For the original choice, there are these possibilities:
123 CGG GCG GGC
So whatever door you choose, your chances are 1/3.
For the second choice, call the door you originally chose 1, and the door Monty opens as 3. Now the possibilities are:
123 CGG GCG
(We know that GGC is not a possibility because Monty showed us a goat behind door 3.)
So if you choose either 1 or 2, you have a 1/2 chance of winning. Which means it doesn't improve your chances to switch.
Now, what's wrong with this argument?
Okay, so let's not renumber. Call the door you originaly pick number 1, and indicate the door Monty opens with parentheses. Then we have the following possibilities:
1 2 3 C (G) G C G (G) G C (G) G (G) C
So, if you stay with 1, your chance of winning is 1/2. If you switch to 2 or 3 (whichever remains unopened), your chances of winning are also 1/2. So it doesn't improve your chances to switch.
Or does it?
1 2 3
C g1 g2 C g2 g1 g1 C g2 g1 g2 C g2 C g1 g2 g1 C
There are three door choices the player can make, yielding 18 possible outcomes:
1 2 3
C g1 g2 C g2 g1 g1 C g2 g1 g2 C g2 C g1 g2 g1 C
1 2 3
C g1 g2 C g2 g1 g1 C g2 g1 g2 C g2 C g1 g2 g1 C
1 2 3
C g1 g2 C g2 g1 g1 C g2 g1 g2 C g2 C g1 g2 g1 C
At this point, the player has "lost" 2/3 of the time. If the player changed the door selection now, the choices would be between two doors, like this:
C g1 C g2 g1 g2 g1 C g2 g1 g2 C
Choosing either column 1 or column 2 still gives a 2/3 chance of losing, so switching doors won't accomplish anything.
Monty now eliminates an un-chosen goat (if there are two to chose from, it doesn't matter which one he chooses, so I'll always choose the rightmost). Note that in 4 out of the 6 cases, Monty is "forced" to give the player a car if he switches, and this is the key -- Monty's choice is not random. The resulting column for a switch is:
C C g1 C g2 C
Here's an expanded chart, where the player's first choice is bolded and the eliminated goat is replaced with a hyphen. I've put a "#" where a switch wins, and a "-" where a switch looses.
1 2 3 win/lose
C g1 - - C g2 - - g1 C - # g1 - C # g2 C - # g2 - C #
1 2 3
C g1 - # C g2 - # g1 C - - - g2 C # g2 C - - - g1 C #
1 2 3
C - g2 # C - g1 # - C g2 # g1 - C - - C g1 # g2 - C -
Well, suppose Monty opens both of the other doors, and they're both goats. Still think you should switch? Doesn't showing both goats behind the other doors mean that your original choice is a sure winner? That the odds for not switching are now 1.0, not 0.33? So why wouldn't showing the goat behind one of the other doors also change the odds?
I know you think this, but why? There are just two doors, and the car is behind one or the other of them. Why doesn't that make the odds 50-50 for the new contestant?
Suppose they don't know which of the two doors I chose. Then aren't their chances 50-50?
BTW, the reason I'm raising these questions is to try to suggest why the Monty Hall problem is a "problem". It wouldn't be a problem if the answer were obvious...
Yes, that is the argument that most convinces me that there is an advantage to switching. But how can we be sure that this result applies to the 3 door case. Certainly the benefit of switching is very much higher in the 1000 door case that in the 3 door case. It obviously decreases as the number of doors decreases. Maybe the advantage vanishes altogether in the 3 door case?
Suppose they choose the door you originally chose. Aren't their chances of winning the same as your chances if you choose not to switch?
Here's a similar problem in the game of Texas hold'em that many people don't believe at first: I offer you your choice among three two-card hands: (1) a pair of fours, (2) an ace and a king of different suits, (3) a ten and a jack of the same suit. After you choose, I choose one of the remaining hands. Then we deal five cards face up, and the player who can make the best five-card poker hand with any combination of his two cards plus the five on the board wins. Which cards do you choose, and what is the expected outcome? It turns out that whoever chooses first loses. If you choose the A-K, I choose the 4-4 and have a slight advantage. If you choose the 4-4, I choose the 10-J and have a slight advantage. If you choose the 10-J, I choose the A-K, and have an advantage!
I take the rather abstract view of an encyclopedia as something which explains everything. Thus I find the article actually quite accessible and helpful, written in a friendly direct way. The perl code is perhaps less useful to someone wanting to understand the problem in an article-reading sort of way, but I think of it as sort of a multimedia addition - a working Monty Hall Problem machine to tinker with. -J
In short, the reason for the above result is that Monty will ALWAYS show the other goat. Look at the following situation with C stand for Car, G for Goat, X for the goat that Monty picked, and with the player always pick the first door:
Initial Monty Result
================================================== CGG CGX Sticker wins and switcher lose. GCG GCX Sticker lose and switcher wins. GGC GXC Sticker lose and switcher wins.
As can been seen above, the chance for the switcher to win is twice that of the sticker.
I believe this is not a compelling argument; it will only convince people who already believe the result or don't follow closely. If I were to criticize the argumentation, I would point out that the case
CGG CXG Sticker wins and switcher losesis missing. So this makes two cases in favor of the sticker and two cases in favor of the loser: 50-50. To counter this objection, one would have to argue lengthily that the four listed cases are not equally likely, that in fact the first and the fourth case both have probability 1/6 while the second and the third both have probability 1/3. But why? And so on.
The crucial and convincing argument is the one given in the article: a switcher wins the car if and only if his original choice was a goat. Only if the reader understands that point will they truly be convinced. AxelBoldt, Wednesday, July 3, 2002
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