1
M2
$
$
$
Expected gain:
E(getting the prize without switching | D)=
1*Pr(N|D) + 0*Pr(M|D)=Pr(N|D)=1/3
Pr(M)
Pr(N)
1/3
1
Expectancy
Expectancy:
E(getting the prize when switching | D)=
0*Pr(N|D) + 1*Pr(M|D)=Pr(M|D)=
2/3
Pr(M)
Pr(N)
2/3
1
Expectancy
Strategy E(Switch|D) E(Stay|D)
1
2/3
1/3
Strategy E(Switch|D) E(Stay|D) Pr(D)
1
2/3
1/3
p=
Pr(D1)
Strategy E(Switch|D) E(Stay|D) Pr(D) E(Switch) E(Stay)
1
2/3
1/3
p=
p*2/3 +
Pr(D1) (1-p)*1/3
1/3
Strategy E(Switch|D) E(Stay|D) Pr(D) E(Switch) E(Stay)
1
2/3
1/3
1
2/3
1/3
easiest
p=
p*2/3 +
Pr(D1) (1-p)*1/3
1
2/3
1/3
1/3
Strategy E(Switch|D) E(Stay|D) Pr(D) E(Switch) E(Stay)
1
2/3
1/3
1
2/3
1/3
Strategy 2: Monty Hall has no privileged
information
D1 irrelevant: Pr(D1 | N)=Pr(D1) (=1)
D2 relevant: Pr(D2 | N) = 1
Pr(D2 | M) = ½
Pr(D2|N)Pr(N)
Pr(N|D)=Pr(N|D2)= ------------------------------Pr(D2|N)Pr(N)+Pr(D2|M)Pr(M)
1*1/3
= ---------------------- = ½=50%
1*1/3 + 1/2*2/3
Strategy E(Switch|D) E(Stay|D) Pr(D) E(Switch) E(Stay)
1
2/3
1/3
1
2/3
1/3
2
1/2
1/2
2/3
1/3
1/3
Strategy 3: If given the opportunity, Monty
Hall always opens a door with the prize. An
empty door is shown only when you've chosen correctly.
D1 irrelevant: Pr(D1 | N)=Pr(D1)
D2 relevant: Pr(D2 | N) = 1
Pr(D2 | M) = 0
Pr(D2|N)Pr(N)
Pr(N|D)=Pr(N|D2)= ------------------------------Pr(D2|N)Pr(N)+Pr(D2|M)Pr(M)
1*1/3
= ---------------------- = 1=100%
1*1/3 + 0*2/3
Strategy E(Switch|D) E(Stay|D) Pr(D) E(Switch) E(Stay)
1
2/3
1/3
1
2/3
1/3
2
1/2
1/2
2/3
1/3
1/3
3
0
1
1/3
0
1/3
Strategy 4: Monty Hall opens an empty door
and offers you the opportunity to switch only
when you've chosen correctly
D1 relevant: Pr(D1 | N)=1, Pr(D1 |M)=0
D2 irrelevant: Pr(D2 | N,D1) = Pr(D2|D1)=1
Pr(D1|N)Pr(N)
Pr(N|D)=Pr(N|D1)= ------------------------------Pr(D1|N)Pr(N)+Pr(D1|M)Pr(M)
1/3*1/3
= ---------------------- = 1=100%
1/3*1/3 + 0*2/3
Strategy E(Switch|D) E(Stay|D) Pr(D) E(Switch) E(Stay)
1
2/3
1/3
1
2/3
1/3
2
1/2
1/2
2/3
1/3
1/3
3
0
1
1/3
0
1/3
4
0
1
1/3
0
1/3
Strategy E(Switch|D) E(Stay|D) Pr(D) E(Switch) E(Stay)
1
2/3
1/3
1
2/3
1/3
2
1/2
1/2
2/3
1/3
1/3
3&4
0
1
1/3
0
1/3
Strategy 5: Monty Hall opens an empty door
and offers you the opportunity to switch only
when you've chosen incorrectly.
D1 relevant: Pr(D1 | N)=0, Pr(D1 | M)=1
D2 irrelevant: Pr(D2 | N, D1) = Pr(D2|D1)=1
Pr(D1|N)Pr(N)
Pr(N|D)=Pr(N|D1)= ------------------------------Pr(D1|N)Pr(N)+Pr(D1|M)Pr(M)
0*1/3
= ---------------------- = 0=0%
0*1/3 + 1*2/3
Strategy E(Switch|D) E(Stay|D) Pr(D) E(Switch) E(Stay)
1
2/3
1/3
1
2/3
1/3
2
1/2
1/2
2/3
1/3
1/3
3&4
0
1
1/3
0
1/3
5
1
0
2/3
1
1/3
Pr(Strategy 1)=Pr(strategy 2)=
Pr(strategy 3 or 4)=Pr(strategy 5)=1/4
1*1/4
Pr(Strategy 1 | D) = ---------------------------------1*1/4+2/3*1/4+1/3*1/4+2/3*1/4
= 3/8=37.5%
Pr(Strategy 2 | D)= Pr(Strategy 5 | D) = 1/4=25%
Pr(Strategy 3 or 4 | D) = 1/8 = 12.5 %
Pr(N | D)=sum Pr(N | Si, D) Pr(Si|D) = 37.5%
Pr(Strategy 2) = 94%
Pr(Strategy 1)=Pr(strategy 3/4)=
Pr(strategy 5)=2%
1*2%
Pr(Strategy 1 | D) = ---------------------------------1*2%+2/3*94%+1/3*2%+2/3*2%
= 3%
Pr(Strategy 2 | D) = 94%
Pr(Strategy 5 | D) = 2%
Pr(Strategy 3/4 | D) = 1%
Pr(N | D) = sum Pr(N | Si, D) Pr(Si|D) = 49%
Pr(Strategy 3 or 4) = 94%
Pr(Strategy 1)=Pr(strategy 2)=Pr(strategy 5)=2%
1*2%
Pr(Strategy 1 | D) = ---------------------------------1*2%+2/3*2%+1/3*96%+2/3*2%
= 5.56%
Pr(Strategy 2 | D)= Pr(Strategy 5 | D) =3.70%
Pr(Strategy 3 or 4 | D) = 87.03 %
PR(N|D) = sum Pr(N | Si, D) Pr(Si|D) = 90.7%