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A QUOTE OF WITTGENSTEIN. 'p' is true = p.  Ludwig Wittgenstein, Remarks on the foundations of mathematics, Part 1, Appendix 1, Section 6. 

Here is the whole of Section 6:

6. For what does a proposition's 'being true' mean? 'p' is true = p . (That is the answer.)

So we want to ask something like: under what circumstances do we assert a proposition? Or: how is the assertion of the proposition used in the language-game? And the 'assertion of the proposition' is here contrasted with the utterance of the sentence e.g. as practice in elocution,---or as part of another proposition, and so on.

If, then, we ask in this sense: "Under what circumstances is a proposition asserted in Russell's game?" the answer is: at the end of one of his proofs, or as a 'fundamental law' (Pp.). There is no other way in this system of employing asserted propositions in Russell's symbolism.

Link: https://archive.org/details/remarksonfoundat0000witt_c4e5/page/n7/mode/2up 

THE MONTY HALL PROBLEM. See: Wikipedia entry and A text of Andrew Vazsonyi 

Setting: We have three doors, one door hides a car, each of the other doors hides a goat. Player wins if she reveals the car.

Game 1: Player picks a door, Host opens another door, revealing a goat. Option 1: Player opens the door she picked. Option 2: Player opens the third door.

Monty Hall Problem: What are the winning probabilities of Options 1 and 2?

I think the Monty Hall problem is a very simple problem made counterintuitive by some red herrings. The main red herring is the fact that the host opens a door. Another red herring is perhaps the fact that the player (instead of a random process) picks a door (this might suggest that the player suspects that this door hides the car).

Claim: The winning probability of Option 1 is 1/3, and that of Option 2 is 2/3.

Game 2: Option 1: Player opens one door. Option 2: Player opens two doors.

Clearly in Game 2 the winning probability of Option 1 is 1/3, and that of Option 2 is 2/3. It suffices thus to show that Game 1 and Game 2 are equivalent in some precise sense. 

Suppose Player plays Game 1 with Option 2. She tells herself: "Suppose I played Game 2 (still with Option 2) instead of Game 1. Let D be the door I would not have opened." Then Player picks door D. We check easily that Player wins Game 1 if and only if she would have won Game 2. This proves that the winning probability of Option 2 in Game 1 is 2/3, as desired. 

A similar argument shows that the winning probability of Option 1 in Game 1 is 1/3, completing the proof of the Claim.

A QUESTION: Do we always have dim End_{End(V)}(V^{**}/V) = 1 ? Explanation: V is a K-vector space and V^{**} is its second dual. In particular V^{**} is an End(V)-module, and so is the quotient V^{**}/V. As such this quotient has a K-algebra A of endomorphisms. And the question is: Is A always one-dimensional? The answer is trivially yes if V is finite dimensional. More on this: https://mathoverflow.net/q/162162/461 

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