Monty Hall or the False Perception of Probability
Thinking in probabilities in uncertain market environments is an absolute necessity for decision makers to run their course though, this is not always as straight as it seems and could even result in missing out of opportunities.
In this article, we give an example how easily this can happen.
The Game
Did you ever hear about the Monty Hall problem?
Most probably not, unless you are more involved in statistical problems.
Anyway, Monty Hall was a TV game show where a player was faced with 3 doors and had to choose one of them. Behind one door, there was a car which the player could win provided that the right door was selected. Behind the other two doors, there was nothing to win (but a symbolic goat).
Every door had the same chance to hide the car. Hence, with 3 doors in the play for each door to hide the car the probability was 1/3 or 33.33%.
Now, after a door was chosen by the player one of the remaining doors was opened (this opened door had to be empty of course, otherwise the game would be over). And with 2 doors left, the player was asked if she wanted to stay with her decision or if she wanted to switch to the other door.
Well, how should the player decide?
Intuitively, with 2 doors left one would say that the chances are 50 : 50 for each of the doors that the car is hidden behind it. Keeping the initially selected door would have a winning probability of 50% and switching to the other remaining door also would have a winning probability of 50% (at least this would have been my first notion).
So, why to bother?!
Because, unfortunately staying with the initial decision gives us a winning chance of 1/3 (or 33.33%). And, switching the door increases the chances to 2/3 (or 66.66%) for winning.
In other words, not getting the situation right in terms of probabilities deprived us from higher winning chances.
Why 2/3 and not 50 : 50?
Before we bend our heads around this logic for too long (as I did), just let’s simulate this.
First of all, we create 1 million incidents of three doors with one door hiding a car and the other two remaining doors just having a goat.
Here are the first few events of this simulation:
- column 1 … door 1
- column 2 … door 2
- column 3 … door 3
- each row … one single situation in which behind one door a car is hidden (represented by “1”) and the other two doors are empty (represented by “0”)
As can be seen, the probability of the car hidden behind one of the doors is approximately 1/3 for each of the doors in this simulation:
At this stage, the player chooses door 1 (with a chance to win of 33.33%).
With this choice made by the player, one of the other 2 doors is opened (which is an empty one of course). In the following, we simulate this process:
As a result, two closed doors remain in the game, i.e. either a combination of door 1 + door 2 or door 1 + door 3. Here are a few results of this stage of the game:
With those two doors left (car_switch[, 1] … door 1, car_switch[, 2] … remaining door 2 or door 3), should the player keep her initial decision or should she switch?
In order to answer this question, we have to take the circumstances into account. Just because 2 doors are remaining in the game does not mean that the car was freshly distributed behind those two doors (in this case, we would have a new probability for each door to hide the car of 50%).
No, we rather have the old distribution which is in 1/3 of the cases the car is still hidden behind door 1. The old distribution stays but new information is washed in, i.e. the person which opened the door knew that this exact door is empty.
And this old distribution combined with the new information leaves us with the following options:
- The car is really hidden behind door 1 which the player initially chose. This is true for 1/3 of the cases. And it does not matter which of the remaining doors is opened (door 2 or door 3) as both of them are empty.
- And in the other 2/3 of the cases? The car is really hidden either behind door 2 or behind door 3. In this case, just one door can be opened, i.e. the empty one and in ANY CASE the door with the hidden car is kept closed and is therefore staying in the game. Hence, in 1/3 of the cases door 2 stays closed and in the game and in 1/3 of the cases door 3 stays closed and in the game. Overall, in 2/3 of the cases the car will be hidden behind the remaining closed door.
In other words, staying with door 1 as initially selected keeps your winning probability at 1/3 or 33.33%. But switching to the remaining door (i.e. door 2 or 3 in our assumption) will increase the winning probability to 2/3 or 66.66%.
Let’s see what our simulation tells us:
The simulation finally confirms what we tried to derive logically. Switching to the other door instead of keeping the initially chosen door increases the chances to win from 1/3 to 2/3 respectively from 33.33% to 66.66%.
This is quite a difference!
Summary
The incorporation of probabilistic thinking in uncertain (market) environments is an absolute necessity. This provides decision makers with a certain guideline how and where to move in insecure frameworks.
Though, embedding probability is not always as straight as it seems.
Sometimes, subtle differences can lead to the missing out of opportunities or the underestimation of risk positions.
References
For a more detailed overview, see:
Monty Hall and the ‘Leibniz Illusion’ by Steven Tijms/ December 13, 2022; published in Chance — Using Data to Advance Science, Education and Society
