Kevin Gong March 17th, 2003
There are numerous web pages devoted to discussion and analysis of the Monty
Hall problem. I hope to add something new and interesting to the discussion with
an analysis of a more interesting problem.
For those of you living in a mathematical cave, I'll briefly explain the
problem and very little on the controversy. In 1991, Parade Magazine
published a column "Ask Marilyn" in which Marilyn vos Savant replies to
a reader's question:
Marilyn replied that the answer was yes, and there was a big uproar
and mathematicians from across the country attacked her. I'll spare
you the details. You can do a web search on that if you're interested.
I'm more interested in the mathematics.
First off, let's answer the original question. Should you switch? The correct answer
is that there is not enough information in the problem as stated to answer the question.
Marilyn assumed that the host always opens a door and shows the contestant a
goat, regardless of whether the contestant has already picked the "right" door or not.
(right meaning the one with the car). So let's first answer the question based on that
assumption. Should you switch? The answer is yes. Some people find this counter-intuitive.
It is especially counter-intuitive if they don't have Marilyn's assumption stated
explicitly. Even with the assumption stated, why wouldn't it just be a 50-50 proposition?
It is pretty easy to get past the counter-intuitiveness with the following simple explanation:
There are basically two cases, you've either picked the right door to start with, or you haven't:
Now, you may be asking yourself: why would the host give you an option to switch if you've
picked the wrong door to begin with? An excellent question. Which leads to the crux of this
analysis. In general, I'd like to analyze the problem given a variety of host behaviors. In
particular, here are the questions I plan to answer:
If the host only gives you the switching option when you've picked the right door, should you switch?
Of course not!
If the host only gives you the switching option when you've picked the wrong door, should you switch?
Of course!
If the host only gives you the switching option 50% of the time, regardless of whether you've picked the right door, should you switch?
Yes.
This is really just a variation of Marilyn's original assumption. In fact, if the host makes the decision
without regard to whether you've picked the right door or not, then you should always switch, no matter what
the percentage is. But for clarity's sake, let's analyze the 50% question. We can break this down into 4 cases:
Since the expected value for the switch strategy (1/2) is greater than the expected value for
the don't-switch strategy (1/3), it is to your benefit to use the switch strategy.
Now let's look at the more general case. Suppose the host gives you the switch option X percent of the time (X is a number from 0 to 100). Then the table
looks like this:
If you follow the don't switch strategy, you'll win 1/3 of the time. If you always switch, your
expected value is (X + 100) / 300, which is greater than 1/3 (unless X = 0, in which case you
are never given the option of switching). Therefore, you should follow the strategy of always
switching when given the opportunity.
If the host only gives you the switching option 100% of the time when you pick the right door, but 2/3 of the time when you've picked the wrong door, should you switch?
Yes.
Let's use the table approach again:
Since 4/9 > 1/3, the strategy of switching whenever given the opportunity is best in this case.
What strategy should the host employ, such that you do not know whether you should switch or not (switching would be a 50/50 proposition)?
This seems like a good question to ask, to keep the game show as entertaining as possible. For the host strategy to work like this,
we want to create a situation where the expected value for the don't-switch strategy is the same as the expected value for the switch strategy.
Look at the above table. If we change the 2/3 probably to something else, we can affect the switch-strategy expected value without affecting
the don't-switch strategy EV. If we give the switch option X percent of the time, we want 1/3 = (2/3) * (X / 100). Solving for X, we get
X = 50. The following table bears that out:
Thus, if we always give the option of switching when the contestant picks the winning door, and give them the option 50% of
the time when they pick the wrong door, then they won't know whether to switch or not.
Are there other host strategies which have the same property? Suppose that if they pick the right door, we'll give them
the switching option X percent of the time; if they pick the wrong door, we'll give them the switching option Y percent of
the time. Then the table looks like this:
We want 1/3 = (100 + 2Y - X) / 300. Solving for this, we find we want X = 2Y. X = 100, Y = 50 is the solution we found above, but any
similar host strategy will work (X = 50, Y = 25, etc.). To maximize the frequency of giving the switch option and keeping the game
entertaining, we'll want the (X = 100, Y=50) solution.
To round out the discussion of this problem, if you know X and Y (suppose you do a statistical analysis of
previous shows), then how do you tell whether to switch or not? Going back to the old table,
it only makes sense to switch if (100 + 2Y - X) / 300 > 1/3. Solving the equation, we find
we only want to switch if X < 2Y. So, for example, if the host gives us the option of switching
60% (X = 60) of the time when we've picked the right door, and 25% (Y = 25) of the time when we've picked the
wrong door, then we shouldn't switch (since 60 is not less than 25*2).
Let's extend this even further. Let's suppose that there are N doors instead of 3 doors (N >= 3),
and there's still just one car and there's a goat behind each of the other N-1 doors. Then what
are the odds? In this case, if we pick the wrong door and switch, then we have a 1 / (N - 2)
chance of picking the right door, since we can eliminate two of the doors (the one we picked,
and the one the host picked). Then the table looks like this:
We should switch if ((100 - X) + (Y(N - 1)/(N - 2))/100N > 1/N. Simplifying the equation,
we want to switch if Y(N - 1)/(N - 2) > X.
I leave it as an exercise to the reader to solve the case for N doors, with cars behind M
of the doors (0 < M < N - 1).
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