The Four Door Problem – in a nutshell.

There are four doors, red, yellow, blue and green.

Three lead the way to dusty death. One leads the way to fame and fortune. They are assigned in order by your evil host who draws four balls out of a bag, coloured red, yellow, blue and green. The first three out of the bag are the colours of the doors that lead to dusty death. The fourth leads to fame and fortune. You must choose one of these doors, without knowing which of them is the lucky door.

Let us say you choose the red door. Since the destinies are randomly assigned to the doors, this means there is a 1 in 4 chance that you are destined for fame and fortune, a 3 in 4 chance that you are destined for a dusty death.

Your host, who knows the doors to death, now opens the yellow door, revealing a door to death. That is part of his job. He always opens a door, but never the door to fame and fortune.

Should you now walk through the red door, the blue door or the green door?

This is a bit like the Monty Hall Problem, and I have labelled it as Monty Hall Plus, to query how generalisable the standard Monty Hall switching strategy is when the number of choices increases.

Common intuition dictates that the chance that the red door leads to fame and fortune is 1 in 4 to start with, because there were four doors to choose from, equally likely to offer the way to fame and fortune. And that’s correct. After the yellow door is opened, that probability must rise. Right? After all, once the yellow door is opened, only three doors remain – the red door, the blue door and the green door. Surely there is an equal chance that fortune beckons behind each of these. If so, the probability in each case is 1 in 3.

Now, the host opens a second door, by the same process. Let’s say this time he opens the blue door, which again he reveals to be a death trap. That leaves just two doors. So surely they both now have a 1 in 2 chance. Take your pick, stick or switch. Does it really matter?

Yes, it does, in fact.

The reason it matters, just as in the standard Monty Hall problem, is that the host knows where the doors lead. When you choose the red door, there is a 1 in 4 chance that you have won your way to fame and fortune if you stick with it. There is a 3 in 4 chance that the red door leads to death. If you have chosen the red door, and it is the lucky door, the host must open one of the doors leading to a dusty death. This is valuable information.

Before he opened the yellow door, there was a 3 in 4 chance that the lucky door was EITHER the yellow, the blue or the green door. Now he is telling you that there is a 3 in 4 chance that the lucky door is EITHER the yellow, the blue or the green door BUT it is not the yellow door. So there is a 3 in 4 chance that it is EITHER the blue or the green door. It is equally likely to be either, so there is a 3 in 8 chance that the blue door is the lucky door and a 3 in 8 chance that the green door is the lucky door. But there is a 3 in 4 chance in total that the lucky door is EITHER the blue door or the green door.

Now he opens the blue door, and introduces even more useful information. Now he is telling you that there is a 3 in 4 chance that the lucky door is EITHER the blue or the green door BUT that it is not the blue door. So there must be a 3 in 4 chance that it is the green door. So now you can stick with the red door, and have a 1 in 4 chance of avoiding a dusty death, or switch to the green door and have a 3 in 4 chance of avoiding that fate.

It is because the host knows what is behind the doors that his actions, which are constrained by the fact that he can’t open the door to fame and fortune, introduces new information. Because he can’t open the door to fame and fortune, he is increasing the probability that the other unobserved destinies include the lucky one.

If he didn’t know what lay behind the doors, he could inadvertently have opened the door to fortune, so when he does so this adds no new information save that he has randomly eliminated one of the doors. If three doors now remain, they each offer a 1 in 3 chance of avoiding a dusty death. If only two doors remain unopened, they each offer in this case a 1 in 2 chance of death or glory. So when the host is as clueless about the lucky door as you are, you might as well just flip a coin – and hope!

 

Appendix

We can apply Bayes’ Theorem to solve the Deadly Doors Problem.

D1: Host opens Red Door (Door 1).

D2: Host opens Yellow Door (Door 2).

D3: Host opens Blue Door (Door 3).

D4: Host opens Green Door (Door 4).

C1: The car is behind Red Door (Door 1).

C2: The car is behind Yellow Door (Door 2).

C3: The car is behind Blue Door (Door 3).

C4: The car is behind Green Door (Door 4).

The prior probability of the game show host finding a car behind any particular door is P(C#) = 1/4,

where P(C1) = P (C2) = P(C3) = P(C4).

Assume the contestant chooses Door 1 and the host randomly opens one of the three doors he knows the car is not behind.

The probability that he will open Door 4, P (D4), is 1/3 and the conditional probabilities given the car being behind either Door 1 or Door 2 or Door 3 are as follows.

P(D4 I C1) = 1/3 … as he is free to open Door 2, Door 3 or Door 4, as he knows the car is behind the contestant’s chosen door, Door 1. He does so randomly.

P(D4 I C4) = 0   … as he cannot open a door that a car is behind (Door 4) or the contestant’s chosen door, so he must choose either Door 2 or Door 3.

P (D4 I C3) = 1/2 … as he cannot open a door that a car is behind (Door 2) or the contestant’s chosen door (Door 1), so he must choose either Door 3 or Door 4.

So, P (C1 I D4) = P (D4 I C1). P (C1) / P (D4) = 1/3 x 1/4 / 1/3 = 1/4

Therefore, there is a 1/4 chance that the car is behind the door originally chosen by the contestant (Door 1) when Monty opens Door 4.

But P (C2 I D4) = P (D4 I C2). P (C2) / P (D4) = 1 x 1/2 / 1/3 = 3/8

P (C3 I D4) = P (D4 I C3). P (C3) / P (D4) = 1/2 x 1/4 / 1/3 = 3/8

So, there is a 3/8 chance the car is behind Door 2 and a 3/8 chance the car is behind Door 3 after Monty Hall opens Door 4. Both are greater than sticking with the original Door (probability of car = 1/4), so it is advisable to switch to either door if offered the opportunity at this point in the game.

If Monty decides to randomly open one of the remaining closed doors, it must be Door 2 or Door 3, as he is not allowed to open Door 1, the door selected by the contestant. It is equally likely that it is behind each, with a probability of 3/8 in each case. Say Door 3 is opened, the probability it is behind the remaining door doubles to 3/4 (the combined probabilities of the three doors which were not selected).

The probability that he will open Door 3, P (D3), is 1/2 and the probability that the car is behind Door 3 is 3/8. The conditional probabilities given the car being behind either Door 1 or Door 2 or Door 3 are as follows.

P(D3 I C1) = 1/2 … as he is free to open Door 2 or Door 3, as he knows the car is behind the contestant’s chosen door, Door 1. He does so randomly.

P(D3 I C3) = 0   … as he cannot open a door that a car is behind (Door 3) or the contestant’s chosen door, so he must choose Door 2.

P (D3 I C2) = 1 … as he cannot open a door that a car is behind (Door 2) or the contestant’s chosen door (Door 1), so he must choose  Door 3.

We know (shown above) that once he has opened Door 4 that P (C1) = 1/4; P(C2) = 3/8; P (C3) = 3/8.

So, P (C1 I D3) = P (D3 I C1). P (C1) / P (D3) = 1/2 x 1/4 / 1/2 = 1/4

Therefore, there is still a 1/4 chance that the car is behind the door originally chosen by the contestant (Door 1) when Monty opens Door 3.

But P (C2 I D3) = P (D3 I C2). P(C2) / P (D3) = 1 x 3/8 / 1/2 = 3/4

So once Monty Hall has opened Door 3 and Door 4, the probability of winning the car rises from 1/4 to 3/4 by switching from Door 1 (the Red Door) to Door 2 (the Yellow Door).

Puzzle Extra: After the host opens one of the remaining three doors (Door 2, Door 3, Door 4), after you have chosen Door 1, you are invited to switch. Assume you do so, to one of the remaining two doors (say Door 2 and Door 3). Say you choose to switch to Door 2. The host now opens Door 3, revealing the path to a dusty death. So fame and fortune car could now lie behind your original choice, Door 1, or your new choice, Door 2. You are invited to stay with your choice of Door 2 or switch back to Door 1. Which should you do?

Clue: By switching from Door 1 to Door 2, the chance of opening the lucky door has increased from 1/4 to 3/8. So the chance it is either Door 1 or Door 3 must be 5/8. The host now opens Door 3, revealing the way to a dusty death. So the chance that Door 1 is the lucky door now 5/8 while the chance that Door 2 is the lucky door is 3/8/ So he should switch back. Right … or wrong?

Exercise

Question 1. You are given the choice of four boxes, coloured red, orange, purple and magenta. Three are empty, a thousand pound prize is inside the fourth. You are asked to select a box. You select the red box.

I know which box contains the prize and must open one of the boxes that I know to be empty. I open the orange box and it is empty.

To maximise your chance of winning the prize, should you switch to the purple box, or switch to the magenta box, stick with the red box, or does it not matter? Why?

Question 2. Suppose in a separate experiment, you are given the choice of four boxes, coloured black, white, grey and brown. Three are empty, a thousand pound prize is inside the fourth. You are asked to select a box. You select the black box.

I do not know which box contains the prize and must open one of the boxes. I open the grey box and it is empty.

To maximise your chance of winning the prize, should you switch to the white box, switch to the brown box, stick with the black box, or does it not matter? Why?

References and Links

Monty Hall Problem. Wolfram MathWorld. http://mathworld.wolfram.com/MontyHallProblem.html

Games and Monty Hall. Statistical Ideas. http://statisticalideas.blogspot.com/2015/06/games-and-monty-hall.html#!/2015/06/games-and-monty-hall.html