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The Deadly Doors Problem – Monty Hall Plus!

November 27, 2014

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 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?

There once was a problem like this, involving three doors, a car, two goats and a game show host called Monty Hall. This has come to be known as the ‘Monty Hall Problem’, and is best known by probability tutors for the number of students who know the answer to the problem without knowing why, but pretending that they do. This is because it featured in a movie called ‘21’, featuring Kevin Spacey as a tutor and some blackjack genius as a student who solves it.

So let me explain.

Intuition dictates that the chance that the red door leads to fame and fortune is 1 in 4 to start with, but must be more than this after the yellow door is opened. 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.

The host now 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 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 is forced to open a deathly door. He knows that. You know that. So he is introducing useful new information into the game.

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 valuable new information. Because he can’t, he is forced to point to a door which is not the door to fortune, 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 a 1 in 2 chance of death or glory. So you might as well just flip a coin – and hope!

 

Appendix

We can apply Bayes’ Theorem to solve this.

D1: Monty Hall opens Red Door (Door 1).

D2: Monty Hall opens Yellow Door (Door 2).

D3: Monty Hall opens Blue Door (Door 3).

D4: Monty Hall 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 Monty Hall 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 Monty Hall 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

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).

 

Further Reading and Links

https://selectnetworks.net

From → Probability, Puzzles

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