Two Plays by Tom Stoppard

I just reread The Real Inspector Hound and After Magritte, two plays by Tom Stoppard.  The Real Inspector Hound, a parody of the "murder at the stately English country home" genre, is much the better.  But my favorite exchange was in After Magritte:

THELMA:  ... All I can say is I'll be glad when it's all over and things are back to normal.  It's making you short-tempered and argumentative.  You contradict everything I say --
HARRIS: (heatedly)  That I deny --

What if, instead of contradicting her, Harris had agreed with Thelma and said "You're right"? 

Stoppard has worked on film scripts as well as theater.  The dystopian film Brazil, directed by Terry Gilliam, is one of my favorites.  I didn't realize that he also contributed much of the dialog for Indiana Jones and the Last Crusade, a Steven Spielberg film that I also enjoyed.

This (substantive) modification of Stoppard's dialog has bothered me:

ALICE:  You contradict everything I say.
BOB:  You're right.

I considered whether this might be some variant of the Liar Paradox similar to:

The next sentence is false.
The previous sentence is true.

That really is a logical paradox:  try to make the first sentence true - it doesn't work;  try to make the first sentence false - it doesn't work either.

But there's really no logical problem with Alice and Bob.  Alice's statement is false and so is Bob's.  Bob's response falsifies Alice's previous statement.  Bob proves Alice wrong by agreeing with her.  However since Bob proved Alice wrong, in a certain sense he actually did contradict her.   Which would seem to make Alice's statement true (or at least it could be true - assuming all of Bob's other statements also contradicted her).  But didn't I just say it was false?  Still it isn't really a paradox in pure logic.  Is it?

While we're at it, let's take a look at this stripped-down version of Stoppard's original Thelma and Harris exchange:

THELMA: You contradict everything I say.
HARRIS: That I deny.

Thelma's statement could be true (at least when Harris contradicts her in all other cases - let's go ahead assume that).  The Harris statement denies what Thelma just said, that denial counts as a contradiction (I suppose) and so the Harris statement is consistent with the meaning of Thelma's assertion.  Hence, the Harris statement supports Thelma's assertion.  Then given that we're assuming Harris otherwise contradicts Thelma, Thelma's statement must be true.  On the other hand, Thelma asserts that Harris contradicts everything she says and Harris' subsequent statement supports that.  If Harris' statement supports Thelma's statement, it can hardly be contradicting her.  So Thelma's statment must be false.  But didn't I just say Thelma's statement must be true?  Oops. 

What do you think?  But please, don't contradict everything I say.

 
"if it was so, it might be; and if it were so, it would be; but as it isn't, it ain't. That's logic."
--- Tweedledee in Through the Looking Glass by Lewis Carroll.

Aristotle's wheel paradox

Embarrassingly I'm still slowly reading Galileo at Work: His Scientific Biography by Stillman Drake.  Aristotle's wheel paradox came up several times in the book so I finally looked it up at Wikipedia.

When the inner and outer attached wheels make a complete revolution, they trace out lines of equal length, yet their circumferences are different.
How can that be?

The Mafia Game

The mafia game (or werewolf or assassin game) is a party game in which the organiser divides the players into two groups: the citizens and the mafia. The mafia are told who the other mafia are, but the citizens don't know. The game alternates between the "day" when all the players decide who to "lynch" and the "night" when just the mafia decide who to eliminate. The game ends when either all the citizens or all the mafia are eliminated.
The preprint A mathematical model of the Mafia game provides an analysis of strategy.

Probability Puzzles

There's an intersting discussion of some confusing Probability Puzzles on John Baez's new blog.

“I have two children. One is a boy born on a Tuesday. What is the probability I have two boys?”

The "born on a Tuesday" information can't possibly make a difference, can it? So the answer should be same same as:

“I have two children. One is a boy. What is the probability I have two boys?”

Shoudn't it?

There's a thorough analysis at Some Thoughts on Tuesday's Child by Greg Egan.

Monty Hall and John Von Neumann walk into a bar ...

I recently read The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser .
Here's my version of the Monty Hall Problem.
The Basic Situation is as follows. There are two individuals: Monty Hall and Alice. There is a car; there are three closed curtains; and the car is behind one of the curtains, which hides it completely. There is nothing behind the other two curtains. The action proceeds as follows:
1. Monty hides the car behind one of the curtains (H); Alice has no idea which one.
2. Alice chooses a curtain (C), but it is left closed. Alice doesn't know yet whether she chose the car or not.
3. Monty opens one of the curtains (S) showing Alice what's behind it.
3a. Monty is not permitted to open curtain Alice's curtain C; S cannot equal C.
4. Alice finally chooses another curtain (F) which can be different than C or the same. Alice gets what's behind the curtain she finally chose. If F=H, Alice wins the car, otherwise Alice gets nothing.

We can nail this down so that it is an exercise in pure logic and probability for Alice by further specifying what Monty does at steps 1 and 3. Here are the Additional Stipulations.
1a. Monty chooses where to hide the car (H) by randomly picking one of the three curtains: each of the three curtains is equally likely.
3b. Monty will only show Alice an empty curtain in step 3. Monty never opens the curtain with the car. S cannot equal H.

Given these Additional Stipulations. the consequences of the Alice's choice in step 4 are completely unambiguous - however the results surprise many people. Alice's two main strategies are Stay (F=C) and Switch (F≠C≠S) Many people guess that Stay and Switch are equivalent and that Alice wins 1/2 the time either way. Surprisingly Stay only wins 1/3 of the time while Switch wins 2/3's of the time.

Suppose Alice elects to follow the following strategy: always choose curtain #1 in step 2 and always Stay with curtain #1 in step 4.
In step 1. Monty hides the car behind curtain #1 1/3 of the time.
In step 2. Alice always chooses curtain #1.
In step 3. Monty will open either curtain #2 or curtain #3. Note that this does not change the actual location of the car, it's still behind curtain #1.
In step 4. Alice will always choose curtain #1 again. Alice wins the car.

In step 1. Monty hides the car behind curtain #2 or curtain #3 2/3's of the time.
In step 2. Alice always chooses curtain #1, which is empty.
In step 3. Monty will open either curtain #2 or curtain #3. Note that this does not change the location of the car, curtain #1 is still empty.
In step 4. Alice will always choose curtain #1 again, which is empty. Alice gets nothing.

So we see that by following the strategy of always chosing curtain #1 both times, Alice only wins the car 1/3 of the time.

Suppose on the other hand Alice uses another strategy: in step 2. she always chooses curtain #1; in step 4. shes always Switches to the only other curtain which is still closed.

In step 1. Monty hides the car behind curtain #1 1/3 of the time.
In step 2. Alice always chooses curtain #1.
In step 3. Monty will always open curtain #2 or curtain #3. Note that this does not change the actual location of the car, it's still behind curtain #1.
In step 4. because Alice always switches she will choose either curtain #2 or curtain #3. But the car is still behind curtain #1 and Alice gets nothing.

In step 1. Monty hides the car behind curtain #2 1/3 of the time.
In step 2. Alice always chooses curtain #1, which is empty.
In step 3. Monty must open curtain #3 - because it is the only door which is empty and is not Alice's. That of course does not change the location of the car - which is still behind curtain #2.
In step 4. Alice always switches to curtain #2 - because it is still closed. Alice wins.

In step 1. Monty hides the car behind curtain #3 1/3 of the time.
In step 2. Alice always chooses curtain #1, which is empty.
In step 3. Monty must open curtain #2 - because it is the only door which is empty and is not Alice's. That of course does not change the location of the car - which is still behind curtain #3.
In step 4. Alice always switches to curtain #3 - because it is still closed. Alice wins.

So Alice always wins if Monty hid the car behind curtain #2 or curtain #3 and Alice always loses if Monty hid the car behind curtain #1. Perhaps suprisingly Alice wins 2/3's the time when she always switches.


There's another formulation of the problem which only uses the facts in the Basic Situation. The Additional Stipulations are not included. Surprisingly Alice can guarentee the same favorable outcome in the Basic Situation that was achievable with the Additional Stipulations! Monty is permitted to choose where to hide the car (H) in step 1. any way he likes. In step 3. he is permitted to choose which curtain to open (S) by any method, as long as he doesn't open curtain C (still forbidden by 3a). It doesn't matter how Monty makes his choices (as long as he obeys 3a), Alice can still win the car surprisingly often.

Here's a paper (in pdf) which explains the Game Theory approach to the Monty Hall Problem: Probabilistic and Game Theoretic solutions to The Three Doors Problem

The Monty Hall Problem: A New Book

Two Doors and a Goat is a review in Science Magazine of The Monty Hall Problem: The Remarkable Story of Math's Most Contentious Brain Teaser by Jason Rosenhouse.

Strategic Pizza

How to eat 4/9 of a pizza

Given two players alternately picking pieces of a pizza sliced by radial cuts, in such a way that after the first piece is taken every subsequent chosen piece is adjacent to some previously taken piece, we provide a strategy for the starting player to get 4/9 of the pizza. This is best possible and settles a conjecture of Peter Winkler.

Let's Beat Monty Hall to Death

There's a series of posts in the New York Times TiernyLab column about the famous Monty Hall problem: Is That a Fair Coin in Your Pocket?.
Partition–Edit–Count: Naive Extensional Reasoning in Judgment of Conditional Probability

Overhang

Overhang

How far over the edge of the table can we reach by stacking n identical, homogeneous, frictionless blocks of length 1? A classical solution achieves an overhang asymptotic to 1/2 ln n. This solution is widely believed to be optimal. We show, however, that it is exponentially far from optimality by constructing simple n-block stacks that achieve an overhang of cn^1/3, for some constant c>0.

With the restriction that only one block can be on top of another, the classical solution is optimal. However with multiple blocks on top of each the solution can be improved significantly.