Hey Steemians,

I wanted to present to you a somewhat paradoxical brain teaser or riddle that is known by the name "The Monty Hall problem" in the anglo-saxon hemisphere but also exists in several other representations in which it may have been posed independently by different authors (amongst them Joseph Bertrand 1889, Martin Gardner 1959 and others).

Let me also preface that this problem was once discussed and solved in a column of the arguably most intelligent person in the world, Marilyn vos Savant, in Parade Magzine where it evoked an outraged debate about the correctness of its result even amongst the most educated of readers. The riddle seems to be so cognitively deceptive that even PhDs and Nobel Prize winners are reported to not only have been fooled by it but also to insist on their wrongness.

Wikipedia

The problem

Suppose you are a contestant of a game show and have the possibility to win a car by picking one of three closed doors, one of which contains the car and each of the other two a goat.
After having made your decision the host opens a door which you didn't pick and which does not contain the car (thereby revealing one of the goats).
Now you are given the option to switch by giving up your initial door and picking the other still unopened one.

The riddle is composed of the two related questions:

• Should you switch to improve your chances?
• What is your probability of winning if you switch?

What do you think?
Statistically speaking, many of you will initially reply that switching does not affect your winning chance at all because the result still depends on your initial choice which had a 1/3 winning probability.
Or that you should switch, because this improves your chances from 1/3 to 1/2 because now you only have 2 doors to choose from.

I'm sorry to disappoint you, while each of the replies contain each one aspect of the truth/explanation, they are stictly speaking both wrong

The Answer

Let me first give you the objectively correct answer, which has been verified both theoretically and experimentally by statistical simulation:

• Yes you should switch
• Switching will improve your winning chance not to 1/2 but to 2/3!

It's okay if you don't believe me though, as I said this riddle is particularly deceptive. But we can dissect it systematically using a tool called probability theory:

The math

First of all let's agree that you have a 1/3 chance to pick the car initially and if you don't switch even the host revealing one goat will not change anything about it.

The interesting case is the second one:
If you do switch you might argue that this should be equivalent to picking from 2 doors with 1 car, 1 goat and uniform distribution of car and goat. Then your winning chance would be indeed 1/2.
But it is not equivalent! The catch is that your initial state after the reveal is still dependent on the probability distribution at the time of your first choice and since "switching" is a deterministic procedure in which you by picking the other door "take the complement of your initial state". And when one stochastical event is conditional on something, then so is its complement.
It is also worth to note that the game host -to your advantage- deliberately adds information to the game because he always reveals a goat and never the car.

We will denote the event of winning the car in your second pick (by switching) by 'A'

So lets assume you picked the car initially (1/3 probability). We denote this event by B1 and it has a probability of P(B1) = 1/3 to occur in your first pick.

Switching will then always cause you to pick a goat and lose, so under the condition of having picked the car, switching has a winning chance of 0.
We write P(A|B1)=0 for the probability for A to happen under the condition B1.

Suppose you picked a goat initially (2/3 probability). We denote this event by B2 and it has a probability of P(B2) = 2/3 to occur in your first pick.
Because you have picked one goat and the host will eliminate the other one, switching will always result in you getting the car because it will be the only thing left behind the other unopened door. Under the condition of having picked a goat, switching has a winning chance of 1.
We write P(A|B2)=1 for the probability for A to happen under the condition B2.

B1 and B2 form a so-called partition of the probability space that is associated with the first pick. This means in your first pick exactly either B1 or B2 will happen. We may then employ the law of total probability (Wiki) to obtain the total probability for A to happen (i.e. to win):

P(A) = P(B1)xP(A|B1) + P(B2)xP(A|B2) = (1/3)x0+(2/3)x1 = 2/3

Alternatively, as we have noted above, you can observe that if you switch, then you will win the car if and only if you chose a goat in your first pick. Since you have a 2/3 chance to pick a goat initially this will also yield the desired answer.

Did any of this convince you? Be sure to let me know in the comments!

Sources:

• https://en.wikipedia.org/wiki/Monty_Hall_problem
• vos Savant, Marilyn (17 February 1991a). "Ask Marilyn". Parade Magazine: 12.
• Gardner, Martin (October 1959a). "Mathematical Games". Scientific American: 180–182. Reprinted in The Second Scientific American Book of Mathematical Puzzles and Diversions