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?
I liked this article, although it was a little hard keeping up with all the numbers. I have a question. The second time you explained it ( which was a little easier to get)
and monty opens 998 for you, and you have two doors left. you might have already explained this (sorry if you did) but how exactly can you tell which one has the prize, or is there no sure way of knowing? :) awesome blog guys
Thaaaank you! <3
So, Monty has 1000 doors. 1 has a prize, the other 999 have goats. You pick one (because you don't know any better). Now, Monty needs to open 998 doors leaving 1 door closed. Now the door that you initially picked either had a goat or did not (logically speaking). Hence, we can analyze the problem in two scenarios:
You had picked a door with a goat: Monty now need to open 998 doors with goats. Note that you already picked a door with a goat and he cannot open that door. Hence, he must open the doors to the remaining 998 goats (we started with 999 goats remember?). Hence, the only remaining closed door is the one with the prize.
You picked a door with a prize: Monty needs to open 998 doors with goats. He picks any 998 of the remaining 999 doors since they all have goats. The remaining closed door still has a goat.
Notice that in the first scenario, if you switch, you win! In the second scenario, if you switch, you lose! What are the chances of you coming upon the first scenario (i.e. randomly picking a goat on your first try)? The chances are equal to the number of goats divided by the total number of doors. 999 goats. 1000 doors. Hence, with 99.9% probability, you will pick a door with a goat. If you switched after the picking this door, you will win.
You will pick the door with the prize on the first try with a 1 in 1000 chance (like a lottery). If you switched doors after picking this door, you lose.
Hence, overall, you have a 99.9% chance of winning, if you always switched your door. Conversely, you have a 0.1% chance of winning if you stayed with your pick and didn't switch. You never know for sure if you will win. But you can guess quite accurately.