November 26th, 2006
Folks,
Marilyn Vos Savant reputedly has the highest IQ ever recorded. She publishes a weekly column in Parade, "AskMarilyn." This Sunday (26 November 2006)she posted the following question and answer:
Q:
You are on a game show with three doors. A car is behind one; goats are behind the others. You pick door No. 9. Suddenly, a worried look flashes across the host's usually smiling face. He forgot which door hides the car! So he says a little prayer and opens No. 3. Much to his relief, a goat is revealed. He asks, "Do you want door No. 2?" is it to your advantage to switch?
W.R. Neuman, Ann Arbor, Mich.
A:
Nope. If the host is clueless, it makes no difference whether you stay or switch. If he knows, switch.
Is Marilyn correct? If not why not?
Cheers,
Dr. Ron
fabrizio:
nothing change if the host don't remember, you still have the advantage if you switch (2/3)
Mr Mean:
Actually, I believe Marilyn is correct. A clueless host means there’s no advantage in switching doors. Conditional probability tells us that P(A|B)=P(A and B)/P(B). For a short explanation see my comments at:
http://www.indium.com/drlasky/entry.php?mode=add&id=506
For a longer explanation see:
http://en.wikipedia.org/wiki/Conditional_probability
Let’s call the the doors G, B1, and B2. We want the probability that we picked G given that the host shows B1 or B2. This is the probability that switching is the wrong thing to do. So, call event A, “we picked door G,” and call event B “the host shows us B1 or B2.”
event B: the host shows us B1 or B2 There are 4 ways this could happen:
1) we pick G (which happens 1/3 of the time) and the host shows us B1 (which happens half of the times that we pick G): (1/3)*(1/2) = 1/6
2) we pick g (which happens 1/3 of the time) and the host shows us B2 (which happens half of the times that we pick G): (1/3)*(1/2) = 1/6
3) we pick B1 (1/3 of the time) and the host shows us B2 (half of the times we pick B1): (1/3)*(1/2) = 1/6
3) we pick B2(1/3 of the time) and the host shows us B1 (half of the times we pick B2): (1/3)*(1/2) = 1/6
So P(B)=4/6=2/3.
event (A and B): we pick door G and the host shows us B1 or B2 If we pick G (which we do 1/3 of the time) the host will always show us either B1 or B2. So P(A and B)=1/3*1=1/3.
This gives P(A|B)=(1/3)/(2/3)==1/2. So if the host shows us the wrong door, we’ve picked the right door half of the time. Therefore, switching has no advantage.
It is very counter-intuitive that this would be different than the case where the host knows the correct door. Doing the calculations might help show why. If we assume the host knows the correct door, and will always show us an incorrect door, we have different probabilities. In this case:
event B: the host shows us B1 or B2
The problem states that the host always shows us an incorrect door. So P(B)=1
event (A and B): we pick door G and the host shows us B1 or B2
This is the same as the unknowing host. If we pick G (which we do 1/3 of the time), the host will show us B1 or B2 every time. So, P(A and B)=1/3*1=1/3
So P(A|B)=(1/3)/1=1/3. This means that if the host shows us an incorrect door, we’ve picked the right door 1/3 of the time. So switching wins the prize 2/3 of the time.
Ron Lasky:
I still think Marilyn is wrong.
Here is a way to look at it that I think makes it more clear.
Let’s say there are 1000 doors. There is one car behind one door and one goat behind each of the other 999 doors.
You are asked to make your selection and you choose door number 137. Let’s now consider that we have two sets. Set #1 consists only of the door you chose, number 137. Set #2 consists of the remaining 999 doors. The chance that the car is in set #1 is 1/1000 and the chance that it is in set #2 is 999/1000.
Assume the host opens all doors in set #2 except door number 641. All of the 998 doors that are open reveal a goat. Whether he knew or not where that car was, there still is a 999/1000 chance that it is in set #2.
So if you are allowed to switch in this case. your chances go from 1/1000 to 999/1000.
Cheers,
Dr. Ron
Julia:
In the question the host does know what is behind the doors. He knows he is opening a door with a goat behind it. It is not a suprise. Switching would give the contestant a better chance of winning.
cxseven:
If you permute what’s behind the doors beforehand, by assumption this will not influence the player or the host. Therefore the probability that the host will pick a particular goat or car is uniform – always 1/3. Likewise the player has an equal probability of choosing the remaining two doors.
That’s probably the simplest argument.
Another argument is to use the conditional probability equation, P(A given B)=P(A and B)/P(B). In this case, let B be the event in which the host opens the door with a goat behind it. An easy mistake to make is to think this has a probability of 1, but in fact it is again 2/3. If the argument in the first paragraph isn’t convincing, think about the two cases which occur after the player picks a door:
Case 1: Player picks the door with a car. Probability that the host picks a goat: 1.
Case 2: Player picks a door with a goat. Probability that the host picks a door with a goat: 1/2.
The first case occurs with 1/3 probability and the second with 2/3 probability, so the total probability for the case in which the host picks a door with a goat is
P(player picks car)*P(host picks goat given player picks car) + P(player picks goat)*P(host picks goat given player picks goat) = 2/3.
To complete the original equation, we need P(player picks car AND host picks goat). But that should be obvious – if the player picks a car, the host necessarily has picked a goat, so it’s 1/3. And (1/3)/(2/3) = 1/2, so the probability that you picked a car GIVEN that the host picked a goat is a half as expected.
This works out even in the new scenario you outlined.
Hope this sufficiently explains things.
cxseven on gmail