A Little Backstory

(NOTE: I briefly describe my motivation for this post; If you want to dive straight into the problem, you can skip to the next section)

I previously posted about the Monty Hall Problem, where I presented this controversial and counterintuitive mind-boggler and presented my simplified solution so that more people could understand and appreciate the apparent paradox within.

Since I’m new to blogging, I decided to borrow my friend’s Reddit account to share my post to the math thread over there and see what people think (I had to borrow an account since Reddit denied my post using a newly created account, thinking I was a spammer).

A user then quickly pointed out that my solution does not extend to the so-called Monty Fall problem. I didn’t know what this was, so I used my understanding of the Monty Hall problem to address it, but eventually found out that it’s a lot more counterintuitive than the Monty Hall Problem. Long story short, I said some incorrect things about the Monty Fall Problem which the user pointed out, and upon realizing that my intuition on this new problem was really off, I was greatly stumped.

I therefore decided to set out on a journey to try and uncover the mystery behind this extremely confusing problem.

Monty Hall vs Monty Fall: Why They’re Very Confusing

Recall that in the Monty Hall Problem, you are presented with three boxes, one of which contains a prize. You are then asked to guess which of three boxes contains the prize. After you have selected a box, the host (Monty) then opens one of the two remaining boxes, making sure that he opens one that is empty. He then gives you a choice: will you switch, or will you stick with your box? The answer to this question is that you should switch, because you double your chances of winning, from 1/3 to 2/3.

Now, the Monty Fall problem is similar, but has one important difference. You’re still presented with three boxes and asked to choose a box, but afterwards, Monty accidentally opens one box, and it happens to be empty. You are then given the choice, as before: should you switch or not? And believe it or not, the answer to this problem is, it doesn’t matter. In this scenario, your chances of winning is 1/2 whether or not you switch (which was the intuitive answer to the Monty Hall Problem to begin with!).

This is really where my mind is blown: how could Monty’s intentions affect the outcome of the game in any way? Shouldn’t his intentions be completely outside the mathematical scope of this problem? I was so convinced that Monty’s intentions in opening the box should be immaterial, because what should matter is that he opened an empty box. But it turns out, I was wrong!

I therefore decided to test whether or not this was indeed the case. I decided to create a simulation to see if the claimed results are really accurate.

The Simulations

I decided to simulate both the Monty Hall and the Monty Fall Problems, so that we can easily compare them side-by-side. I decided to present my simulations using HTML and JavaScript, so that anyone can take a look at the results using only a browser, without ever needing a separate compiler. Let me briefly detail the outline of my code, so you can check the validity of my simulations:

For the Monty Hall:

Initialize counter wins to 0 Set three boxes labelled 1, 2 and 3 Randomly pick one box, and assign it as the Prize Box Randomly pick another box, and assign it as Initially Selected Box Randomly pick a box that is neither the Prize Box nor the Initially Selected Box, and assign it as Monty’s Opened Box (if the Prize Box is different from the Initially Selected Box, then only one option is left, so the “random” picking loses its meaning) Pick the box that is neither the Initially Selected Box nor Monty’s Opened Box, and assign it as the Switched Box Check if the Switched Box and the Prize Box are the same: if they are, add 1 to wins Repeat 1-7 N times, and the win percentage is wins/N

For the Monty Fall:

Initialize counters totalValid and wins to 0 Set three boxes labelled 1, 2 and 3 Randomly pick one box, and assign it as the Prize Box Randomly pick another box, and assign it as Initially Selected Box Randomly pick a box that is not the Initially Selected Box, and assign it as Monty’s Opened Box If Monty’s Open Box and the Prize Box are the same, then declare this trial as void and proceed to 9; otherwise, add 1 to totalValid and proceed to 7 Pick the box that is neither the Initially Selected Box nor Monty’s Opened Box, and assign it as the Switched Box Check if the Switched Box and the Prize Box are the same: if they are, add 1 to wins Repeat 1-8 N times, and the win percentage is wins/totalValid

Here’s the link to my simulations: Monty Hall vs Monty Fall Simulations

[WARNING: Don’t enter a very large N, because your browser may crash. Try it out with small values first, like 100, to test if your browser can handle it.]

Indeed, the results of my simulations agree with the accepted results: switching has a 2/3 chance of winning for the Monty Hall, and a 1/2 chance of winning for the Monty Fall. For N = 10000, for instance, I got 67.3% for Monty Hall and 50.5% for the Monty Fall.

The Math

Of course, I couldn’t just settle with the simulations; I had to find an easily understandable explanation as to why this is so. I now present my solution.

Note that, with my previous post on the Monty Hall problem, I used illustrations and very simple concepts to explain. Here, however, I would need to get slightly more technical. In this approach, we invoke a few concepts of probability.

The Monty Hall

For the Monty Hall Problem, note that it shouldn’t matter which box you select initially because the three boxes should be identical and all three boxes have a 1/3 chance of containing the prize. So, let’s assume that you chose Box 1.

There would then be the following possibilities:

You Chose… Prize is at… Monty opens… You switch to… Result 1 1 2 or 3 3 or 2 LOSE 1 2 3 2 WIN 1 3 2 3 WIN

Therefore, we see that indeed, switching results in a 2/3 chance of winning.

The Monty Fall

For the Monty Fall, again it shouldn’t matter which box you choose at the start, so we again assume that you choose Box 1. Noting that Monty has no idea where the prize is, he now can open any of the two remaining boxes that you didn’t choose. However, we would only count those scenarios where he accidentally opens an empty box. The following would be the possibilities:

You Chose… Prize is at… Monty opens… You switch to… Result 1 1 2 3 LOSE 1 1 3 2 LOSE 1 2 2 – VOID 1 2 3 2 WIN 1 3 2 3 WIN 1 3 3 – VOID

Clearly, if we don’t count the void, then we only have 4 valid scenarios, 2 wins and 2 losses. Therefore, switching indeed leads to a 1/2 chance of winning.

The Cause of the Discrepancy

Note that, if we list out only the valid scenarios for the Monty Fall case, we will arrive at 4 scenarios as opposed to Monty Hall’s 3. Specifically, we see that whereas in the Monty Hall, we only have a single entry for the “2 or 3” case, in the Monty Fall, we have an entry for each.

This is because for the Monty Hall problem, for each possible position of the prize, Monty only has to select one box to open: if the prize is at Box 2 or 3, then it’s easy, because he would select Box 3 or 2, respectively. When the prize is at Box 1, then he would have to pick at random between either 2 or 3.

Now, for the Monty Fall problem, for each position of the prize, Monty can now pick two boxes: any of the two remaining boxes that you did not pick. So clearly, if the prize is at Box 1, then he can pick Box 2 or Box 3, each with a probability that’s equal to the probability of each of the remaining 4 scenarios, of which 2 is void. Therefore, if we want each entry in the table to have equal probabilities of happening, then these are necessarily two separate entries in the table, as opposed to one in the Monty Hall case.

Now, if we instead decide to separate the “2 or 3” case for the Monty Hall problem into two entries, one for “2” and another for “3,” then we might conclude that the probability becomes 1/2 as is the case for the Monty Fall. However, this would not be valid, because the entries in the table no longer have equal probabilities of happening: each of the cases which we just separated has half the probability of happening compared to the other two cases, because they’re just two possibilities for one occurrence.

Conclusion

We have just seen that indeed, Monty’s intentions do matter. It’s something I find really absurd, but is true nevertheless. The Monty Hall and the Monty Fall are two distinct problems that result in two different outcomes, and the difference lies in Monty’s head.

I hope you found my post useful in comprehending these ridiculously baffling problems, or at the very least, interesting.

If you have any thoughts, corrections, clarifications or violent reactions, please let me know!

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