The Monty Hall Problem

Sources and figures in this page are adapted from:  

        1-Workshop week 5 - Lecturer: Dr Stephen Woodcock

        2-Keith Ellis' website: "The Monty Problem'

What is the Monty Hall Problem about?

At the UTS workshop week 5, Woodcock (2016) introduces the story of the Monty Hall problem in which it is considered as a famous problem in terms of decision making: Switch to a new decision or stick to the original one when dealing with a choice for a hope with a better outcome.

The following is the problem about:

The problem is named from the name of the host of the American game show 'Let’s Make a Deal' - Monty Hall.

In the show, the game consists of three doors. Behind two of these doors, there is a goat [or something of non-prize]. Behind the rest door, there is a prize. [in this scenario, there are two non-prizes and one prize]

The game is described, for example you are on the Monty Hall show as a player, as follows: 

- You are given your first selection to choose one of the three doors

- The host, Monty Hall, then will open one of the remaining doors, and announce that it is [always] a goat or a non-prize.

- You are then offered two options: to stick with the original choice or to change to the other unopened door.

Should you change the door, or stick to your original choice, or does it make no difference?

Obviously, there will be four outcomes:

    Stick and win

    Stick and lose

    Change and win

    Change and lose

Interestingly and amazingly for the solution, Woodcock (2016) provides the information that on average, you are twice as likely to win the prize by changing after your first selection rather than by sticking with it. In detail, Woodcock explains the solution as below:

1-Although there are two options left, it is not a 50% /50% chance between these two.

2-Importantly and more simply, the only way you win by sticking is if you were correct with your first selection, as there were 3 equally likely doors initially, this occurs with probability 1/3. Similarly and visually with an image, Ellis (1995) exhibits the probability of three doors in the following diagram.

Further, in confronting these three doors, Ellis indicates that the probability of one door hiding a prize is 1/3, and the probability of non-prize-hiding doors is 2/3.  It means that behind any two doors considered together, the probability of being a prize is 2/3; hence, the pictures of a modified diagram from the above image, and an explained snapshot are offered as follows (Ellis, 1995).

3-If you were wrong originally (probability 2/3) then you will win by switching. The reason is that the initial choice has only one chance in three of being right, and therefore the remaining door has two chances in three to win (Ellis, 1995; Woodcock, 2016). See an image below.

Why does the Monty Hall Problem Matter?

How is the Monty Hall Problem applied in Practice?

References

Ellis, K.M. (1995). The Monty Hall Problem. Retrieved 8 May 2015 from http://www.montyhallproblem.com/

Woodcock, S. (2016). Lecture notes PowerPoint-UTS workshop week 5. Autumn 2016

Useful links

Relevant links  1  2  3  4  5  6  7  8  9