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 = p_3 = p_4 = ... = 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 <= b <= 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 <= b <= 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) = G_2(1-b, b) = b G_3(b) = G_3(1-2b, b, b) = 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 onlyM >= 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 = q_3 = ... = 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 playR(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 <= b <= 1 / (M - 1)

which implies the weaker inequality

(M - 3) / (M^2 - 3M + 1) <= b <= 1 / (M - 1)

and since for the constraint-

1? compliant G_M()p_j = b or p_j = (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 >= 1/M and M >= 3. (This G_M-1? also happens to satisfy constraint1? 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

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 = q_3 = ... = (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 <= 1/(M-1): otherwise p_1 = (1 + b - Mb) is negative; hence we obtain

c = (1-b)/(M-2) >= (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) = G_M(1/M, ..., 1/M) = 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
65k1sa1sblyx2scla1lc/-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)