2 Goats 1 Prize
Consider this scenario:
You have almost won at a game show! Your host, Monty, gives you one final task. He lifts the curtains to reveal three closed doors, labeled A, B, and C. He informs you that behind the doors, in some order, there are two goats and a grand prize. The doors are identical and there is no way to tell what is behind them. He asks you to pick a door, which you do, picking one at random (let’s say A). Monty turns around and, in a fit of generosity, tells you that he will help you out by opening one of the unselected doors that contains a goat (let’s say B). Having taken one of the doors out of the running, he now gives you a choice. “Would you like to switch your selection to C or stay with A?”, he asks.
As you stand there, what should you do to have the best chance of winning the grand prize? Is there a systematic way to find out?
You would be best of if you chose to always switch. If you always switched, you would win 2 out of 3 times.
If you said that it doesn’t make a difference what you choose, don’t fret! This problem is not intuitive. At the very beginning, it made no difference which door you picked. Each door was just as likely as any of the others to have a goat behind it. There were three doors, one with the prize you wanted. Hence, the odds of you picking the door which had the grand prize was 1/3. Then things get interesting. The host shows you a door with a goat. Note that he must show you a goat. He cannot accidentally open a door that contains the prize. He very deliberately opens a door that contains a goat. This has given you information. The problem has changed. Let us do a case analysis to show this:
- Let us assume that the door you picked indeed does have the prize behind it. The other two doors contain goats. Monty opens one of the remaining doors (it doesn’t matter which; they both hide goats). At this point, if you stay, you win the prize. If you switch, you lose.
- Let us assume that the door you pick has a goat behind it. One of the other two doors contains a goat. Monty is forced to open this door. This implies that the remaining unopened door must contain the prize. At this point, if you stay, you get a goat. If you switch, you win.
There is a 1/3 chance that the first case occurs and a 2/3 chance that the second case occurs because, as discussed before, the odds of initially picking a door with the prize is 1/3. If you always stayed with your choice, you win in the first case and lose in the second case. Hence, there is a 1/3 chance of you winning and a 2/3 chance of you losing. If you always switched, there is now a 1/3 chance of you losing (if you initially managed to pick the winning door) and a 2/3 chance of you winning (if you managed to pick a door with a goat behind it). Hence, you stand a better chance of winning if you always switched.
This problem is a well known problem in discrete mathematics to illustrate conditional probabilities and how they change in ways that aren’t intuitive to someone who isn’t formally trained in statistics and probability theory. It is known as the Monty Hall problem.
To make the problem more understandable intuitively, let us consider the case with 1000 doors with 999 goats and 1 grand prize. Monty lets you make an initial pick and then open 998 doors to show you goats. Should you switch or stay? That one remaining door has a very high chance of containing the grand prize. After all, every other door was opened instead of it. As you can see, this illustrates why switching is better. The 3 door situation is just 997 doors fewer than the above mentioned situation. Can you evaluate your chances of winning if you switched in the 1000 door scenario? How about for an ‘n’ door scenario for any ‘n’ greater than 2?