Monty Hall Problem Pt 3/3: The “Jason Hall Problem”

The point: Change the option of the “Monty Hall Problem” from “switch one door for another” to “switch TWO doors for one door” in order to eliminate the “preference bias noise” of the original version of the problem and make the correct solution even more counter-intuitive.

The rant that goes with it:

In Parts 1 & 2 of this 3-part series, I demonstrated how “Monty Hall Problem” can be intuitively explained, the implication that it has on the reliability of intuition and how such faith in one’s false convictions can expose you to predators…like me. In this Part 3, I propose an alternative version of the problem that I cheekily call the “Jason Hall Problem” (there are those, including my wife, who will tell you that this is merely one in a long list…).

In the “Monty Hall Problem”, the premise of trading one door for one other door often confuses people into thinking the problem is about preference as opposed to statistics. In other words, “If it doesn’t matter whether I stay or switch, staying is better because I stick to my commitments”. Even though these people believe that statistically it doesn’t make a difference, when asked what is the best thing to do they insist that it is either “better” to keep their original choice (if they tilt towards the psychotic) or to switch doors (if they tilt towards the neurotic). The commitment to one or the other over principal alone is very strong, a fact made clear in this anecdote:

While doing the 10-card demonstration (ref Part 2, the trials with my boss) for one of my colleagues, he insisted repeatedly that it was in his best interest to stay with his original choice, even as I continued to explain to him that the actual statistics gave him a 90% chance of winning if he switched. Then I switched tactics: I put $5 on the table and told him that the money was his if he ended up with the winning card. At that point, he said it was obvious that he should switch, which he did. I then asked him, before telling him that he had won the $5, “and if there were no money involved?” His response, amazingly, was, “then I wouldn’t switch”. I concluded from this experiment that some people are really, really messed up beyond saving…and that preference bias is strong, even can be bought-off.

Changing the problem so that the correct answer is to trade away one number of doors in exchange for a different number of doors eliminates this “preference noise” of the original problem. Furthermore, to emphasize counter-intuitive properties of this problem in the revised version, the correct answer must (of course) be one where the player must give away more doors than he gets back in order to have the best statistical advantage. As impossible as this seems, if you truly understand the original version of the problem, it is not so difficult to achieve, as I will explain.

In my version, there are 5 doors rather than 3.

5 closed doors

Mr. Hall (that would me ME this time around) asks the contestant to select any 2 doors from the 5…

5 closed doors, 2 selected

…and then he (I) open 2 of the unselected doors with booby-prizes (hee hee, that’s funny again).  All other parameters of the problem are the same as the original “Monty” 3-door version.

5 doors, 2 selected, 2 of the unselected opened

At this point in the 5-door version, the contestant has selected closed doors #1 & #2, the host (that’s me again) has opened doors #4 & #5 to reveal crap prizes and the unclaimed #3 door remains closed.

The question is this: Will the contestant keep BOTH of the doors he picked OR will he trade BOTH of his doors for the ONE closed door that he did not pick? What would YOU do? (please just PRETEND that you don’t know that the problem is engineered so that the least-intuitive answer is the correct one!).

The answer:  It is in your best interest to switch (of course)

5 open doors

Although the intuitive response of someone unfamiliar with the statistics involved would most likely choose to keep his two doors on the basis that he apparantly has a 2/3 chance of winning, the corrrect answer is to switch his two doors for the one unselected door because in fact he would be trading a 2/5 chance for a slightly better 3/5 chance of winning, as shown in this “replay” where each guitar pick represents a 1/5 chance of the door hiding the prize:

10 unscratched tickets
10 unscratched tickets, 1 selected
10 tickets, 8 scratched, 1 selected and 1 unselected, each unscratched
10 scratched tickets

In the “Jason” case, the chances of winning don’t double like they do in the “Monty” case (making it harder to demonstrate as clearly through trials), but the advantage of switching still exists, because:

One in the bush is worth (more than) two in the hand!

Comments are welcome and encouraged.  However, before you post, please read my Moderation Policy, which I’ve adopted to control Spam.  Basically, if you link to another website AND you do NOT refer to some specific detail about my post or another commenter’s post, your comment will be trashed before it appears, even if you are kind enough to say only, “I like your blog”.  Sorry ’bout that, but the spam-bots have wrecked it for all of us.

2 thoughts on “Monty Hall Problem Pt 3/3: The “Jason Hall Problem”

  1. Hey, no fair, I just came up with exactly this game last week!

    The fun thing is that just like in the original, you can continue increasing the number of doors for ever-wackier results. For example, say we start with 110 cards and you get to take 10. I remove 99 duds and keep the one remaining card. Do you keep your 10 or trade with my 1? The odds are actually the exact opposite of what “non-initiates” would think! (Assuming all the usual caveats.)

    One has partially achieved enlightenment when one realizes that it makes zero difference how many doors Monty opens so long as he always offers you all the “leftovers”. Taking the offer means that, in a sense, you are going to “receive” all those empty doors anyway! (One could make a koan out of that…)

    • Ha! The beauty of publishing: I get to “plant the flag” in an idea! No, I’m sure many others have thought of the 5-card version, but strangely I didn’t come across any others while I wasn’t looking for them :) Besides, as Monty is my name-sake, there is something poetic about naming an alternate version after ME, right?

      Yes, it does scale up as you say…and that is very cool, but the beauty of the lower numbers is in its apparant simplicity. I believe there is an inverse relation between the size of the numbers and the elegance of the problem. People can get their head around 5 cards (or doors), which is 2 less than an American phone number. As such, with several cards the “mark” might either…

      (a) get confused and start using logic that doesn’t even make sense to them, or
      (b) think they are being tricked and don’t respond in the way you need them to (like just being contrary for the sake of it and “lucking” into the right answer) in order to make your point to him or the audience, or
      (c) lose patiance as you spend 1/2 hour flipping over 99+ cards.

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