Monty Hall Problem Revisited Again Again!

Cecil's Columns/Staff Reports
81

Incidentally, an invaluable mathematically tool for solving these problems is Bayes’ Theorem [I’ve basically stuck to English descriptions, but maybe a little math wouldn’t be too scary?]:

I shall write p(A) to denote the probability of fact A being true, before we’ve learnt anything new. And I’ll write p(B | A) to denote the probability of fact B being true, after learning that A is true.

Well, clearly, p(A and B) = p(A) * p(B | A). But, also, the same way, p(A and B) = p(B) * p(A | B).

Therefore, p(A) * p(B | A) = p(B) * p(A | B). Writing this in a different form, we get that p(B | A) = p(B) * p(A | B) / p(A). This is called Bayes’ Theorem. It will save your ass on many occasions.

How can we apply it to this case? Well, instead of B and A, we’ll use “Door X wins” and “Monty reveals Door Y”. We get that p(Door X wins | Monty reveals Door Y) = p(Door X wins) * p(Monty reveals Door Y | Door X wins) / p(Monty reveals Door Y).

Note how p(Door X wins | Monty reveals Door Y) isn’t necessarily just equal to p(Door X wins). This means that learning that Monty reveals Door Y can change our calculated probability that Door X wins. Specifically, it does so by multiplying it by p(Monty reveals Door Y | Door X wins) / p(Monty reveals Door Y). What this fraction equals depends, of course, on how we are probabilistically modelling Monty’s door-revealing proclivities. But the fraction won’t equal 1 except in the case where p(Monty reveals Door Y | Door X wins) = p(Monty reveals Door Y) [the case where “Door X wins” is probabilistically independent of “Monty reveals Door Y”].

If learning “Door X wins” would increase the probability of “Monty reveals Door Y”, then, conversely, it must also be true that learning “Monty reveals Door Y” will multiply the probability of “Door X wins” by a factor greater than 1.

Similarly, if learning “Door X wins” would decrease the probability of “Monty reveals Door Y”, then, conversely, it must also be true that learning “Monty reveals Door Y” will multiply the probability of “Door X wins” by a factor less than 1.

82

Uhh, although as far as I can see you’re correct Indistinguishable, I’m not sure where Bayes theorem would help you with this problem. Using mathematical notation is always a good way to keep things correct though. I’m going to do the same here, although i’ll try to explain as I go along for those of you who don’t understand mathese ;). Before I start working it through I’m going to label each event with a letter, to reduce the amount I have to type, and you have to read, and show a little bit of maths.

First up we have the event that I initially pick the door with the car behind it. I will label this event R.
We also have the event that the door I pick does not have the car behind it. As one of either this event or R has to occur, and only one, this is the complementary event of R. Hence I label this Rc (NB normally the c would be superscript, but I don’t know how to make it superscript here).
Next we have the event that the door that I am offered to swap to is has the car behind it. I’ll label this event S. The complementary event of this is that the door I am offered does not have the car behind it, so is event Sc.
The probability of an event X happening is written P(X), so that the probability of the door I initially pick not having the car behind it is P(Rc).

I could give seperate labels to the event of me winning and losing, and to me taking the swap or not. However, if you think about it you should realise that the probability of me winning if I do not swap is equal to P®; If I guessed correctly and do not swap I win. Similarly, the chance I win if I take the swap is P(S).

There are a few important bits of maths to know. Firstly, the chance of X occuring if you know Y has occured is different than if you don’t know whether Y has occured or not; we call this the probability of X given Y, in shorthand P(X|Y).
It is true in all cases that:
P(X) = P(X|Y)*P(Y) + P(X|Yc)*P(Yc)
Lastly, if event A occuring means that event B cannot occur then:
P(B|A) = 0

Right, now I’ve done all that I can start on the working out the problem. I’ll start with finding P®. This is simple, as picking a door is the first thing that happens. This makes P® equal to the number of prizes over the number of doors, in the normal case 1/3.
Hence the chance of winning if you do not swap is (in the normal game) 1/3.
This does not change.

So far, so simple. P(S) however, will take a bit more work. First, i’ll use the equation above to say that:
P(S) = P(S|R)*P® + P(S|Rc)*P(Rc)
However, as the car cannot be behind both the door I picked and the door I am offered as a swap S cannot occur if R has occured. Therefore:
P(S) = P(S|Rc)*P(Rc)
as P(S|R)=0
We know the probability that I do not pick the correct door, it is the number of doors minus one over the number of doors. This means with 3 doors we get
P(S) = 2/3 * P(S|Rc)

Up to this point we have not had to consider whether or not monty’s memory is any good. It is at this point that it could possibly make a difference. However, as I’m going to show, this is only if monty has to directly pick which door I am being offered to swap with, instead of having to eliminate one door and give you a choice to swap with the last.

If we assume monty’s memory is correct then in both cases it is easy to see that P(S|Rc) is equal to 1, giving us the result that (with 3 doors) the probability of winning if I swap is 2/3. This result seems to be generally accepted by everyone.
Now, if monty has to choose which door to offer us, and he cannot remember which door has the car behind it, as the car is not behind my door he has a half chance of choosing the door with the car behind it, so P(S|Rc)=1/2, giving P(S) to be 1/3. This is the result that people have obtained, and are mistakenly using to say that the probability of winning by switching is equal to that of winning by not switching. That is because this result is only true whilst we do not see what is behind the last door.

If monty, as he does on the show, picks one of the other two doors and reveals what is behind it the situation changes. This is exactly the same as if he picked one door to offer me and then shows us what is behind all the others except the one I picked and the one he has picked, which in the 3 door case means he shows us what is behind one door.
The only probability I need to find is the probability that he has picked correctly given I picked wrongly. If I picked wrongly, then there must be a goat behind my door. This means that if monty opens all the doors except mine and his and reveals goats behind all the opened doors, as the only door that is left that we don’t know has a goat behind it is monty’s door, then his door has the car behind it every time. Hence, assuming we do not see a car when he opens the doors, with 3 doors P(S|Rc)=1, and P(S)=2/3.
This means that if monty does not know (or has forgotten) which door the car is behind it does NOT change the probability of winning if you switch, it is still favourable to switch. This argument also works if you increase the number of doors in the competition, as long as monty opens all but yours and his, but I’m not sure it would work if you increase the number of prizes.

83

Because it illustrates how probabilities should be updated when receiving new information.

Let’s take a starting point where I’ve already chosen Door 1, and currently think each door has probability 1/3 of being the winner and 2/3 probability of instead containing a goat. [Note that even this depends on certain assumptions being in my probabilistic model, though we tend to gloss over them].

Now, I receive the information that “Monty opens Door 2 and it contains a goat”. What does this do to my probabilities?

Well, we can break it into two updates. First, I learn that “Door 2 contains a goat”. Then, I learn that “Monty opens Door 2”.

Ok, what does the “Door 2 contains a goat” information update do for me? By Bayes’ Theorem, p(Door Y wins | Door 2 contains a goat) = p(Door Y wins) * p(Door 2 contains a goat | Door Y wins) / p(Door 2 contains a goat). Well, as I said, my current p(Door Y wins) is 1/3 and my current p(Door 2 contains a goat) is 2/3; furthermore, we know that p(Door 2 contains a goat | Door Y wins) is 0% if Y is 2, but 100% otherwise. Thus, p(Door Y wins | Door 2 contains a goat) = 1/3 * (0% if Y is 2, but 100% otherwise) / (2/3) = 0% if Y is 2, but 50% otherwise. Thus, I arrive at the position where I’m confident Door 2 won’t win, but am 50-50 split between Doors 1 and 3.

Great. Let’s denote this new probability distribution with q() instead of p(). Now let’s add the remaining information “Monty opens Door 2”. How does Bayes’ Theorem tell us to incorporate this new information? Well, q(Door Y wins | Monty opens Door 2) = q(Door Y wins) * q(Monty opens Door 2 | Door Y wins)/q(Monty opens Door 2).

The details from hereon out depend on how Monty chooses to open doors. Let’s say Monty opens doors by picking one at uniform random from the ones I haven’t chosen which contain a goat; thus, q(Monty opens Door 2 | Door Y wins) = 50% if Y is 1, but 100% if Y is 3. From this, it follows that q(Monty opens Door 2) = q(Door 1 wins and Monty opens Door 2) + q(Door 2 wins and Monty opens Door 2) + q(Door 3 wins and Monty opens Door 2) = 50% * 50% + 0 + 50% * 100% = 75%.

Thus, going back to q(Door Y wins | Monty opens Door 2) = q(Door Y wins) * q(Monty opens Door 2 | Door Y wins)/q(Monty opens Door 2), we have that q(Door 1 wins | Monty opens Door 2) = q(Door 1 wins) * q(Monty opens Door 2 | Door 1 wins)/q(Monty opens Door 2) = 50% * 50%/75% = 1/3, while q(Door 3 wins | Monty opens Door 2) = q(Door 3 wins) * q(Monty opens Door 2 | Door 3 wins)/q(Monty opens Door 2) = 50% * 100%/75% = 2/3.

Alternatively, if Monty opens doors at complete random, then q(Door Y wins | Monty opens door 2) = q(Door Y wins) * q(Monty opens Door 2 | Door Y wins)/q(Monty opens Door 2) = q(Door Y wins) * (1/3)/(1/3) = q(Door Y wins), and nothing changes. I remain thinking Door 1 and Door 3 each have probability 50% of winning.

You see? Bayes’ Theorem makes it easy to figure out how to update probability distributions upon receiving new information. Of course, if you’re able to solve a problem another way, that can be good too; but you’ll almost always be able to use Bayes’ Theorem if you want.

84

It clicked for me. Monty randomly (without knowledge) selecting one of the doors is equivalent, I believe, to my always starting by choosing door #1, and Monty always opening door #3 and hoping for the best. So, when I said this…

…and applying it to the above, my error was assuming that switching from door #1 gave me a 2/3 chance of winning. It does only when Monty always reveals a goat, which only occurs when he knows what’s behind the doors; in that case, switching always leads to the car if I haven’t already chosen it; there is no “I picked wrong initially” scenario where it’s not to my advantage to switch to the other unopened door.

If Monty always just opens door #3, though, I’ve lost the ability to “fix” my wrong choice of door #1 in instances where he reveals a car behind #3, permitting me to switch to door #2 if I want to (why would I?). So, limiting my perspective to only those instances where door #3 is a goat, half the time I’ll have selected the car already, the other half I won’t have. Makes no difference if I switch.

I realize multiple people have already tried to explain this exact thing in some form to me–just sharing how it clicked for me, in case it helps anyone else.

85

Bingo. :slight_smile:

86

Had a long conversation over dinner last night about this topic.

We concluded that, even though you have 3 choices and you must pick one of them, your chances are really 1 in 2, not 1 in 3. Those are your chances before Monty eliminates a door, because you know he’s going to be eliminating a door with a goat.

87

Since there were at least two of you, why didn’t you just play the game a bunch of times?

88

Your analysis would be correct if Monty chose his door first–in that case, he’d be reducing your choice from 1/3 to 1/2. However, Monty is restricted in that he chooses second, which gives a different result.

As ZenBeam says: just play the game a bunch of times and see what the result is.

89

Obviously, your chances of winning the state lottery are 1 in 2 as well, because you know the two results are either “winning” or “losing.”

/sarcasm

90

Quoth Gangster Octopus:

Point of order: You’re assuming (as I do as well), that if Monty opens the car door, you’re still allowed to choose that door, even though it’s already open. In this case, your odds of winning are the same regardless of whether Monty knows where the car is, no matter how many doors there are or how many are revealed. However, it appears that most other posters assume that you cannot choose an open door, and that therefore, if Monty opens the car door, you’re screwed.

91

I know you’ve since “gotten it”, but I wanted to reply anyway. =)

If all decisions are made randomly (winning options in bold):

[ol]
[li]You choose the car door, Monty reveals one of the goats, you switch, you lose: 1/3 * 1 * 1/2 = 1/6 probability[/li][li]You choose the car door, Monty reveals one of the goats, you don’t switch, you win: 1/3 * 1 * 1/2 = 1/6[/li][li]You choose the car door, Monty reveals the car, you switch, you lose: 1/3 * 0 * 1/2 = 0[/li][li]You choose the car door, Monty reveals the car, you don’t switch, you lose: 1/3 * 0 * 1/2 = 0[/li][li]You don’t choose the car door, Monty reveals the other goat, you switch, you win: 2/3 * 1/2 * 1/2 = 1/6[/li][li]You don’t choose the car door, Monty reveals the other goat, you don’t switch, you lose: 2/3 * 1/2 * 1/2 = 1/6[/li][li]You don’t choose the car door, Monty reveals the car, you switch, you lose: 2/3 * 1/2 * 1/2 = 1/6[/li][li]You don’t choose the car door, Monty reveals the car, you don’t switch, you lose: 2/3 * 1/2 * 1/2 = 1/6[/li][/ol]
Note that the probabilities add up to one, and it doesn’t matter whether you switch or not; one of the two win-scenarios requires you to switch, and the other requires you to not switch, and both have the same probability.

However, if Monty knows where the car is, and guarantees he won’t open the door with the car, then the probabilities for options 7 and 8 drop to 0. Those two 1/6-probabilities have to go somewhere. Where do they go? Well, in the case where you don’t choose the car, Monty only has one choice, so the probabilities of options 5 and 6 go up:

  1. You don’t choose the car door, Monty reveals the other goat, you switch, you win: 2/3 * 1 * 1/2 = 1/3
  2. You don’t choose the car door, Monty reveals the other goat, you don’t switch, you lose: 2/3 * 1 * 1/2 = 1/3

Now, since two losing options have been eliminated, two other scenarios have doubled in probability. Options 5 and 6 are each now twice as likely, but one of them is a winning scenario! You’ve taken a 1/6 losing option (old option 7) and a 1/6 winning option (old option 5) and replaced them with a single 1/3 winning option (old option 5).

Now, you still have two options that lead to winning, and there’s still one that requires a switch and one that requires no switch. The difference is that the one that requires a switch is now twice as likely to occur, so you’re much better off switching.
Powers &8^]

92

Correct! When you have the initial three doors you will be correct only one third of the time. When you have the two doors you will be correct one half of the time. The reason we have trouble seeing this is because we think like real world people who have only one shot at the car and not like mathematicians who want to run 1000 or more choices.

93

You have just said correct and then said something entirely different, contradicting the point Little Nemo made.

The expected answer is that you have a 1/3 chance of winning when you are given 3 doors, Monty reveals one and gives you a second choice, and now you have a 1/2 chance of being right. This logic is incorrect, assuming Monty knows where the car is and will never reveal the car.

Rather, you have a 1/3 chance of being right, and Monty reveals a known goat, you still have a 2/3 chance of being wrong on your first choice.

94

Look at it this way. Forget about the goat. Just imagine there are three doors and behind one of them is a prize. Now pick one door. Monty then offers you a chance to trade whatever is behind the one door you picked for whatever is behind both of the other doors. Do you now see how switching doubles your chances of winning the prize?

95

I believe msean is contrasting the initial odds in the case where you start with three doors with initial odds in another case where you start with two doors.

96

zut, I disagree. That’s not how his statement is worded, so if that is what he means, he did not communicate well.

Note that he says “When you have the initial three doors”, and then says “When you have the two doors”. When you have, not if you have. He’s contrasting two points in the game, not two different games.

He is saying that when Monty eliminates a door, your odds go from picking one door in three to picking 1 door in 2. In other words, you remake your choice from the remaining two doors (stay or switch).

That seems reasonable, but the reality is that is just restating the condition as if Monty picked his door at random, so you started with 1/3 chance of being right, Monty had 1/3 chance of getting the prize, and the remaining door had 1/3 chance of having the prize. Monty was wrong, eliminating 1 door and leaving 2 more doors with equal chance of having the prize.