This is a classic!

Consider you’re on the game show Let’s Make a Deal!. The game is simple: there are 3 doors presented to you. One of the them has a fabulous prize behind it, while the other two hide nothing but shattered dreams.

When you initially select a door, the host (Monty Hall) will reveal one of the doors which hides your shattered dreams. Because Monty knows which doors conceal what, he will always reveal one door that is empty, while always leaving 1 empty and 1 prize door.

* If you initially select the winning door, Monty will reveal an empty door.
* If you initially select an empty door, Monty will reveal the other empty door.

Left with two doors (one with the prize, the other with nothing), you have the opportunity to either stick with your initial selection, or vacillate entirely and change your mind, selecting the other remaining door.

So the question is simple: Should you stick or change?

The answer, quite counter-intuitively, is that you should most definitely change.

WHAT?? How does this make any sense? If there are two doors left, surely the probability of winning or losing is equivalent, right?

Nope. But why?

Let’s look at this in steps.

Initially (unconditioned), the probability of selecting any given door is as follows:

L1 – 1/3
L2 – 1/3
W – 1/3

Obviously, with 3 doors, you have a 1/3 chance of selecting any particular one.

Now, let’s remove one of the loser doors at random:

L2 – ?
W – ?

Instinctively, it feels like the probability distribution here should be 1/2 – 1/2, right? Ah, but such is not the case!

This is because, in effect, the probability of choosing the correct door remains 1/3 through the entire game. Even when a randomly selected losing door is revealed, there is still a 2/3 probability that the contestant has chosen a losing door. However, because one of the losing doors has been removed, the player can reverse the odds, for there is an equal 2/3 probability that the remaining, unselected door conceals the prize.

Example:

:: Test 1 ::

* Select L1 (1/3): Switch: Win
* Select L2 (1/3): Switch: Win
* Select W (1/3): Switch: Lose

:: Test 2 ::

* Select L1 (1/3): Stick: Lose
* Select L2 (1/3): Stick: Lose
* Select W (1/3): Stick: Win

Therefore, you should clearly always switch, given that you have a 2/3 probability of wining, rather than the 1/3 chance that your initial, unchanged selection is correct.

This is a bit easier to conceptualize if you drastically increase the number of doors. For our example, let’s say there are 50 doors, only one of which conceals a prize. Unconditioned, you have a 1/50 chance of selecting the correct door. So let’s say that you select a door, and Monty Hall runs around opening 48 losing doors. Faced with the remaining 2 doors, would you still feel that your odds of winning with your original selection are 50%? Or, given that choosing the winning door out of 50 is a rare occurrence, would you not rather switch without hesitation, knowing that the greater probability (49/50) remains the same, and is now wholly contained within the other remaining door that you did not initially choose?

Select L1 (1/50): Switch: Win
Select L2 (1/50): Switch: Win
Select L3 (1/50): Switch: Win
Select L4 (1/50): Switch: Win
…
Select L49(1/50): Switch: Win
Select W (1/50): Switch: Lose
———————————————-

Select L1 (1/50): Stick: Lose
Select L2 (1/50): Stick: Lose
Select L3 (1/50): Stick: Lose
Select L4 (1/50): Stick: Lose
…
Select L49(1/50): Stick: Lose
Select W (1/50): Stick: Win

Finally, a super-simple way to think about this is by treating the initially removed door and the final, selected door as one unit. When you stick with the initial selection, it’s equivalent (in terms of probability, at least) to saying that you would rather have a shot at choosing from one door (1/3 probability) than from two (2/3 probability). If you switch, however, you gain the benefit of the combined probability of 2 doors, increasing your chances of winning dramatically.