<pre>
<verbatim>
The fair price for large N is $0.6321205588285576784; I will offer
a proof along with optimal strategies.

Denote the game as G_N().  After (N-M) rounds of play, the game will have
the same form as G_M().  Depending on the strategies each of the M boxes
will have a probability p_i of containing the dollar.  Let Waldo lock
the M'th box (renumbering the boxes if necessary).  Denote the game and
Waldo's expected winnings in the game by
		G''M(p''1, p''2, ..., p''M)
where
		p''1 + p''2 + ... + p_M = 1

When
		p''2 <code> p''3 </code> p''4 <code> ... </code> p''M
we adopt the abbreviation
		G''M(b) = G''M(1 + b - Mb, b, b, b, b, ..., b)
and note that since probabilities are never negative:
		1 + b - Mb >= 0, or
		0 <<code> b <</code> 1 / (M-1)

Various G''M(p''1, p''2, ..., p''M) have difficult solutions but we are asked
only to solve G_M(1/M) and it turns out we can accomplish this by
considering only the games
		G_M(b)  where	1/M  <<code>  b  <</code>  1/(M-1)		[[1]]
Games of this form will be said to satisfy constraint [[1]].

Induction is used for one of the theorems, so we'd better solve G''2 and G''3
for our basis.
		G''2(p''1, p''2) = max (p''1, p_2)
		G''3(p''1, p''2, p''3) = max (p''1 + p''2, p_3)
since after Monty opens box #1, box #2 will have probability (p''1 + p''2)
and vice versa.  When the probabilities satisfy constraint [[1]]:
		G''2(b) <code> G''2(1-b, b) </code> b
		G''3(b) <code> G''3(1-2b, b, b) </code> 1 - b

The proof of Theorem 1 is based on the probability p_i that box #i
contains the dollar.  (Of course this is Waldo's perceived probability:
Monty's probability would be 0 or 1.)  It may seem wrong for Waldo to
"forget" the game history and remember only the computed p_i.  For
example he may have previously locked some but not all of the boxes
and this fact is ignored except in the calculation of p_i.  Or Monty may
have some higher level "plan" which mightn't seem to translate directly
into probabilities.  But probability algebra obeys some simplifying
linearity rules and, if Monty keeps a "poker face", the probability model
is the only thing Waldo has to act on.

Especially paradoxical is the derivation of Waldo's p_i in his trivial
strategy below: he can adopt inferior but "correct" p_i to simplify the
analysis.

Theorem 1)
	If   b >= 1/M   then
	G''M(b) = G''[[M-1]]( (1-b) / (M-2) )

Proof)
	We will show that Monty and Waldo each have a strategy in G_M(b) to
	reduce the game to G_[[M-1]](b, q, q, ..., q) where q = (1-b) / (M-2)
	and where the boxes have been renumbered so that box #1 was box #M
	(the one Waldo locked) from the prior round and the new box #(M-1)
	is the one Waldo locks next.  Note that if Monty indeed arranges
	the probability mixture G_[[M-1]](b, q, q, q, q, ..., q) it won't
	matter which box Waldo locks (Box #1 has the only non-equal
	probability but Waldo cannot lock the same box twice in a row);
	this is a typical property of "saddlepoint" strategy.

	We will derive the necessary and sufficient condition for Monty to
	reduce any game G''M(p''1, p''2, p''3, ..., p_M) to a single game with
	the form G''[[M-1]](b, q, q, ..., q).  Using the numbering of G''M()
	let R(i,j) be the joint probability that Box #i contains the dollar
	and Box #j is opened by Monty in G_M().  We need consider only
		M >= 3
	Clearly,
		R(i, j) >= 0
		R(i, i) = 0
		R(i, M) = 0, i < M
		sum''over''j R(i,j) = p_i
	and to achieve q''2 <code> q''3 </code> ... = q''[[M-1]] in G''[[M-1]],
		R(1, j) = R(k, j)
			for 1 < j,k < M and j ===== k
====
		R(2, 1) = R(k, 1)
			for 2 < k < M
	and to make G_[[M-1]] be independent of Monty's play
		R(M, j)/R(1, j) = R(M, 2)/R(1, 2)
			for 2 < j < M
		R(M, 2)/R(1, 2) = R(M, 1)/R(2, 1)

	The above have a simple unique solution:
		R(i, j) = (1 - p''M) / (M - 2)  - p''j		[[2]]
			for i,j < M and i ===== j
====
		R(M, j) = p''M - p''j * p''M / (1 - p''M)		[[3]]
			for j < M
		p''j * (M-2) + p''M <= 1				[[4]]

	For the theorem we are given that G_M(b) satisfies constraint [[1]]
		1 / M  <<code>  b  <</code>  1 / (M - 1)
	which implies the weaker inequality
		(M - 3) / (M^2 - 3M + 1)  <<code>  b  <</code>  1 / (M - 1)
	and since for the constraint-[[1]] compliant G_M()
		p''j <code> b   or   p''j </code> (1+b-Mb)   for all j
	the inequality [[4]] follows directly.

	Hence Monty can transpose G''M(b) into G''[[M-1]]( (1-b) / (M-2) )
	whenever b ><code> 1/M and M ></code> 3.  (This G_[[M-1]] also happens to
	satisfy constraint [[1]] as needed for the next theorem.)

	It should be easy to argue that this strategy is optimal for Monty,
	but we want to derive Waldo's best strategy anyway and if it
	guarantees the same value we know we're at the "saddlepoint".
	If Waldo knows Monty has a non-optimal strategy he can take
	advantage of it, but we will just derive a strategy good enough
	to achieve the saddle-point value.

	Monty must transform G_M(b) into some
		G''[[M-1]](b, q''2, q''3, ..., q''[[M-1]])
	where Waldo has the choice of locking any of boxes #2 through #(M-1).
	If Waldo locks each of the (M-2) available boxes with probability
	1/(M-2) it is easily seen that the average probability that he
	locks the box with the dollar is (1-b) / (M-2) and the probabilities
	q''2, q''3, ..., q_[[M-2]] will also have the average value (1-b)/(M-2).
	If Waldo pretends to "forget" which box he picked before, he can
	take q''2 <code> q''3 </code> ... = (1-b)/(M-2) thereby constructing the same
	game Monty constructed with his saddlepoint strategy====
====

	In the above Waldo in effect "degraded" the accuracy of his
	probability estimates with the substitutions
		q''2' = (q''2 + q''3 + ... + q''[[M-1]]) / (M - 2)
		q''3' = (q''2 + q''3 + ... + q''[[M-1]]) / (M - 2)
		  et cetera
	If Waldo "knows" more than this, he can pretend he doesn't====
====
	For example he can ask Monty to secretly shuffle the boxes.

	Thus either player can reduce
		G_M(b), b >= 1/M
	to
		G_[[M-1]]((1-b)/(M-2))
	so this must be the saddlepoint.
	Q.E.D.

Theorem 2)
	If   b >= 1/M   then
	G_M(b) = 1 - 1/2==== + 1/3! - ... - (1-b)(-1)^M/(M-2)!
====
		= - sum (-1)^i/i==== - (1-b)(-1)^M/(M-2)!
====
	where the sum is over i = 1, 2, 3, ..., M-3

Proof)
	The proof is by induction.  We know the theorem holds for M = 3
	and we will assume it holds for (M-1).  Set
		c = (1-b) / (M-2)
	We noted earlier that b <<code> 1/(M-1): otherwise p_1 </code> (1 + b - Mb)
	is negative; hence we obtain
		c <code> (1-b)/(M-2) ></code> (1 - 1/(M-1)) / (M-2)
	or simply
		c >= 1/(M-1)
	Thus the condition of the inductive hypothesis is satisfied and
		G_[[M-1]](c) = 1 - 1/2==== + 1/3! - ... + (1-c)(-1)^M/(M-3)!
====
	But from Theorem 1
		G''M(b) = G''[[M-1]](c)
	and from the definition of c,
		c/(M-3)==== = (1-b)/(M-2)!
====
	which establishes the theorem.

Theorem 3)
	G''M(1/M) <code> G''M(1/M, ..., 1/M) </code> 1 - 1/2==== + 1/3! - ... -(-1)^M/M!
====

Proof)
	This follows directly from Theorem 2 and the observation that
		(1/M)/(M-2)==== = 1/(M-1)! - 1/M!
====

For large M, G_M(1/M) approaches (1 - 1/e).  It will be a little bigger
when M is odd and a little smaller when M is even.  I've appended the
numeric values below.

% dc
[[Solution for M =]]Plb1+pdsb]sy
65k1sa1sblyx2sc[[la1lc/-dsaplclyx'''scla1lc/+dsaplclyx'''sclzx]]dszx
Solution for M =2
0.50000000000000000000000000000000000000000000000000000000000000000
Solution for M =3
0.66666666666666666666666666666666666666666666666666666666666666666
Solution for M =4
0.62500000000000000000000000000000000000000000000000000000000000000
Solution for M =5
0.63333333333333333333333333333333333333333333333333333333333333333
Solution for M =6
0.63194444444444444444444444444444444444444444444444444444444444445
Solution for M =7
0.63214285714285714285714285714285714285714285714285714285714285714
Solution for M =8
0.63211805555555555555555555555555555555555555555555555555555555556
Solution for M =9
0.63212081128747795414462081128747795414462081128747795414462081129
Solution for M =10
0.63212053571428571428571428571428571428571428571428571428571428572
	. . .
Solution for M =52
0.63212055882855767840447622983853913255418886896823216549216319831

P. S. There are related unsolved problems:
(a) what about G''M(p''1, p''2, ..., p''M) that do not fit the pattern used
in the above solution?
(b) what if two boxes contain dollars?  (first, what should the rules be?)

-- james@crc.ricoh.com (James Allen)
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