[dba-Tech] The Three Doors Problem

Arthur Fuller artful at rogers.com
Fri Aug 26 19:52:03 CDT 2005


You have just changed the game entirely, Lembit, and I repeat my former
invitation. Should you care to risk some dollars or Euros in the game of
backgammon, I would love to play. Your analysis of this problem suggests
that if we play 20 hours, I will make some serious money.
So. Now that I have switched the topic to backgammon, I will present the
simplest possible problem in backgammon. In case you do not play the game,
there is an object called the doubling cube which entitles both players to
double the stakes (but only when it is their turn and before they roll the
dice. If the cube is at 1 then a game is worth 1 unit (Euro, dollar,
whatever). Should the loser be gammoned, then 2 units; backgammoned then 3
units.
A double is like a raise in poker. If I do not accept the double then I fold
and you win the game. If on the other hand I do accept the double, then I
own the cube and you cannot double -- only I can. (This rule does not play
into the following problem, but I thought that I should explain the rules
anyway.)
One more important rule (important for this problem, anyway). Should a
player roll any double such as 4-4, that means the player can move 4 moves
of 4. In other words, if I roll 1-1, I em entitled to move any checker one
point four times: that could be one checker four points or four checkers one
point each or two checkers two points each or 1 checker one point and
another checker three points.
We are in the final rolls of the game. Let us call you player A and me
player B. A and B both have 2 checkers remaining. A has checkers on points
two and three. B has checkers on points one and two. It is A's move. Given
the foregoing, this problem can be reduced to, if A rolls a one, he loses
(because any possible roll takes both of B's checkers off, so if B gets a
chance, he wins).
1. Should A double?
2. Much more interesting: should B accept the double? (i.e. play on for two
points).
In this situation, the facts are:
a) number of possible rolls = 36 (6 sides * 6 sides)
b) the 36 possible rolls break down into 6 doubles (i.e. 1-1) - each of
which can occur exactly once, plus 15 pairs of disparate rolls (x-y). To
visualise this, imagine that one die is red and the other green. Therefore
you could roll 2-1 in two ways: Red 1 green 2 or green 2 red 1. As for the
doubles, there is only one possible way to roll 1-1: red 1 green 1. So the
chances of rolling any double are precisely 1 over 36. However, the chances
of rolling a 1 are different, since any roll containing a 1 counts.
c) The number of rolls containing a 1 is 11 out of 36, approximately 30.55
percent. (The five pairs 1-2 1-3 1-4 1-5 1-6, plus 1-1, = 11.)
At this point, I will raise a conjecture which some of you may dispute. In
my opinion there is only one sensible way to attack this problem, which is
to assume the 'multiple-universes' strategy. That is, I am playing 36 games
at once, and in each I get one unique occurrence of the dice. To put it
another way, I am playing 36 games at once, and subtracting the losses from
the wins in each universe... the result dictating what I choose to do in
THIS universe. I am concerned with winning the money. I am aware that we
will play lots of games. I am convinced that if I make the correct
(arithmetically correct) play in most of the games, I will come out a
winner. This is the principle upon which casinos are founded, and it seems
to work for them. The fact that one player in a million puts one token in
and wins millions is Good Advertising, not a problem.
Back to the problems as expressed above:
Problem 1 reduces to, How many ways does A fail? The answer, as illustrated
above, is 11 out of 36. Therefore A should clearly double the bet, since he
wins about 2 in 3 games. So if the stake is $100 he should eagerly double.
Problem 2 is more interesting. In the 36 virtual games, if B folds then A
wins 36 points ($3600). On the other hand, if B accepts the double and the
game continues, then A wins 25 games worth 2 points each but B wins 11 games
worth 2 points each... 50 - 22 = $2800 lost -- B would have lost $3600 had
he folded 36 times. 
This difference is called the per-game equity. By folding, B throws 8 over
36 per game, even though he has only a 31% chance of winning. By quitting, B
is throwing money away. By accepting the double, B has a much greater chance
of losing less money. This is not to say that B will be a winner. This
problem is not about who wins. It is about How to lose less money.
This is the simplest possible problem in backgammon analysis. It gets more
complex rapidly.
I am by no means a backgammon expert but I have played in several world
championships and hundreds of other tournaments. I am quite eager to have my
arithmetic defeated. I might just learn something!
I will venture this. What actually happens in one game is irrelevant. It is
all about playing LOTS of games. I have played backgammon against serious
competition for 3 days without sleep, at $100 US a point. Stamina plays a
part. Logic plays a part. Bluffing plays a part. At the end of the day, all
you can rely on is arithmetic, IMO. If the arithmetic says, Accept the
double, then Accept the double -- does not matter how good or bad your
opponent is, or how much money you are ahead or behind... all that is noise
as opposed to signal. The signal is the arithmetic correctness of your
analysis. It is quite possible that only 2 rolls in 36 will save the ass of
your opponent -- and if you double and he accepts and then rolls one of
those two numbers, such is life. Do not whine, do not belly-ache... accept
that the correct play was not the one that worked in this actual universe.
So what.
In a game such as backgammon or poker or even the 3-doors problem, I suggest
that we must not focus on any particular universe. Instead we must use a
multi-universe algorithm and calculate the number of universes in which we
win. Subtract the multi-universe wins from the losses, add the results
across universes, and the result tells you what to do.
A,

-----Original Message-----
From: dba-tech-bounces at databaseadvisors.com
[mailto:dba-tech-bounces at databaseadvisors.com] On Behalf Of Lembit Soobik
Sent: August 26, 2005 11:24 AM
To: Discussion of Hardware and Software issues
Subject: Re: [dba-Tech] The Three Doors Problem

I see what you mean, but thing is that you have changed the rules during the

game.
now assume after you revealed the one non-winning door
a third person comes in
what are his chances to select the winning door from the two which are left 
over?

Lembit

----- Original Message ----- 
From: "Scott Marcus" <marcus at tsstech.com>
To: "Discussion of Hardware and Software issues"
<dba-tech at databaseadvisors.com>
Sent: Friday, August 26, 2005 5:00 PM
Subject: RE: [dba-Tech] The Three Doors Problem


> When you originally picked the door, you had a 1/3 chance that the door
> was the winning door. If the host reveals a non-winning door that isn't
> the door you picked, that door you picked still has only a 1/3 chance of
> being correct.
>
> Lembit, lets play a game. Pick a number between 1 and 1,000,000. Only
> one number will be a winning number. After you pick your number, I'm
> gonna reveal 999,998 doors as non-winning doors so that only your door
> and one other door remains (of which one is the winning door). Are your
> chances still 50/50?
>
> So really pick a number. I'll have a third party monitor the winning
> number so that no cheating can be involved. We'll do this 1000 times and
> I bet you that if you switch doors that you will win way more than 50%
> of the time (it will be very close to if not 100% of the time).
>
> Scott Marcus
> IT Programmer
> TSS Technologies Inc.
> www.tss.com
>
>
> -----Original Message-----
> From: dba-tech-bounces at databaseadvisors.com
> [mailto:dba-tech-bounces at databaseadvisors.com] On Behalf Of Lembit
> Soobik
> Sent: Friday, August 26, 2005 10:45 AM
> To: Discussion of Hardware and Software issues
> Subject: Re: [dba-Tech] The Three Doors Problem
>
> which leaves one winning and one non-winning door
> and since you do not know which one the winning door is,
> you have a 1 out of 2 chance.
>
> The host has simply changed the game from 1 out of 3 to 1 out of two
> the new situation is not dependent on your previous choice.
>
> how could your making a choice at the beginning change the situation
> after the
> 3rd door was removed?
>
> Lembit
>
> ----- Original Message ----- 
> From: "Scott Marcus" <marcus at tsstech.com>
> To: "Discussion of Hardware and Software issues"
> <dba-tech at databaseadvisors.com>
> Sent: Friday, August 26, 2005 4:36 PM
> Subject: RE: [dba-Tech] The Three Doors Problem
>
>
>> John,
>>
>> The problem was stated originally that the host revealed a non-winning
>> door.
>>
>> Scott Marcus
>> IT Programmer
>> TSS Technologies Inc.
>> www.tss.com
>> _______________________________________________
>> dba-Tech mailing list
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>> Website: http://www.databaseadvisors.com
>>
>>
>> -- 
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>>
>
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