From: james@crc.ricoh.com (James Allen)
Subject: Re: 52-card Monty
Organization: RICOH California Research Center
Date: Sat, 21 Nov 92 20:02:05 GMT


I lost the original posting by Jim Saxe; here is the problem reworded:

===============================================================
Monty and Waldo play a game with N closed boxes.  Monty hides a
dollar in one box; the others are empty.  Monty opens the empty boxes
one by one.  When there are only two boxes left Waldo opens either box;
he wins if it contains the dollar.  Prior to each of the N-2 box
openings Waldo chooses one box and locks it, preventing Monty from opening
it next.  That box is then unlocked and cannot be so locked twice in a row.

What are the optimal strategies for Monty and Waldo and what is the
fair price for Waldo to pay to play the game?
===============================================================

I posted a poor effort on this earlier; let me try to make amends.
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)				[1]
In the solution we will need consider only the case
		p_2 = p_3 = p_4 = ... = p_M
We abbreviate this case by
		G_M(b) = G_M(1 + b - Mb, b, b, b, b, ..., b)	[2]
Of course the probabilities must sum to one; also since
probabilities are not negative:
		1 + b - Mb >= 0, or
		0 <= b <= 1 / (M-1)				[3]
In fact we will need to consider only the case
		b >= 1 / M					[4]
We will refer to constraints [2] and [4] below.

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.  With constraints [2], [4]
		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) )			[5]

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.

	The following applies to any G_M(p_1, p_2, p_3, ..., p_M).
	To achieve the saddlepoint Monty need simply assure that each box
	available for Waldo to lock has equal probability of containing
	the dollar.  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().  Clearly,
		R(i, j) >= 0
		R(i, i) = 0			[ M equations ]
		R(i, M) = 0, i < M		[ M-1 equations ]
		sum_over_j R(i,j) = p_i		[ M equations ]
	and to achieve constraint [2] (nearly all probabilities equal)
	in G_[M-1],
		R(1, j) = R(k, j)		[ M^2 - 5M + 6 equations]
			for 1 < j,k < M and j != k
		R(2, 1) = R(k, 1)		[ M-3 equations ]
			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)  [ M-3 equations ]
			for 2 < j < M
		R(M, 2)/R(1, 2) = R(M, 1)/R(2, 1)  [ 1 equation ]

	Viewing R(i, j) as M*M unknowns, the above gives M*M linear
	equations so there is a simple unique solution:

		R(i, j) = (1 - p_M) / (M - 2)  - p_j		[6]
			for i,j < M and i != j
		R(M, j) = p_M - p_j * p_M / (1 - p_M)		[7]
			for j < M

	Monty can arrange for this solution if and only if R(i,j) >= 0;
	as seen that is equivalent to
		p_j * (M-2) + p_M <= 1  , from [6]
		p_j + p_M <= 1          , from [7]

	When constraint [2] is satisfied (p_M = b and p_j = b or (1+b-Mb) )
	this becomes,
		b  <=  1 / (M - 1)
		b  >=  (M - 3) / (M^2 - 3M + 1)  >=  1 / M
	The first inequality was shown above and the second is somewhat
	weaker than constraint [4] ( b >= 1/M ).  Hence Monty can transpose
	G_M(b) into G_[M-1]( (1-b) / (M-2) ) whenever b >= 1/M and M >= 3.

	And of course
		R(2,1) / sum R(i,1) = (1-b) / (M-2)
	so after the next round the result is always
		G_[M-1]( (1-b) / (M-2) )

	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.

	In the game G_[M+1](p_1, p_2, ..., p_M, Q) Waldo has locked
	box #(M+1) which has probability Q.  After Monty's next move
	Waldo has the choice of (M-1) boxes; he should simply lock
	each box with probability 1/(M-1).  That box will contain the
	dollar with mean probability (1-Q)/(M-1).  The other (M-2)
	available boxes must have the same average probability (1-Q)/(M-1)
	since the box Waldo just unlocked still has probability Q.
	Thus Waldo can take the view that he has constructed the game
		G_M(Q, r, r, r, ..., r),  for r = (1-Q)/(M-1)
	which is precisely the game Monty constructed with his optimal
	strategy!

	In the above Waldo set his probability estimates to each be
		p_2 = (p_2 + p_3 + ... + p_[M-1]) / (M - 2)
		p_3 = (p_2 + p_3 + ... + p_[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)!		[8]
	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)
	and note that since b <= 1/(M-1) from [3], we have
		c = (1-b)/(M-2) >= (1 - 1/(M-1)) / (M-2)
	or simply
		c >= 1/(M-1)
	so 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.

Obligatory puzzle:  What property does this poem have?

		Ouija ionic, cameo aeons,
		Usual ennui there.
		Miaou.

James

% 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
^C