Understanding the Monty Hall Problem
Three Simple Choices for Obtaining Perspective
Let’s Make a Deal ended its initial run 40 years ago after more than a decade run as a popular TV game show in the United States. Its co-creator, Monty Hall was immortalized 25 years ago when the Monty Hall Problem was first coined in Parade magazine. The problem had previously been titled the Three Prisoners problem 30 years earlier in Scientific American, but goats and Canadians have proven more popular than death and inmates…
The problem is simple. An individual has chosen one of three doors knowing that behind only one of those doors is the item they want. On Let’s Make a Deal this was typically a new car (our prisoners were hoping for a pardon). The host (or warden) then opens one of the unselected doors revealing one of the two unwanted items, goats or death. This reveal is NOT done at random as both the host and warden know where the winning item is.
On Let’s Make a Deal, Monty Hall would then ask the contestant if they would like to change their choice of door to the other remaining door. Should they?
Surprisingly, the answer is yes.
Changing selections results in getting the car two-thirds of the time. This is quite counter-intuitive to most people who believe the odds have just changed to 50–50 for the remaining two doors.
If you want a detailed breakdown, Wikipedia has a fairly lengthy article available here. But for a more concise perspective, read on.
The Solution behind Door #1 — Seeing is believing
If you are struggling with this concept, try it. Find a friend and play out the game or build a simulation in excel or another easy language of your choice. Assuming your sample size is high enough and your code is right, you will quickly learn that keeping your original choice results in getting the car only one-third of the time.
The Solution behind Door #2 — Use a Little Math
Bayes Theorem can be quite helpful here. Bayes allows us to calculate how the odds change when new information is revealed. Only, in this case, it doesn’t!
Running the Bayes equation, we learn that when we start with a 1/3 chance and add information that has independent odds of 50/50, our new probability remains 1/3. We gained no additional certainty for our original choice!
The key to properly modeling both #1 & #2 is that Monty’s reveal is NOT random. In Bayes equation, his ‘choice’ has a 100% probability. If is didn’t, our odds would be 50/50 OR the game would have already ended (1/3 of the time).
The Solution behind Door #3- Your Inner Contrarian
Assume for a second that you chose Door #1. Monty revealed a goat behind Door #2 and you are now left with the choice of whether to switch to Door #3. Now stop that!
Change your perspective, assume you picked the opposite. Not Door #3, the opposite would have been BOTH doors #2&3. Clearly choosing two of the three doors would have given you a 2/3 chance. Monty’s reveal only acts to clean away the door that clearly wasn’t the car.
The probability changes because Monty knows which door contains the prize and because he purposely chooses a door that does not. When he offers to let you change doors, he is giving you the option to select two rather than just one.
Final Perspective
The Monty Hall Problem is a great case study of a veridical paradox and the implications of dependent probability. It teaches that us that misclassifying new information can lead you to a false understanding.
And far better to learn this lesson with goats than death…
Quintessentially is an article format created by Corsair’s Institute to increase the reader’s comprehension of key concepts in a quick and engaging fashion. For more articles from Perspective, Quintessentially — click here.
For more information on the author visit his profile on LinkedIN — George Earl
