statistics/odds problem?

Question

Ok this is a question that i cannot figure out. Say there are 3 doors and a prize behind 1 of the doors and u get to choose any of the 3 doors. lets say u choose door 1, your odds are 1/3 so 33% of getting it right. but what if i open up door 2 and show u there is nothing behind it, then i give u the option of switching your original choice (1) to door 3. Mathematically this would not make your odds 50% but 66% to get it right. Is this correct?

The best way i can explain it is this. Say there are 1 million doors and i say to choose any door in between. you choose door 100. i then open ever door to show it's empty except door 100 and door 257,652. now are your odds still 50% between the 2 doors? do you really think you picked the right door on the very first try. I believe your odds of switching doors if given the option would go up somewhere around 99.9% in this case. Can anyone back me up on this or better explain if im right or wrong, preferably in case 1 with the 3 doors. thanks!

Answer 1

This is the old "Monty Hall" problem, because the sceario is supposed to take place on the show "Let's Make a Deal".

People still debate it to this day, but the fact of the matter is yes, your odds DO change. This is because Monty 1) always knows what's behind the doors, and 2) will ALWAYS open a door that is empty (or as it's usually described, has one of the two "bad prizes" just like the real show; usually a goat) after you pick another door. This is known as a "conditional probability" because odds depend on certain other conditions.

Think of the proabilities this way. You have a door with a brand new car behind it, one door with a goat (G1) and another door with a goat (G2). There are only three different things that can happen:

1) You pick the door with G1. This forces Monty to open the door with G2.
2) You pick the door with G2. This forces Monty to open the door with G1.
3) You pick the door with the car. Monty then opens either G1 or G2.

Your chances of winning if you stay with your original choice is 1/3, because only in the third analogy would you win. However, there is a 2/3 probability that the situation at hand is either #1 or #3, in which case switching would make you a winner. In short, there's a 2/3 chance of choosing a BAD door initially, in which case the switching choice will be a win.

The million-door analogy you give does indeed hold true. After Monty opens the other 999,998 doors, that means either 1) you actually picked the correct door, and he decided to open all the doors except for #257,652, or 2) you did NOT pick the correct door, and Monty Hall is opening all doors except for the one you picked and the one with the prize. Clearly, the odds of #1 are extremely slim, so #2 is more likely what's going on. What makes the original problem misleading is that with 3 doors, opening "all doors except 2" means opening just one door.

On a side note, Monty Hall himself has said that the probelm is amusing but wouldn't work in the real world, because the person simply isn't given the option to switch doors (or in the 80s version of the show, curtains). They do however reveal one of the bad prizes, just to add to the suspense and comedy.

Answer 2

This is a classic problem that shows how conditional probabilities are usually not intuitive. Our strategy for this problem is to pick a door, then switch after one empty door is opened.

Lets say that the prize is in door #1.

1.........2..........3
(Prize) (Empty) (Empty)

If we pick door #1 and switch, we have a 0% chance of winning the prize.

If we pick door #2, that means that door #3 is opened, we switch to door #1 and win the prize.

If we pick door #3, that means that door #2 is opened, we switch to door #1 and win the prize.

Because these are the only 3 possibilities, and they each occur with equal probability (1/3), that means we will win the prize 2/3 times, or 66% of the time.

Answer 3

There is a 2/3 chance of being right if you switch. It's not purely a random problem because the knowledge of the person opening the doors must be taken into account. That person will never open the door that has the prize. So it's not a purely random problem.

Check out the site below. I found the probability/decision tree to be the most helpful.

Answer 4

Yes. You have 2/3 chance of getting it right if you switch everytime. Simply stated, this is because you have a 2/3 chance of picking the wrong one initially. So if you switch, you will turn this into success.