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Three door problem is a remarkable probability puzzle or a brain teaser, especially due to its counterintuitive solution. Sometimes it is known as Monty Hall problem according to the host of a popular US television game show Let's make a deal.

**The host of the game show, Monty Hall, offers a choice of three doors to a competitor.** There is *a car* as a main prize behind one of the doors and *two goats* as a consolation prize behind the remaining doors. The task of the competitor is to pick one door only.

After that Monty Hall steps in and opens one of the remaining two doors with a goat. We assume that the host is perfectly honest and that he always opens the door with a goat behind it. He lets the competitor see. **The point of the game is that now the host offers to the competitor that he can (but does not have to) change his initial door choice.**

Let us be more specific for a better understanding and mark the doors by numbers 1, 2 and 3. The competitor chooses e.g. door number 1. Then Monty Hall opens the door number 3 and lets the competitor see that there is a goat behind it. Now he asks the competitor, whether he wants to stay with the door #1 or whether he wishes to change his mind and choose the door #2 instead. The competitor has no other clue at his disposal.

The question is: *"Shall he stay with his first choice or change his initial choice or it does not matter at all?"*

Do you think that there are only two doors left and thus the odds are 50:50 – in other words it does not matter whether the competitor keeps his choice or changes it? Then you think in a way as the most of population, however this answer is not correct. How come?

Although it may look absurd at the first sight, **the competitor's best decision is to change the doors.** How do we arrive at this conclusion?

We have an idea that, on the beginning of the game, there is a 1/3 probability that the competitor chooses the right door (there is one car behind one door only). On the contrary there is a 2/3 probability that the competitor picks a bad door.

Again we may be more specific and use our previous example: the probability of the car being behind the door #1, which is chosen by our competitor, is 1/3. The probability of the car being found behind the doors #2 or #3 is 2/3. This is very important for an optimum decision during the next course of game! And it also partly indicates a correct solution.

As we know, now the host comes into play and shows to the competitor that there is a goat behind the door #3 and he asks him the question: will you stay with your initial choice (door #1) or will you change your mind and go for the door #2. We can feel that **the game consists of two following steps** and that is *crucial for a good assessment of the puzzle* (conditional probability).

**The best decisions of the competitor is a change!** The fact that the host shows to the competitor *another door without the car behind it does not reduce this puzzle to two equal choices.* The probability that the car could be (from the very beginning of the game) behind the doors #2 or #3 does not disappear by that! Staying with the initial door (#1) means staying with the probability 1/3. Changing the decision means increasing the probability of winning to 2/3 and therefore it is the best decision. Does it still seem unbelievable?

The right solution of the Monty Hall problem can be better understood by **adding more doors.** With three doors only the solution may dive and the differences may not be apparent.

Therefore *let us try to increase the number of the doors to one hundred.* Other rules are the same. Once again the competitor shall choose only one door and there is one car behind one door only. The probability of winning the car, that is of choosing the right door, is 1/100. There are 99 doors remaining. So far so clear. Monty Hall comes into game and opens 98 doors with a goat in front of the sight of the competitor.

Now Monty Hall asks the competitor, whether he wishes to keep his choice or to change it. **Can you feel the difference compared to the three doors only?** You have to admit that you would have to be extremely lucky to pick the right door on the beginning of the game. Since *it is much more probable that the car is found behind of the 99 doors,* despite or therefore the host showed you 98 doors. Thus it is best to change the door! (as the solution of the puzzle does not shrink to 50:50 neither with three nor with one hundred doors).

You may argue that the host offers to change the door only if you were successful. Two things. (1) The puzzle assumes that the host is honest. (2) Let us replace the host with a machine that randomly shows all remaining doors except for one door – left for your choice. The essence of the correct solution of this puzzle not only remains the same, but it is highlighted by that – the best decision is to change the door.

Note: the correct solution of the three door problem can be calculated or verified precisely by conditional probability (there it matters what happened "before" a specific event). The three door problem is sometimes called as Monty Hall paradox, however it is not an appropriate label as the correct solution can be reached by means of math. The solution to this puzzle is just a bit counterintuitive.

Where to go next:

→ Probability Puzzle of Chevalier de Mere (17^{th} century);

→ Two beagles Probability Riddle;

→ All Articles on Probability in Gambling and Betting.

Based on the original Czech article: Pravděpodobnostní problém tří dveří (Monty Hall).

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