Deal Part Deux: A Simple Explanation

When Overthinking Produces the Simplest Explanation

In my previous post on the Monty Hall Problem, I walked through the logic and statistics behind the 66% probability of winning by switching your answer when given the option to do so. Math, examples, and code stuffs — including examples of tests I had performed trying to break the system — were provided. And yet, while I grasped the answer, I wasn’t 100% happy with the explanation. It just wasn’t simple enough, and it felt like it could be more straightforward. Well, tonight I had a second “a-ha!” moment — not one of understanding, but figuring out a way to explain the problem in such a way that you should be able to understand it even if you don’t fully get all the statistics behind it.

And I was *so* close to getting there the first time around too!

One of the tests I performed was making my algorithm switch labels from whatever the numeric values were to “A” and “B” instead. The lower number of the remaining two doors would be A, and the higher number would be B, and the initial guess and the correct answer were relabeled accordingly. In the end, I stored the values in a list with the winning answer, the guess, and the two doors remaining in a list that I could use to evaluate things.

The function that performed the relabeling magic (see previous post if this doesn’t make sense…)

The output looked like this (first 20 observations out of 1,000 trials shown):

When I ran an experiment to measure the percentage of times the initial guess value matched the winning value, the results lined up with the statistical explanation — roughly 33.33% of the time.

Making the Correct Answer the Obvious Answer

Here is where another thought exercise makes the correct outcome the obvious outcome, no fancy math required. Let’s start from a simple set of 100 observations. Given three doors and one winner, what percentage of the time do we expect the guess to match the winning door?

That one’s easy — your odds are 1 in 3, so 33.33% probability. No one’s arguing here (I don’t think…)

So let’s create a list where we make those guesses and record the winning doors.

A, B, C vs. 1, 2, 3 doesn’t matter here.

Our list starts out as being 33.33% accurate. With me so far?

Now, remove a door that doesn’t matter. Does the list magically become more accurate?

No.

The numbers don’t change (much…)

And that’s the simple explanation. You start off being 33.33% accurate. If you don’t change your answer, you will still win only 33.33% of the time. However, of the two choices you have in the second part of the game, the other one is now going to be accurate 66.66% of the time.

Going back to our A / B example: Say you always picked door 1, so your answer was always going to be converted to A. A will be accurate 33.33% of the time. Thus, if B is the only other possible option, and A is accurate 33.33% of the time, B **must** be accurate the remainder, meaning 66.66% of the time.

Click?

Nothing beyond basic math and probability calculations required to get to that conclusion. Removing the door that doesn’t matter does not change how often you were right to begin with. Thus, you will always be better off — statistically speaking — by switching your answer when given the chance.

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Jason Eden

Jason Eden

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Data Science & Cloud nerd with a passion for making complex topics easier to understand. All writings and associated errors are my own doing, not work-related.