## Sunday, September 19, 2010

### Probability

There is a fairly common probability mind bender that continues to keep problem solvers deeply divided. It comes it many different variations, with goats, money, envelopes, and doors. There are only possible answers to the problem. I used to believe one answer but the more I think about the problem, the more I believe the other answer. I will present you with both the problem and both answers with their respective reasonings.

The Problem: Four identical sealed envelopes are presented to you on a table. One envelope contains \$100, while the other three are empty. You select one envelope. Two of the three remaining envelopes on the table are then opened to show you that they are empty. You then have the option of either keeping your selected envelope or exchanging it for the remaining unopened envelope on the table. Should you exchange envelopes?

One variation of the puzzle involved a Let’s Make A Deal scenario, with four doors. Behind one door was a car, and behind the other three doors were goats. You could really change it however you wish.

The First Answer: The current thinking of this problem tells us that the answer is to exchange envelopes. I used to subscribe to this answer because the reasoning was intuitive and simple. It goes like this. When you selected an envelope, you had a 25% chance of selecting the \$100. That means there was a 75% chance the \$100 was in one of the remaining envelopes on the table (duh). But then two of the three envelopes on the table were removed, leaving one envelope. Intuition would then dictate that there is still a 75% chance the \$100 is on the table. But since there is only one envelope left on the table, there must be a 75% chance the \$100 is in it. Therefore, you have a 75% chance of getting \$100 by exchanging envelopes. Sounds simple, right?

The Second Answer: The other intuitive way of looking at this is to simply say the when there are two envelopes left in play, one must contain the \$100. Two empty envelopes have been removed, so they do impact the initial calculation of odds. (This is similar to why the results of a double blind medical research study are tainted if anyone at all looks at the unblinded results before the study is complete.) So if two envelopes remain, and only one has \$100, your odds are 50% that you hold the \$100 in your hand.

Not convinced by these glib intuitive reasons? I sympathize with you.

So now let us look at the problem the way a mathematician might approach it. There are four ways the envelopes could be set up.

1 \$ -- -- --
2 -- \$ -- --
3 -- -- \$ --
4 -- -- -- \$

Each scenario gives the same outcomes, so we need only examine one of them. Let us use the fourth. By selecting an envelope you create four combinations of envelopes remaining on the table.

1 -- -- \$
2 -- -- \$
3 -- -- \$
4 -- -- --

Now if two empty envelopes are removed, the first three scenarios can only result in the \$100 being left on the table. There is only one possibility for these three.

1 -- -- \$ → \$
2 -- -- \$ → \$
3 -- -- \$ → \$

The fourth scenario, however, has more than one possible outcome from the removal of two empty envelopes. Because each empty envelope is actually different, you can remove two empty envelopes three different ways. I have labeled each empty envelope to help you keep track.

4 --a --b --c → --c
4 --a --b --c → --b
4 --a --b --c → --a

So, after you have selected an envelope, there are six ways to remove two empty envelopes. Three result in the \$100 remaining on the table, and three result in an empty envelope being left on the table. That would indicate even odds, and so it makes no difference whether or not you exchange envelopes.

Is there a lesson to all of this? Sure. The lesson is to always write out the possibilities when some arrogant bastard tries to give you an intuitive answer but cannot back it up with true math, because chances are he is an executive somewhere coordinating technical projects using the only “people person” skills he has to offer.