Solve the Monty Hall Problem using Logic and Mathematics


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The Monty Hall Paradox: Image by bzo

Monty Hall, the host of “Let’s Make a Deal”, asks a contestant to choose the one door to a valuable prize, rejecting the other two doors leading to junk. The host then opens one of the rejected doors, revealing a junk prize. The “Monty Hall problem” now is: should the contestant stay with the original choice, or switch to the other unopened door?

Setting the Stage for the Monty Hall Problem

Let’s label the doors ‘A’, ‘B’ and ‘C’, even though the game may have called them ‘1’, ‘2’ and ‘3’. We will see why in the next sub-section.

For a science-oriented experiment, let’s ensure that Monty Hall has no prior knowledge about what is hidden behind which door. That way, he cannot reveal anything to an observant contestant.

While we’re stipulating conditions, let’s also say that there is neither cheating nor ESP: the contestant has no prior knowledge, supernatural insight or extra assistance to help with the selection process.

The final condition is that Monty Hall and his team do not cheat by moving the prizes around after the contestant has made a choice… especially not after the second choice!

To Simplify Discussing the Monty Hall Problem

For the sake of this discussion, the contestant will always choose door ‘A’ first. This convention simplifies the discussion, because ‘A’ really means “the door that the contestant first chose.” Monty Hall will then reveal a door with a junk prize.

Probability Theory, Logic and the Monty Hall Problem

It is obvious that the contestant has a one-third chance of making a correct guess at the start. The Monty Hall solution, however, depends on whether the probability changes after one junk prize has been revealed.

So, the contestant chose door ‘A’. Monty Hall has revealed door ‘C’ with its junk prize. What should the contestant do to maximize the chances of winning the valuable prize? Can math or logic be of more help?

Two Incorrect Approaches to the Monty Hall Problem using Logic

Let’s first approach the Monty Hall problem using logic and reasoning, saving the mathematics for later.

An Immobile Prize Leaves the Odds the Same

The prize did not move, so each door, and the contestant, keep the one-third chance of being correct. According to logic, there is nothing to gain by switching, since each remaining door retains its one-third chance.

The conclusion is wrong. Although the prize did not move, the contestant now has more information: one “losing” door has been eliminated.

The Odds Have Become Even

Now the contestant knows that the prize is behind one of the two remaining doors. Therefore, there is a fifty-fifty chance, or “even odds”, that either door has the valuable prize. Therefore, according to logic, there is no advantage to switching.

This reasoning is also faulty. It treats the situation as if the contestant had not made the first choice at all. In truth, the fact that the contestant has made one choice means that even more information is now available.

A similar fallacy that leads to the correct behaviour accepts the false notion that the odds have become even. A correct conclusion from that fallacy is “switch because now the contestant has a 1/2 rather than 1/3 chance of being correct”.

Correctly Solving the Monty Hall Problem using Logic

The contestant choses door ‘A.’ Monty Hall cannot reveal door ‘A’ without ending the game prematurely. He also cannot reveal the valuable prize, so when Monty Hall opens door ‘C’, it must reveal junk. What choices does the game’s host have?

There are two situations. If the contestant chose ‘A’ correctly the first time, then either ‘B’ or ‘C’ became equally available. The contestant had 1/3 odds that ‘A’ had the valuable prize. As was stated before: sticking to the original choice does not change that chance of success. Thus, the strategy of “always switch” inherits that 1/3 chance… of choosing incorrectly.

In the second situation, the contestant’s original choice of ‘A’ led to junk. If door ‘B’ had the prize, then Monty Hall had no choice but to open door ‘C’. The contestant had a 2/3 chance that door ‘A’ had junk, but the consequence is a 100% certainty that the only other “junk” door would be opened. Therefore, there is a 2/3 chance that switching results in the correct door.

Another correct way of understanding the situation is that the original choice has a 1/3 chance of success. Suppose Monty Hall were to then ask the contestant, “Would you rather switch to select the other two doors, and give me back the prize you don’t want”? A wise contestant would switch to the two other doors, and have the 2/3 chance of success. This is essentially the choice Monty Hall does give the contestant, by opening the non-selected junk prize door.

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