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# Analysis of the Monty Hall Problem

 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:

 "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 other 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 take the switch?"

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:

• Case 1: You've picked the right door to start with. If you switch, you'll lose.
• Case 2: You've picked the wrong door to start with. The host shows you the other wrong door, so if you switch, you'll clearly win.
Now, Case 1 only happens 1/3 of the time, and Case 2 happens 2/3 of the time. Therefore, if you follow the strategy of always switching, you'll win 2/3 of the time. Whereas, if you never switch, you'll of course get the right door only 1/3 of the time. If that's not clear enough, take a look at the following table. In this and all tables in this analysis, "EV" refers to "expected value."

Case
Original door choice
Probability
Don't Switch Outcome
Switch Outcome
Don't Switch EV
Switch EV
1
Right
1/3
Win
Lose
1/3
0
2
Wrong
2/3
Lose
Win
0
2/3
Total
1
1/3
2/3

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:

Case
Original door choice
Host gives switch option
Probability
Don't Switch Outcome
Switch Outcome
Don't Switch EV
Switch EV
1a
Right
Yes
1/6
Win
Lose
1/6
0
1b
Right
No
1/6
Win
Win (can't switch)
1/6
1/6
2a
Wrong
Yes
1/3
Lose
Win
0
1/3
2b
Wrong
No
1/3
Lose
Lose (can't switch)
0
0
Total
1
1/3
1/2

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:

Case
Original door choice
Host gives switch option
Probability
Don't Switch Outcome
Switch Outcome
Don't Switch EV
Switch EV
1a
Right
Yes
X / 300
Win
Lose
X / 300
0
1b
Right
No
(100 - X) / 300
Win
Win (can't switch)
(100 - X) / 300
(100 - X) / 300
2a
Wrong
Yes
2 * X / 300
Lose
Win
0
2 * X / 300
2b
Wrong
No
2 * (100 - X) / 300
Lose
Lose (can't switch)
0
0
Total
1
1/3
(X + 100) / 300

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:

Case
Original door choice
Host gives switch option
Probability
Don't Switch Outcome
Switch Outcome
Don't Switch EV
Switch EV
1a
Right
Yes
1/3
Win
Lose
1/3
0
1b
Right
No
0
N/A
N/A
0
0
2a
Wrong
Yes
4/9
Lose
Win
0
4/9
2b
Wrong
No
2/9
Lose
Lose (can't switch)
0
0
Total
1
1/3
4/9

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:

Case
Original door choice
Host gives switch option
Probability
Don't Switch Outcome
Switch Outcome
Don't Switch EV
Switch EV
1a
Right
Yes
1/3
Win
Lose
1/3
0
1b
Right
No
0
N/A
N/A
0
0
2a
Wrong
Yes
1/3
Lose
Win
0
1/3
2b
Wrong
No
1/3
Lose
Lose (can't switch)
0
0
Total
1
1/3
1/3

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:

Case
Original door choice
Host gives switch option
Probability
Don't Switch Outcome
Switch Outcome
Don't Switch EV
Switch EV
1a
Right
Yes
X / 300
Win
Lose
X / 300
0
1b
Right
No
(100 - X) / 300
Win
Win (can't switch)
(100 - X) / 300
(100 - X) / 300
2a
Wrong
Yes
2 * Y / 300
Lose
Win
0
2 * Y / 300
2b
Wrong
No
2 * (100 - Y) / 300
Lose
Lose (can't switch)
0
0
Total
1
1 / 3
(100 + 2Y - X) / 300

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.

 If the host gives the switch option X percent of the time when the contestant originally picks the right door, and 2X percent of the time when the contestant originally picks the wrong door, then the contestant will not know whether or not to switch.

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).

 If the host gives the switch option X percent of the time when the contestant originally picks the right door, and Y percent of the time when the contestant originally picks the wrong door, then the contestant should only switch if X < 2Y.

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:

Case
Original door choice
Host gives switch option
Probability
Don't Switch Outcome
Switch Outcome
Don't Switch EV
Switch EV
1a
Right
Yes
(X / N) / 100
Win
Lose
(X / N) / 100
0
1b
Right
No
((100 - X) / N) / 100
Win
Win (can't switch)
((100 - X) / N) / 100
((100 - X) / N) / 100
2a
Wrong
Yes
(Y(N - 1) / N) / 100
Lose
Win
0
(Y(N - 1) / N(N - 2)) / 100
2b
Wrong
No
((100 - Y)(N - 1) / N) / 100
Lose
Lose (can't switch)
0
0
Total
1
1 / N
((100 - X) + (Y(N - 1)/(N - 2))/100N

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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