D. Samet I. Samet D. Schmeidler

S2S Two sums of money, S and 2S are put in two envelopes. Probability 1/2: the double sum is in the blue… To switch or not to switch?

S2S Probability 1/2: the double sum is in the red. 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…

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 argument for switching. 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. X An envelope is selected at random and handed to you.

To switch or not to switch? An argument for switching. Suppose there is X in the red envelope. X The same argument holds if you receive the blue envelope. No matter what envelope you get switch without opening it! Mind blowing !

John E. Littlewood (Ervin Schrödinger) 1953 Kraitchik, M Zabell, S.L Nalebuff, B Christensen, R., Utts J 1994 Jackson, F., P. Menzies, G. Oppy Castel, P., B. Diderik Sobel, J. H Brams, S. J., D. M. Kilgour Chae, K. C. Broome, J Bruss, F. T. 1997Arntzenius, F., D. McCarthy Merryfield, K.G., N. Viet, S. Watson Scott, A. D., M. Scott McGrew, T., D. Shier, H. S. Silverstein 1998 Norton, J.D Clark, M., N. Shackel, Horgan, T Blachman, N. M., D. M. Kilgour 2002 Chase, J. Chalmers, D.

荫鋙 舷聪 I write two real numbers x < y on two slips of paper, and put each in one of the envelopes. Which is the larger one? 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 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. Can you guarantee that your expected payoff is positive? your expected payoff is 0.

What can you guarantee? A threshold strategy: b Fix a number b (for big) If observed number ≥ b If observed number < b. bet it is the larger. bet it is the smaller.

.. What can you guarantee? b..... Case 1 b ≤ x < y xy Observing either x or y you bet it is the larger. Your expected payoff is 0.

.. What can you guarantee? b..... Case 2 x < y < b xy Observing either x or y you bet it is the smaller. Your expected payoff is 0.

.. What can you guarantee? b..... Case 3 x < b ≤ y xy Observing x you bet it is the smaller. You gain 1 for sure! Observing y you bet it is the larger.

What can you guarantee? xy Or any distribution that assigns positive probability to non-trivial intervals. A mixed strategy: choose b from a normal distribution.

What can you guarantee? xy With probability p b ≤ x < y. Your expected payoff is 0. p q r With probability q x < y < b. Your expected payoff is 0. With probability r, x < b ≤ y. Your gain is 1 for sure. This mixed strategy guarantees that your expected payoff is positive.

What can you guarantee? You can guarantee that your expected payoff is positive. Really??? Therefore, I cannot guarantee for myself zero expected payoffs.

Why cannot I guarantee 0? P is a probability over pairs of numbers x 1 and x 2 in the red and blue envelopes. x1x1 x2x2 x 2 > x 1 x 1 > x 2

Why cannot I guarantee 0? x1x1 x2x2 Property 1: Given any set of values A of x 1, the probability of x 2 > x 1 and x 1 > x 2 is 1/2. x 1 > x 2 x 2 > x 1 A /21/2 1/21/2 P(x 1 > x 2 ) | A) = 1/2 P(x 2 > x 1 ) | A) = 1/2 P is a probability over pairs of numbers x 1 and x 2 in the red and blue envelopes.

B Why cannot I guarantee 0? x1x1 x2x2 Property 2: Given any set of values B of x 2, the probability of x 2 > x 1 and x 1 > x 2 is 1/2. x 1 > x 2 x 2 > x 1 P(x 1 > x 2 ) | B) = 1/2 P(x 2 > x 1 ) | B) = 1/ /21/2 1/21/2 P is a probability over pairs of numbers x 1 and x 2 in the red and blue envelopes.

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. There is no probability P with these two properties. A contradiction!

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 x 1 > x 2 x 2 > x 1.. Q1Q1 Q2Q2 A1A1 A2A2. P(Q 1 )+ P(A 1 )= P(A 2 ) P(Q 2 ) = P(A 1 )  P(A 2 )  P(Q 2 ) = 0 P(Q 1 ) = 0 P( ) = 0

Property 2: Given any set of values B of x 2, the probability of x 2 > x 1 and x 2 < x 1 is 1/2. There is no probability distribution P over pairs of numbers x 1 and x 2 such that, Property 1: Given any set of values A of x 1, the probability of x 2 > x 1 and x 2 < x 1 is 1/2. We have shown the following... We now show that the following generalization holds...

There is no probability distribution P over pairs of numbers x 1 and x 2 such that, Property 1: Given any set of values A of x 1, the probability of x 2 > x 1 and x 2 < x 1 is 1/2. We allow the events x 2 > x 1, x 2 < x 1 and also x 2 = x 1 to have any probability. Property 2: Given any set of values B of x 2, the probability of x 2 > x 1 and x 2 < x 1 is 1/2.

There is no probability distribution P over pairs of numbers x 1 and x 2 such that, Property 1: Given any set of values A of x 1, the probabilities of x 2 > x 1, x 2 = x 1, and x 2 < x 1 are p, q, r. Property 2: Given any set of values B of x 2, the probability of x 2 > x 1 and x 2 < x 1 is 1/2.

There is no probability distribution P over pairs of numbers x 1 and x 2 such that, Property 1: Given any set of values A of x 1, the probabilities of x 2 > x 1, x 2 = x 1, and x 2 < x 1 are p, q, r. Property 2: Given any set of values B of x 1, the probabilities of x 2 > x 1, x 2 = x 1, and x 2 < x 1 are p, q, r. This is wrong if any of p, q, r is 1.

There is no probability distribution P over pairs of numbers x 1 and x 2 such that, Property 1: Given any set of values A of x 1, the probabilities of x 2 > x 1, x 2 = x 1, and x 2 < x 1 are p, q, r, all  1. Property 2: Given any set of values B of x 1, the probabilities of x 2 > x 1, x 2 = x 1, and x 2 < x 1 are p, q, r, all  1.

The one-observation theorem X 1, …, X n are n random variables. Y = f (X 1, …, X n ) is their order. If Y depends on (X 1, …, X n ), then it depends on at least one of the variables. e.g. X 7 < X 2 = X 4 < X 2 …

Interactive epistemology: a reminder states partitions knowledge common knowledge common prior posteriors 1/10 2/10 3/10 1/10 2/10 1/10 0 2/3 1/3

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 Is agreeing to agree possible? E. Lehrer and D. Samet

1’s profits 2’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.

1’s profits 2’s profits 1’s profits 2’s profits E E E P - a common prior, symmetric w.r.t. the axes. The posteriors of E in each state are 1/2. It is common knowledge that the posteriors coincide. There is no common prior for which the agents agree to agree on nontrivial posteriors for E.... There is a common prior for which the agents agree to agree on nontrivial posteriors for E.

1’s profits 2’s profits E E At each state the agents are ignorant of E: They cannot tell whether or not E. F – a 4-state event.... When F is added to the agents’ information they are still ignorant of E.

1’s profits 2’s profits E E At each state the agents are ignorant of E: They cannot tell whether or not E.... There is a finite event F which after being added to the agents’ information, they are still ignorant of E.

1’s profits 2’s profits At each state the agents are ignorant of E: They cannot tell whether or not E. E 1’s maximal profit in F 1 cannot tell not- E 2 cannot tell E no

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. 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.

2’s profit Shift-rotate left this line an irrational distance E The state space: the four thick lines in the four unit-squares The prior: the uniform distribution 1’s profit The posteriors of E: 1/2 in each state It is possible to agree to agree on nontrivial posteriors of E. there is no finite F as required.