You’re on the game show Let’s Make a Deal. You are shown three doors, and each has a prize behind it. One has a new car and the other two have “zonks.” If you choose the correct door, you get the car. You choose door #1. Monty Hall then reveals what’s behind door #3. It’s a zonk. He then tells you that you can either stick with door #1, or switch to door #2.
Why is there an advantage to switching? I don’t get the math as to why that’s the best decision.
You could start with Cecil’s explanation.
First, keep in mind that Monty knows which door has the zonker.
How about this:
Let’s say there were 100 doors, and you pick the one you think has the prize. There’s a 99% chance you picked the wrong door.
Monty then opens 98 zonkers, leaving you with one door still closed, and the door you picked.
There’s still a 99% chance that you picked the wrong door at the beginning.
I’d advise a switch.
Works the same with three doors, only there’s a 66 2/3% chance you still picked the wrong one, when there are the two remaining.
Earl etc. has the right idea. The probability that the prize is behind the door you picked after Monty opens his door(s) is the same as the probability that you chose the right door in the first place.
Sticking with the same door after Monty opens a zonk door is desireable if you picked the right door initially. You will pick the right door initially 1/3 of the time.
Switching doors after Monty opens a zonk door is desireable if you picked the wrong door initially. You will pick the wrong door initially 2/3 of the time.
Thus, 2/3 of the time switching doors will get you the prize.
There have been posts on this before, but here’s my understanding of it:
In your initial choice, you have a 1 in 3 chance of picking the car. After the reveal, you still only have a 1/3 chance of having picked the car. And door #3 has a 0/3 chance of having the car (since you’ve seen the zonk). That means door #2 has a 2/3 chance of having the car, since the sum of the probabilities of all three doors has to equal 1 (since the car is behind one of them).
Here’s a page with a few different variations on analogies to explain the issue.
The answer does depend on what Monty Hall’s policy is.
If Monty Hall always opens a door that you have not chosen, then (as has been explained), you should always change doors, to go from a 1/3 chance to a 2/3 chance.
If Monty Hall opens a second door if you have chosen the right door, and doesn’t open a second door if you have chosen a wrong door, then you should stick with your original choice if he opens a door, and change if he doesn’t. This is one clear case where you should stick with the original door if he opens another, because if he opens another door you have a 100% chance by staying with the original choice.
If Monty Hall opens a second door if you have chosen a wrong door, and doesn’t open a second door if you have chosen the right door, then you should stick with your original choice if he doesn’t open a door, and change if he does. This is a very clear case for changing: you will go from a 0% to a 50% chance.
Monty Hall might have other, more complex policies, and then the calculations will get a bit more complex.
I was just about to address Giles’s point. I ran through this before, and I seem to remember that as long as the prize is equally likely to be behind any door, or you have no preference for choosing doors, it’s still to your advantage to switch. I’ll have to go back and do it again for more details.
All good answers, but this is the one that makes sense to me.
But the point is well taken, if Monty wants to mess with your head, all bets are off.
Empiricism is the answer to this one.
Get a deck of cards.
Deal out three, face down.
You look, and you know which card is the highest.
You let the guest choose
You are Monte. You don’t reveal the jackpot. (highest card of the three.) You reveal whichever remaining card is not the jackpot, after the player makes his choice. Evidently that is the way they did it on the show.
Try it out for half a dozen hands, and you will quickly learn how it works.
Tris
I was going to post something like what Giles just did. In the way the situation was explained in the OP, Money was not constrained always to have to show you a non-winning door, therefore you don’t know what his motivations are, therefore you can’t really say that it’s better to switch. In order for the problem to have the 2/3 - 1/3 answer that’s always given, you would need to explicitly state that Monty must show you a non-winning door every time you play.
I’m not saying anything here that hasn’t already been explained, but what helped the problem make sense to me was simply to visualize the possible situations.
There are three possible arrangements, shown below (X is the car; O is a zonk). Let’s say you choose Door 1, on the left:
X O O -Monty opens either 2 or 3; if you switch you lose
O X O -Monty opens door 3; if you switch to 2 you win
O O X -Monty opens door 2; if you switch to 3 you win
You can see that switching wins 2 out of 3 times.
Ironically, I think this works out confusingly because it goes against the grain of what gets shoved into you when you learn statistics. Specifically, “Each roll of the dice is independant of all others.” This is still true here, but people tend to get confused because they already rolled the dice when they made their first selection. It certainly confused me.
Does the problem work out any differently if the host doesn’t know which doors contain zonks? As in, you pick a door, he randomly selects a door (not yours) that just happens to have a zonk. Is it still in the best interest to switch?
Well, I’ve explained this problem a jillion times to people and, without resorting to probabilities or Bayes’ theorem or anything technical, the simplest explanation is the following … Assuming that Monty always opens a door that does not have the car, the choice of switching from your selected door (say, it’s door #1) to the other door is equivalent to choosing both doors #2 and #3 (regardless of which of these have been revealed to be non-winner since Monty will always show you a non-winner). So which is the better choice: choosing 1 door (not switching) or 2 doors (switching)? The answer is obvious.
Now that I understand the answer, I can address this.
If Monty opens a random door, it may well turn out to be the car. In which case you lose. But if he picks a zonk, the result is the same as if he knew. A Monty-bot could choose. Your odds are still 33/66.
I totally didn’t get this until I sat down to write short program to simulate it. When I got the point in the code where Monty would reveal the door, and I couldn’t think of what the code should do at this point to change the data, it dawned on me that the reveal made no difference.
I convinced a coworker (who thought the odds were equal, change or not) by having him imagine the experiment with a million doors. Also, imagine the chooser always sticking with his original choice, and the experiment being done a million times. If my coworker was right, then half the time the chooser would pick the good door, out of a million choices, half the time.
Oh, the Monty Hall Problem has been beaten to death on this board before. Look here.
You chose #1. There was a 1/3 chance that it was #1.
There was also a 1/3 chance that it was #2, or #3.
Once #3 is eliminated, your odds are 50/50 no matter which way you go. Eliminating one of the three (reducing the probablity to zero) just means there are two possibilities, and you’ve got one.
Think of it this way:
You are playing the daily number. Your number is 123. Your odds of winning: 1/1000 (1/101/101/10). Yolanda says the first number is 1. For the next two seconds, you will have a 1/100 chance of winning (1/11/101/10). The 2 ball in the second container and the three ball in the third container are not going to become less likely to rise to the top just because you have a 1. After all, somewhere there’s a guy with 135.
You may now proceed to pick apart my analogy as being completely inappropriate.
Nope. It’s not 50-50, it is to your advantage to switch. I assume there will be a jillion posts here in a second so for brevity’s sake I’d suggest that you read Cecil’s original column on this problem (link in post #2).