Making the Best Choice

Jan submits this challenge for those of you up to the task.

You meet a man on the street who holds ten envelopes full of money. You get one of the envelopes. The rules are that you can look in the first one and accept or reject it, then look in the second one, and on and on. Once you reject an evelope, you cannot go back to it, it’s gone forever. You can only see later envelopes when you have rejected the current one. When you accept an envelope, the game is over.

According to this article, you should reject the first envelope and accept the next envelope you see that has more money than the first. It says:

According to Colley’s Rule, never accept the first offer, but instead note its qualities and accept the first subsequent offer which beats it. It was recently proved that this both increases the chances of making the best choice, while minimising the chances of making the worst.

In his book, Innumeracy, John Allen Paulos discusses this very game. He doesn’t mention Colley’s Rule and his results are different. Paulos proves that you should reject the first 37% of the choices and take the best one after that. Slightly different from just rejecting the first.

Your task, should you choose to accept it, is to create a simulation in Excel to see who, if either, is correct.

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7 thoughts on “Making the Best Choice

  1. In your comment about Paulo’s method, you state that “Paulos proves that you should reject the first 37% of the choices and take the best one after that. Slightly different from just rejecting the first.”

    If we were given each envelope individually, then how could we determine if it was the “best one after that”? Should we accept the first one that is greater than the maximum of the content of the first four envelopes?

  2. According to the Workbook that I made (using the logic from my above post, and assuming that either person would pick the last envelope if they had not already chosen one), I came up with the following results based on several thousand runs:

    Paulos would “win” nearly half the time (around 47%).
    Colley would “win” only about a quarter of the time (roughly 24-27%).
    They would tie nearly a quarter of the time (again, roughly 27%).

    Therefore, I would definitely have to agree with Paulos’ method.

    I can submit the Excel file if you’d like.

  3. Andy, I went back and looked up that Paulos thing. After the first 37%, he says to pick the one that’s better than all previous ones. So what you said is correct, greater than the maximum. He doesn’t really say if the last one is the default, but I think that’s a valid assumption. Email the workook to me, if you don’t mind.

  4. Paulos method is far more effective.

    Amidst around 390 data, Paulos selection method returned around 90% better results.

    I am sending the workbook. You can run the recorded macro by keeping the cursor at cell M4 & pressing Ctrl and “q”.

  5. Hey there, long time listener, first time caller.

    I started this problem and compared my results with the other posters, when it occurred to me that we’re all doing it wrong. The object is not to determine which method prevails more often, but rather which method yields the greatest prize.

    I ran 20,000 reps of both methods, including the Paulos scenarios where the first three are automatically rejected, the first four are automatically rejected,…,the first n are automatically rejected.

    I found the best method to be to skip the first two and then start the bidding. The Paulos method yields slightly higher returns when either two or three are automatically rejected, ties when four are rejected, then loses after that.

    In terms of winning percentage, skipping two, three, four, or five yields a higher percentage than loses after that.

    I wouldn’t take a bullit defending my numbers, but they seem reasonable enough and I’m convinced we want to win more money, not more often.

    If anyone wants to see the code or the results let me know.

  6. Interesting BUT if we are concerned with the $ value rather than the number of times rounds they won out then I think Colley wins.
    I accumulated the $ values taken by each method over the number of trials and ran this a large number of times. Colley won out on the $ value almost every time.

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