I was perversely flipping through the Parade section of my Sunday newspaper when I stumbled upon Marilyn vos Savant’s “Ask Marilyn” column. Even more perversely, I read it. It wasn’t a total loss, though, because it appears she made another mistake, even worse than the one you pointed out in a very entertaining column a few months ago. Here’s the question:
Suppose you’re on a game show and you’re given the choice of three doors: Behind one door is a car; behind the others, goats. You pick a door, say No. 1, and the host, who knows what’s behind the doors, opens another door, say No. 3, which has a goat. He then says to you, “Do you want to pick door No. 2?” Is it to your advantage to switch your choice?
ANSWER: Yes; you should switch. The first door has a one-third chance of winning, but the second door has a two-thirds chance. Here’s a good way to visualize what happened. Suppose there are a million doors, and you pick door No. 1. Then the host, who knows what’s behind the doors and will always avoid the one with the prize, opens them all except door #777,777. You’d switch to that door pretty fast, wouldn’t you?
Correct me if I’m wrong, but aren’t the odds equal for the remaining doors–one in two?
This “riddle” has been debated now for several years. Even Cecil got in on the act a while back.
I have studied the problem in detail, and have read the comments and conclusions from many folks.
And here’s the final answer (drum roll, please): Marilyn is correct. If you randomly pick a door, and the host opens another door (with no prize, obviously), and you switch to the other unopened door, your chance of winning is exactly 2/3. If you do not switch, your chance of winning is exactly 1/3.
Once you have completed this homework assignment, if you have any more questions, you can come back and post them here at your own peril (be prepared for the possibility of an acrimonious discussion.)
Hmm, but your original odds were 1 in 3. What if you picked the door with the prize on your first try? Sure it’s a 1 in 3 chance, but for every 3 people who play, an average of 1 will pick correctly on the first try. My thinking is that even though the host opens a door, and offers to let you exchange, your odds are still 1 in 3 that you picked the prize winning door to begin with.
Now you start a whole new game with only two doors. If you change your mind, and choose the other door, you have a 50/50 chance of winning the prize. If you stay with the door you chose originally, you still have a 1 in 3 chance of it being the correct door, because you made that choice when all three doors were closed, right?
Arnold Winkelried is right; this is an old, beaten topic. Anyway…
> Hmm, but your original odds were 1 in 3.
Correct.
What if you picked the door with the prize on your first try? Sure it’s a 1 in 3 chance, but for every 3 people who play, an average of 1 will pick correctly on the first try.
Correct.
> My thinking is that even though the host opens a door, and offers to let you exchange, your odds are still 1 in 3 that you picked the prize winning door to begin with.
That’s correct. But if you automatically switch, your odds increase to 2/3. The point of this hopeless exercise is that you should always switch, no matter what.
> Now you start a whole new game with only two doors…
You can’t play this game with two doors. The host always opens a door (without a prize) before you make you final selection. Thus the minimum number of doors with which you can play this game is three.
I think we’ve got enough legitimate threads on this subject without having one to titillate a known troll and sock puppet. I’m closing this thread. Apologies to all who contributed.