3 Doors, A Car and 2 Goats - Test your math and reasoning skills

[Sgt_Kickaxe]

Pretend that in front of you are 3 doors. Behind one door is a car and behind the other two doors are goats. (You want the car)

- Go ahead and pick a door. Remember the door you picked. You now have a 1 in 3 chance (33%).

- One of the other two doors opens and it's a goat. You now have a 1 in 2 chance of getting the car (50%)

- Next you are given the option of switching from the door you picked earlier to the other remaining door.

Question: Should you accept and switch doors to increase your odds of winning the car or does it not matter because the odds will still be 1 in 2 of getting the car if you switch or not? Explain your decision.

[ronin]

Always switch.

In step two, it looked like one of the other two doors was opened to reveal a goat.

But, actually, it was all of the remaining doors except one which were opened - and they were all opened by a party who knew which door the car is behind.

This is harder to see when it's three doors, but easier when it's ten:

- You choose a door. You have a 1 in 10 chance (10%).

- Eight of the other nine doors are opened and it's eight goats.

- The door you chose is still a 1 in 10 chance (10%). But the other door now represents a 9 in 10 chance (90%).

When we look at it this way we can now see that in the the three-door setup, when one of the three doors is opened, you don't have a 1 in 2 chance of getting the car. You still have your original 1 in 3 chance (33%). The other closed door represents a 2 in 3 chance (66%).

In either setup, when it's just your originally-chosen door and the other door, you should always switch. You need to appreciate the value of what the other party knew - ie. all the doors where the car wasn't => the door behind which the car is.

By taking advantage of this knowledge in the first set up you're doubling your odds of winning the car. In the second you're nine times more likely to win.

[Sgt_Kickaxe]

This is the Monty Hall problem, named after the old game show. It made many extremely intelligent mathematicians get downright hostile for even suggesting that switching improves your odds. The accepted reasoning was that if 1 door has a car and 1 door has a goat, you have a 50/50 chance, PERIOD (their words).

But do you? Let's work out the problem.... for the sake of this explanation, assume the car is behind door #1 and that a door with a goat was revealed each time.

- If you chose door #1, and then switched, you LOSE
- If you chose door #2, and then switched, you WIN
- If you chose door #3, and then switched, you WIN

So if it's a 50/50 chance, how come you win 66.6% of the time if you always switch?

Ronin was close, but what the person making the offer to switch knows is irrelevant. Hint: there are many theories, but no agreed upon solution yet.

If you can come up with a provable theory other than the ones already offered, you could become famous.
[en.wikipedia.org...]

My contribution - common sense tells me the odds don't change if you switch, and that the problem is in the math trying to explain it. You were going to be shown where a goat was no matter what you did. 2 doors remain, odds are 50/50. The car will ALWAYS be behind 1 of 2 doors....

So why does the math not work out? In my mind, it does. You cannot run the example I did above with 3 options AFTER you eliminated one of the options, leaving only two. I'm not sure why, but the mathematicians haven't pointed that out, yet. Read all the effort that was put into solving this in that wikipedia link above. It describes the theorems better than I can.