The Monty Hall Problem Revisited

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raydpratt

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In what has come to be known as "The Monty Hall Problem," a gameshow host, Monty Hall, offers a contestant the opportunity to choose any one of three doors in search of the Grand Prize. The other two doors hide low-value prizes like goats.

After the contestant makes an initial choice, but before the door is opened, Monty Hall invariably opens one of the remaining two doors if it is not the Grand Prize door, and then Monty Hall offers to let the contestant change the contestant's initial choice to the other closed door.

The question posed in The Monty Hall Problem is whether a contestant would gain any greater probability of success in choosing the Grand Prize door by abandoning the initial choice and switching to the other closed door.

Marilyn Vos Savant ( http://www.marilynvossavant.com/ ), a well-known genius with the highest recorded I.Q. (intelligence quotient) in the entire world, said yes, that switching from the initial choice would increase a contestant's random chance of choosing the Grand Prize door from 1/3 before switching, to 2/3 after switching. This conclusion has been verified by mathematicians, computer scientists, and other members of academia all over the world.

The logic and math are somewhat as follows:

Each initial choice has a 1/3 chance of success because there is only one Grand Prize randomly hidden behind one of the three doors. However, after an initial choice is made, the probability that the Grand Prize actually resides behind one of the other two remaining doors is 2/3. This is true because each of the two remaining doors also have a 1/3 probability of being the correct choice, and so the sum of their probabilities equals 2/3. As a consequence, as the argument goes, if one of those two unchosen doors is opened by Monty Hall and shown to have a goat, then the 2/3 probability of hiding the Grand Prize now resides entirely in the other unchosen door that has not been opened. Therefore, switching from one's initial choice to the other closed door will increase a contestant's probability of success from 1/3 to 2/3.

I disagree. Although the Grand Prize was initially hidden behind one of three doors, such as to give each door a 1/3 probability of being the correct door, the opening and elimination of a goat door leaves the Grand Prize hidden behind only one of two doors in every game, thus giving each of the two remaining doors a 1/2 probability of being the correct door. Switching choices is meaningless because both of the unopened doors have increased from a 1/3 probability of being correct to a 1/2 probability of being correct.

As proof, let's break it down and count all the possibilities:

First, there are six possible configurations of doors and prizes:

A. Door #1 (Stupid Goat), Door #2 (Fartsy Goat), Door #3 (Grand Prize)

B. Door #1 (Fartsy Goat), Door #2 (Stupid Goat), Door #3 (Grand Prize)

C. Door #1 (Stupid Goat), Door #2 (Grand Prize), Door #3 (Fartsy Goat)

D. Door #1 (Fartsy Goat), Door #2 (Grand Prize), Door #3 (Stupid Goat)

E. Door #1 (Grand Prize), Door #2 (Stupid Goat), Door #3 (Fartsy Goat)

F. Door #1 (Grand Prize), Door #2 (Fartsy Goat), Door #3 (Stupid Goat)

As to any one of these configurations, there are six exhaustive ways to change your mind about one door, after being shown an open door, and to choose the other door. However, two of the six ways of choosing will not be allowed in each configuration of doors and prizes because the door being opened and eliminated would be the Grand Prize door. You will see that there are two wins and two losses for each configuration of doors and prizes with all the allowed patterns of choice:

A. Door #1 (Stupid Goat), Door #2 (Fartsy Goat), Door #3 (Grand Prize)
......Opened.........................Chosen....................Changed Mind..........= Lose
......Chosen........................Opened.....................Changed Mind..........= Lose
......Opened.......................Changed Mind..............Chosen..................= Win
......Chosen.......................Changed Mind..............Opened..................= Not Allowed
......Changed Mind...............Opened......................Chosen..................= Win
......Changed Mind...............Chosen......................Opened..................= Not Allowed

B. Door #1 (Fartsy Goat), Door #2 (Stupid Goat), Door #3 (Grand Prize)
......Opened.........................Chosen....................Changed Mind..........= Lose
......Chosen........................Opened.....................Changed Mind..........= Lose
......Opened.......................Changed Mind..............Chosen..................= Win
......Chosen.......................Changed Mind..............Opened..................= Not Allowed
......Changed Mind...............Opened......................Chosen..................= Win
......Changed Mind...............Chosen......................Opened..................= Not Allowed

C. Door #1 (Stupid Goat), Door #2 (Grand Prize), Door #3 (Fartsy Goat)
......Opened.........................Chosen....................Changed Mind..........= Win
......Chosen........................Opened.....................Changed Mind..........= Not Allowed
......Opened.......................Changed Mind..............Chosen..................= Lose
......Chosen.......................Changed Mind..............Opened..................= Lose
......Changed Mind...............Opened......................Chosen..................= Not Allowed
......Changed Mind...............Chosen......................Opened..................= Win

D. Door #1 (Fartsy Goat), Door #2 (Grand Prize), Door #3 (Stupid Goat)
......Opened.........................Chosen....................Changed Mind..........= Win
......Chosen........................Opened.....................Changed Mind..........= Not Allowed
......Opened.......................Changed Mind..............Chosen..................= Lose
......Chosen.......................Changed Mind..............Opened..................= Lose
......Changed Mind...............Opened......................Chosen..................= Not Allowed
......Changed Mind...............Chosen......................Opened..................= Win

E. Door #1 (Grand Prize), Door #2 (Stupid Goat), Door #3 (Fartsy Goat)
......Opened.........................Chosen....................Changed Mind..........= Not Allowed
......Chosen........................Opened.....................Changed Mind..........= Win
......Opened.......................Changed Mind..............Chosen..................= Not Allowed
......Chosen.......................Changed Mind..............Opened..................= Win
......Changed Mind...............Opened......................Chosen..................= Lose
......Changed Mind...............Chosen......................Opened..................= Lose

F. Door #1 (Grand Prize), Door #2 (Fartsy Goat), Door #3 (Stupid Goat)
......Opened.........................Chosen....................Changed Mind..........= Not Allowed
......Chosen........................Opened.....................Changed Mind..........= Win
......Opened.......................Changed Mind..............Chosen..................= Not Allowed
......Chosen.......................Changed Mind..............Opened..................= Win
......Changed Mind...............Opened......................Chosen..................= Lose
......Changed Mind...............Chosen......................Opened..................= Lose

So, we can see and count and verify all 24 of the allowed outcomes for all the possible configurations of doors and prizes and all the possible and allowed patterns of choice, and we can see that there are a total of 12 wins out of 24 allowed outcomes. Thus, the probability of choosing the correct door after switching from your initial choice is 1/2, not 2/3; and, thus, there is no net gain from switching choices.
 
Because Monty always reveals a goat behind a door which you did not chose, the probability of winning goes up from 1/3 if you don't switch to 2/3 if you do switch.

There are 3 possibilities of where the prize is, PL. You start with 3 door to choose from, P1. Monty (who has information) has a choice of 1 doors to choose from if P1 = PL and 2 doors if P1 != PL, M. You need to have a strategy to switch to remaining door or not, S.

Weight 1/9 PL=1 P1=1 M=2or3 S0 -> Win S1 -> Lose
Weight 1/9 PL=1 P1=2 M=3 S0 -> Lose S1 -> Win
Weight 1/9 PL=1 P1=3 M=2 S0 -> Lose S1 -> Win
Weight 1/9 PL=2 P1=1 M=3 S0 -> Lose S1 -> Win
Weight 1/9 PL=2 P1=2 M=1or3 S0 -> Win S1 -> Lose
Weight 1/9 PL=2 P1=3 M=1 S0 -> Lose S1 -> Win
Weight 1/9 PL=3 P1=1 M=3 S0 -> Lose S1 -> Win
Weight 1/9 PL=3 P1=2 M=2 S0 -> Lose S1 -> Win
Weight 1/9 PL=3 P1=3 M=1or2 S0 -> Win S1 -> Lose
-- Summary:
Weight 1/3 PL=Whatever P1=PL M=!PL S0 -> Win S1 -> Lose
Weight 2/3 PL=Whatever P1=!PL M=!PL,!P1 S0 -> Lose S1 -> Win

So Stratgey S0 -- do not switch, has odds of winning of 1/3 exactly as expected.
Strategy S1 succeeds exactly when S0 fails, so its probability of winning is exactly 2/3.

The choice of strategy does not depend on any information revealed to the player, so it doesn't matter that if was chosen before play began or after Monty reveals an unworthy choice. If you'll look at your own table, the only time that the switching strategy loses is when the original pick was the winning door prize, which happens with probability 1/3. Since Monty's "Choice" between the goats is made from a position of complete information, and when he has two choices (as only happens when the player has picked the winning door at the start), we don't care which door he picks. Effectively he is flipping a coin between two choices we don't care about, and on your table, these cases should each be weighted to 1/2. Otherwise, you aren't recovering the correct probabilities of either S0 or S1.
--
If you assume Monty is making a random choice between the two unopened doors, and just by happenstance doesn't open the door with the grand prize, then the probability may indeed be 1/2 as you calculate -- but this assumption seems contradicted by the long history of the game. As you say, he invariably opens a door where there is no grand prize. So he has information.
 
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raydpratt said:
I disagree. Although the Grand Prize was initially hidden behind one of three doors, such as to give each door a 1/3 probability of being the correct door, the opening and elimination of a goat door leaves the Grand Prize hidden behind only one of two doors in every game, thus giving each of the two remaining doors a 1/2 probability of being the correct door. Switching choices is meaningless because both of the unopened doors have increased from a 1/3 probability of being correct to a 1/2 probability of being correct.

As proof, let's break it down and count all the possibilities:
Click to expand...
Your analysis is flawed in that you are looking at the number of permutations of Monty's possible actions but not considering the frequency of that action occurring.
In your analysis you see that Monty opens the doors to Goat 1 and Goat 2 twice each within the four allowable options, and you have assigned the end result (Win or Loss) equal footing.
But the frequency of Monty opening doors to Goat 1 and Goat 2 that lead to Loss is half that of the frequency that lead to Win.
And you have not taken this frequency into account.
Thus the analysis is flawed.


The set-up is as follows:
If a goat (G1) is behind door X, another goat (G2) behind door Y and the Star Prize (S) is behind door Z then there are 3 options for the contestant.

On the occasions when the Player is asked if they want to swap:
1) Player chooses G1, G2 is opened by Monty and S is selected - WIN.
2) Player chooses G2, G1 is opened by Mondy, and S is selected - WIN.
3) Player chooses S, either G1 or G2 is opened, and either of the goats is selected - LOSS.

From a Player perspective each option (1 to 3) has 1/3 chance of occuring, so you can see that if they don't swap then the probability of winning is 1/3 (i.e. they need to choose S initially). But if they do swap then 2 of those 3 have a chance of WIN - so the probabiliy of winning when swapping increases to 2/3.

So, to reiterate, your analysis does not take into account that the two options (in your analysis) that lead to LOSS only occur at half the frequency each compared to the WINs.
 
I don't know what happened to my post, but I tested the Monty Hall problem in a computer program, and swapping was an advantage just like it is supposed to be.
 
Well this computer program is about 8 years old, it's written when I was learning this new language, so it's not very well written, but here it is anyway...
Code: 
Rem Change Choice
Amount = 1000000

Randomize Timer()

For n = 1 to Amount

Car = rnd(2)+1
choice = rnd(2)+1
Goat = car+1
if Goat = 4 then Goat = 1
if goat = choice then goat = goat + 1
if Goat = 4 then Goat = 1
change = goat + 1
if change = 4 then change = 1
if change = choice then change = change +1
if change = 4 then change = 1
if change = Car then inc Changewin
next n

Rem Stick to Choice

For n = 1 to Amount

Car = rnd(2)+1
choice = rnd(2)+1

if choice = Car then inc Choicewin
next n

Print "Change and win = "; Changewin

Print "Stick to First choice = "; Choicewin

Suspend for key

It produces the 1/3 advantage.
 
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I wrote a program to solve this once. It's correct. You have a better chance if you switch doors. This is because Monty is not randomly opening a door.

Under normal circumstances you have a 1/3 chance of being correct and a 2/3 chance of being wrong. Since Monty never opens a good door, there is still a 2/3 chance you are wrong. Best to swap with him.
 
The difference is that Monty KNOWS. His action is not random - he NEVER opens the good door.
 
The problem with the way all the possibilities are counted in the OP is that they're not all equally likely. For example, if the arrangement happens to be "A", then
raydpratt said:
A. Door #1 (Stupid Goat), Door #2 (Fartsy Goat), Door #3 (Grand Prize)
......Opened.........................Chosen....................Changed Mind..........= Lose
Click to expand...
this happens 1/6[sup]th[/sup] of the time (1/3 chance that the player picks the door with the prize behind it, then 1/2 chance that the host opens the first door).
......Chosen........................Opened.....................Changed Mind..........= Lose
Click to expand...
This also happens 1/6[sup]th[/sup] of the time.
......Opened.......................Changed Mind..............Chosen..................= Win
Click to expand...
This happens 1/3[sup]rd[/sup] of the time.
......Changed Mind...............Opened......................Chosen..................= Win
Click to expand...
This also happens 1/3[sup]rd[/sup] of the time.

So you recover the 2/3 chance of winning with this strategy.
 
The easiest way to understand the power of information given to the contestant by Monty opening a losing door is:

Contestant is presented with 100 doors, and chooses one. Monty then opens 98 remaining losers, with a single door remaining unopened. Contestant's first choice is still 1/100 to be a winner, but does anyone believe the remaining door is also 1/100? No, the remaining door is a winner 99/100 of the time.
 
gmilam said:
The difference is that Monty KNOWS. His action is not random - he NEVER opens the good door.
Click to expand...

Right, precisely.

The Monty Hall problem really got me going when I first encountered it. Like many people, I didn't think that it would make a difference whether I switched or not, that after Monty opened a door and showed that there was nothing behind it, the odds would be 50-50 for the remaining two doors.

But as you say, Monty wasn't just opening a door at random. He already knew that the door he opened was one of the two wrong doors.

When you initially chose the door you wanted opened, you had a 1/3 chance of being right, and there was also a 1/3 chance that the prize was behind either of the other two doors, adding up to a 2/3 chance that the prize was behind one or the other of the two doors you didn't select.

Now Monty helpfully shows you which of those two doors definitely doesn't have the prize. That doesn't change the initial 1/3 odds of your first choice having been correct. It just means that the initial 2/3 chance that the prize is behind one of the other doors must be associated with that one remaining door.

So however counterintuitive it seems at first, it's always going to be rational to switch doors.

BTW, the Mythbusters TV show recently had a very interesting segment on the Monty Hall problem and explained it very well, I thought.

It's interesting that almost everyone gets this wrong at first, even some professional mathematicians. I'm speculating that the human mind has some built in subroutines to estimate common-sense probabilities, and people just naturally assume (in this case incorrectly) that the one-in-three and one-in-two situations are both independent random distributions that are independent of each other.
 
Moderator note: Multiple identical threads merged. Do not cross-post.
 
Anybody else think the mathematical approach that refutes the intuitive no-benefit-to-switching-solution is messed up somehow? Maybe it is that the events are actually independent and can't rightly be mathematically linked.
 
Imagine the problem with 100 doors where Monty proceeds to open 98 of the remaining all to show goats. Would you still say the door you chose is as likely to house the car as the one that Monty left?
 
Yeah, but Steve, imagine the problem with 100 doors where Monty proceeds to open 98 of the remaining all to show goats. Would you will say the door you chose is as likely to house the car as the one that Monty left??

Oh, I thought we were playing "repeat the same point someone else made a couple of posts ago". haha :D
 
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