In an old game show titled Let’s Make a Deal, host Monty Hall presented contestants with three doors.
One of the doors contained a prize while the other two did not.
Monty would ask the contestant to choose which door they believed contained the prize.
After the contestant selected a door, Monty would then open one door that did not contain the prize.
The prize remained in one of the two unopened doors.
Monty would then ask the contestant if they would like to switch doors.
While it might be hard to believe, it turns out that switching doors at this point in the game actually gives you a higher probability of winning.
Keep reading to find out why!
The Monty Hall Problem Explained Visually
To illustrate why switching doors gives you a higher probability of winning, consider the following scenarios where you pick door 1 first.
Scenario 1: You pick door 1 and the prize is actually behind door 1.
In this case, Monty will open either door 2 or 3 and show you that nothing is behind one of those doors. If you stay with door 1, you win.
Scenario 2: You pick door 1 and the prize is actually behind door 2.
In this case, Monty must open door 3 and show you that nothing is behind it. If you stay with door 1, you lose.
Scenario 3: You pick door 1 and the prize is actually behind door 3.
In this case, Monty must open door 2 and show you that nothing is behind it. If you stay with door 1, you lose.
These are all of the possible outcomes if you pick door 1. Notice that if you stay with door 1, you only win one-third of the time. But if you switch, you win two-thirds of the time.
The following table summarizes all of the possible scenarios in this game show along with the outcomes associated with staying and switching:
From the table we can see that you win 33% of the time when you stay, but you win 66% of the time when you switch.
Thus, switching doors increases the probability that you will win the prize.
Although this doesn’t seem to make sense intuitively, the math doesn’t lie.
Additional Resources
Check out this video for a nice explanation of the Monty Hall Problem by the Numberphile YouTube channel:
And refer to this interactive online simulator to simulate the Monty Hall Problem yourself: Monty Hall Problem Simulator.
Wonderful explaination! Can I please print out this article?
May I ask you for a clarification please? In your neat little table you say that you have laid out all the possible scenarios, and so you have, but you have not given each scenario the same weighting. On lines 1,5 and 9 you have lumped 2 scenarios together yet have only given 1 result. If the table is re-written so that lines 1,5 and 9 are expanded to treat the opening of different doors as being different scenarios (which I think they are) , the table looks rather different. It would have 12 lines and the “result if contestant stays/switches” columns would both be 50% / 50%. Help me please – am I talking rubbish?
Hi Anthony…You’re not talking rubbish at all. You’ve noticed something important about the way the table is weighted.
If you split the cases where the host has a choice into separate lines, it’s true that you’ll get more total rows. But those rows don’t all have the same probability. That’s the catch that often leads people to think the result is 50/50.
Let’s walk through the probabilities if you pick Door A.
1. Car is behind A (probability 1/3).
The host can open B or C, each with a probability of 1/2.
* Car A, host opens B: probability 1/6
* Car A, host opens C: probability 1/6
2. Car is behind B (probability 1/3).
The host must open C.
* Car B, host opens C: probability 1/3
3. Car is behind C (probability 1/3).
The host must open B.
* Car C, host opens B: probability 1/3
Now, if the host opens B, the possible cases are:
* Car A, host opened B: probability 1/6
* Car C, host opened B: probability 1/3
Normalizing gives:
* Probability car is behind A if host opens B: 1/3
* Probability car is behind C if host opens B: 2/3
The reason you don’t get 50/50 is that when you initially choose correctly, the “split” cases each have half the probability of the forced cases where you initially choose wrong. If you weight all rows equally, you flatten this difference and the real advantage of switching disappears.
The quick way to remember it is: switching wins whenever your first choice was wrong, and that happens two-thirds of the time under the standard Monty Hall rules.