D. Samet
I. Samet
D. Schmeidler
To switch or not to switch?
Two sums of money, S and 2S
are put in two envelopes.
Probability 1/2: the double sum is in the blue…
S
2S
To switch or not to switch?
Two sums of money, S and 2S
are put in two envelopes.
Probability 1/2: the double sum is in the blue…
Probability 1/2: the double sum is in the red.
2S
S
To switch or not to switch?
An envelope is selected at random and handed to you.
You can take the money in the envelope . . .
. . . or take the money in the other envelope.
The situation is symmetric. Why switch?
To switch or not to switch?
An envelope is selected at random and handed to you.
An argument for
switching.
X
Suppose there is X
in the red
envelope.
In the blue: 2X or ½X, each with probability 1/2.
The expected sum in the blue:
½ (2X) + ½ (½ X) = 1¼ X
This is true for any X.
Switch before you see the amount in the red envelope.
To switch or not to switch?
An envelope is selected at random and handed to you.
An argument for
switching.
X
Mind
blowing!
Suppose there is X
in the red
envelope.
The same argument holds if you receive the blue envelope.
No matter what envelope you get
switch without opening it!
John E. Littlewood (Ervin Schrödinger)
1953
Kraitchik, M.
1988
Zabell, S.L.
1989
Nalebuff, B.
1992
Christensen, R., Utts J
1994
Jackson, F., P. Menzies, G. Oppy
Castel, P., B. Diderik
Sobel, J. H.
1995
Brams, S. J., D. M. Kilgour
Chae, K. C.
Broome, J.
Which is the larger one?
I write two real numbers x < y on two slips of paper,
and put each in one of the envelopes.
An envelope is selected at random and you observe the
number in it.
You can bet that the number is the larger of the two, or
the smaller.
If you are
right
I pay you 1
荫鋙
舷聪
if you are
wrong
you pay me 1.
What can you guarantee?
Always bet the number you observe
is the larger
The probability you are right is 1/2.
your expected payoff is 0.
Can you guarantee that your expected payoff is
positive?
What can you guarantee?
A threshold strategy:
Fix a number b (for big).
If observed number < b
bet it is the smaller.
.....
If observed number ≥ b
bet it is the larger.
.
.
b
.....
What can you guarantee?
Case 1
b≤x<y
Observing either x or y you bet it is the larger.
Your expected payoff is 0.
.....
.
b x
.
y
.....
What can you guarantee?
Case 2
x<y<b
Observing either x or y you bet it is the smaller.
Your expected payoff is 0.
.....
.
x
.
y
.....
b
What can you guarantee?
Case 3
x<b≤y
Observing x
you bet it is the smaller.
Observing y
you bet it is the larger.
You gain 1 for sure!
.....
.x
.
b y
.....
What can you guarantee?
A mixed strategy:
choose b from a normal distribution.
Or any distribution that assigns
positive probability
to non-trivial intervals.
.....
.x
.
y
.....
What can you guarantee?
With probability r, x < b ≤ y.
Your gain is 1 for sure.
With probability
p
b ≤ x < y.
Your expected
payoff is 0.
With
probability q
x < y < b.
Your expected
payoff is 0.
p
.....
.x
r
.
y
q
.....
What can you guarantee?
You can guarantee that your
expected payoff is positive.
Therefore, I cannot guarantee for myself
zero expected payoffs.
Really???
Why cannot I guarantee 0?
P is a probability
over pairs of numbers
x1 and x2
in the red and blue
envelopes.
x2
x2 > x1
x1
x1 > x2
Why cannot I guarantee 0?
P is a probability
over pairs of numbers
x1 and x2
in the red and blue
envelopes.
x2 > x1
..
..
..
x2
Property 1:
Given any set of values A of
x1, the probability of
x2 > x1 and x1 > x2
is 1/2.
1/2
1/2
A
P(x1 > x2) | A) = 1/2
P(x2 > x1) | A) = 1/2
x1
x1 > x2
Why cannot I guarantee 0?
P is a probability
over pairs of numbers
x1 and x2
in the red and blue
envelopes.
x2
Property 2:
Given any set of values B
of x2, the probability of
x2 > x1 and x1 > x2
is 1/2.
..
..
..
x2 > x1
1/2
..
..
..
B 1/2
P(x1 > x2) | B) = 1/2
P(x2 > x1) | B) = 1/2
x1
x1 > x2
Why cannot I guarantee 0?
When I use the mixed strategy P:
Given any information you get about one of the
numbers, the probability it is the larger is 1/2.
Your bet must result in zero expected payoff.
A contradiction!
There is no probability P with
these two properties.
Explaining puzzle 1
An assumption is made:
No matter what envelope you hold
and which number you observe,
The probability you have the bigger number
is 1/2
This assumption presupposes a probability distribution P over
pairs of sums which was shown, by puzzle 2, not to exist.
A direct proof
A2
Q2
P(Q2) = P(A1)  P(A2)
A1
P(Q2) = 0
.
Q1
P(x2 > x1 ) = 0
P(Q1) + P(A1) =P(A2)
P(x1 > x2 ) = 0
P(Q1) = 0
There is no probability
distribution P
over pairs of numbers
x1 and x2
such that,
We now show that the following
We have shown the following...
generalization
holds... Property 2:
Property 1:
Given any set of values A
of x1, the probability of
x2 > x1 and x2 < x1
is 1/2.
Given any set of values B
of x2, the probability of
x2 > x1 and x2 < x1
is 1/2.
We allow the events
x2 > x1, x2 < x1 and also
There is no probabilityx2 = x1 to have any
distribution P
probability.
over pairs of numbers
x1 and x2
such that,
Property 1:
Given any set of values A
of x1, the probability of
x2 > x1 and x2 < x1
is 1/2.
Property 2:
Given any set of values B
of x2, the probability of
x2 > x1 and x2 < x1
is 1/2.
There is no probability
distribution P
over pairs of numbers
x1 and x2
such that,
Property 1:
Given any set of values A
of x1, the probabilities of
x2 > x1, x2 = x1, and x2 < x1
are p, q, r.
Property 2:
Given any set of values B
of x2, the probability of
x2 > x1 and x2 < x1
is 1/2.
This is wrong if
any of p, q, r
is 1.
There is no probability
distribution P
over pairs of numbers
x1 and x2
such that,
Property 1:
Given any set of values A
of x1, the probabilities of
x2 > x1, x2 = x1, and x2 < x1
are p, q, r.
Property 2:
Given any set of values B
of x1, the probabilities of
x2 > x1, x2 = x1, and x2 < x1
are p, q, r.
There is no probability
distribution P
over pairs of numbers
x1 and x2
such that,
Property 1:
Given any set of values A
of x1, the probabilities of
x2 > x1, x2 = x1, and x2 < x1
are p, q, r, all  1.
Property 2:
Given any set of values B
of x1, the probabilities of
x2 > x1, x2 = x1, and x2 < x1
are p, q, r, all  1.
The one-observation theorem
X1, … , Xn are n random variables.
Y = f (X1, … , Xn) is their order.
If Y depends on (X1, … , Xn) ,
then
e.g. it depends on at least one of the
variables.
X7 < X2 = X4 < X2 …
Interactive epistemology: a reminder
states
partitions
knowledge
common knowledge
common prior
posteriors
. .
. . .
. .
1/10
0
2/3
2/10
3/10
2/10
1/3
1/10
1/10
Agreeing to disagree is impossible.
(Aumann, 1976)
Having common knowledge
Having different posteriors for an
event.
Is agreeing to agree possible?
E. Lehrer and D. Samet
2’s profits
1’s profits
Each point is a state that
specifies the profits of the firms.
Each firm knows only its
own profit.
1’s partition – vertical lines.
2’s partition – horizontal lines.
2’s profits
2’s profits
E
...
E
1’s profits
1’s profits
There is
a common prior
for which the agents
P - a common prior, symmetric
agree to agree
w.r.t. the axes.
on nontrivial
The posteriors of E in each
for E.
stateposteriors
are 1/2.
...
There is
no common prior
for which the agents
agree to agree
on nontrivial
posteriors for E.
E
It is common knowledge that
the posteriors coincide.
...
At each state
the agents are
ignorant of E:
They cannot tell
whether or not E.
2’s profits
E
1’s profits
E
...
F – a 4-state event.
When F is added to the agents’
information they are still ignorant of E.
...
At each state
the agents are
ignorant of E:
They cannot tell
whether or not E.
2’s profits
E
1’s profits
E
...
There is a finite event F which after
being added to the agents’ information,
they are still ignorant of E.
At each state
the agents are
ignorant of E:
They cannot tell
whether or not E.
1 cannot tell not- E
2’s profits
E
2 cannot tell E
1’s profits
1’s maximal profit in F
There is no
a finite event F which after
being added to the agents’ information,
they are still ignorant of E.
For countable partitions,
it is possible to agree to agree on
nontrivial posteriors for E
iff
there exists a finite F which after being
added to the agents’ information,
they are
ignorant of E.
Shift-rotate
2’s profitleft this line an
irrational distance
E
there is
no finite
F as
required.
1’s profit
The state space:
the four thick lines
the on
four
unit-squares
It is possible to agree to in
agree
nontrivial
The prior:of E.
the uniform distribution
posteriors
The posteriors of E: 1/2 in each state