Wednesday, 4 March 2015

El Monty's Three Amigos

Please note that since I wrote this article, I have been persuaded that the argument it relates to is wrong (meaning that chrysics and Mathematician and irishsultan and ChalkboardCowboy were all right from the start and I should have listened to them rather than arguing with them).  Fortunately, I didn't because, for me at least, this little intellectual journey has been far more interesting than it would have otherwise been.

The correct answer for the scenario as it is worded is not 1/2 but rather 1/3 (meaning that the likelihood of winning as a consequence of staying is 2/3).


One of my interlocutors recently encouraged me to learn more of conditional probability.  While I was pretty sure that I knew a bit about it, this didn’t stop me from trying to expand my knowledge.  While I was fossicking around the internet I came across an interesting and related little scenario, one that I have modified below:

Say that Dusty, Lucky and Ned are booked to perform for El Monty, who is having a Very Bad Day.  At the end of the performance, El Monty calls over one of his henchmen and says: “I like these guys, these are funny guys!  Just kill two of them!”

The henchman asks which two and El Monty points out that he doesn’t care: “Just pick two of their names at random.  But don’t tell them which ones!  We don’t want them to be sad on their last night, do we?”

The Three Amigos are then led away into the dungeon and thrown into a cell.

Because Lucky is played by the better comedian, they all agree that he should survive.  They come up with a plan to swap nametags once they have wheedled some information.

Lucky then wanders up to the cell door and strikes up a conversation with the guard who promptly refuses to tell Lucky whether he is going to get shot in the morning.  However, after Lucky performs some great gags and signs an autograph for the guard’s niece, the guard agrees to provide the name of one of the other Amigos who is definitely going to be executed in the morning.

Lucky goes back to the other guys and tells Ned the sad news.

Should Lucky and Dusty swap nametags?  Why?

Before they get the chance to decide themselves, however, El Monty is suddenly at the door: “¡Stupido!  ¿Did I not tell you, do not tell them?”

El Monty then takes out a gun and shoots Ned through the forehead, killing him instantly.

Should Lucky and Dusty swap nametags?  Why?

Now, how is this related to the Reverse Monty Hall Problem?

Lucky knows that there is a 2/3 likelihood of being executed (equivalent to a pre-existing likelihood of 2/3 of having a goat behind any door)

The guard refuses to tell Lucky anything about his fate (making Lucky equivalent to the unselected door and making Dusty and Ned equivalent to the selected doors)

The guard identifies Ned as one of the two to be executed (equivalent to being told that a goat is behind a specific door)

El Monty kills Ned (making the probability that Ned will be executed 100%, equivalent to knowing with absolute certainty that there is a goat behind the door, because it's been opened)

Lucky can swap nametags with Dusty (equivalent to switching doors)

So, put it another way, if you are Lucky – do you switch your nametag with Dusty (who is eager to die so that you may live)?  What is the likelihood of you surviving if you do switch?

If you think that this is somehow not equivalent to The Reverse Monty Hall Problem, feel free to explain how.


  1. Hi

    I agree that this is equivalent to the Reverse Monty Hall problem, but I think your analysis as to why is a bit off.

    The two notes pulled from the bag is equivalent to the two doors chosen. You don't know which names are on the notes, which are equivalent to the fact that you don't know what is behind the doors.

    The fact that the guard refuses to say anything about Luckys fate makes Lucky equivalent to the car, as the host in the Reverse Monty Hall problem will never reveal the car.

    This makes Ned and Dusty equivalent to the goats.

    The rest seems fine.

    The probability should not change just because El Monty kills Ned. Either Luckys name is on the other note, or it's not. Unless El Monty decides to re-randomize who gets to live, the fact that Ned dies changes nothing.

    So the answer should be the same as the Reverse Monty Hall problem. There's a 2/3 chance that Luckys name is on one of the two original notes, so by swapping with Dusty he should reduce his chances of dying to 1/3.

    "But", I can hear you say, "But, there would also be a 2/3 chance that Dusty's name was on one of the two first notes. Why do Luckys probability 'trump' dustys?".

    The guard refuses to speak about Luckys future, but makes no such promise about Dustys. This is what is important here. We can again write up all possibilities:

    Let N be Ned, L be lucky, D be Dusty. DN implies Dusty and Ned will die, etc. DN is the same as ND.

    There are 3 possible outcomes:


    After speaking to the guard, the guard will randomly blurt out one of the victims, UNLESS that victim is Lucky, because he refuses to talk about Luckys fate. The possibilities are

    ND : Guard blurts out Ned. Probability 1/3 * 1/2 = 1/6
    ND : Guard blurts out Dusty. Probability 1/3 * 1/2 = 1/6
    NL : Guard blurts out Ned. Probability 1/3
    DL : Guard blurts out Dusty. Probability 1/3

    Now we remove the impossible situations and we're left with an original 1/6 chance of Dusty being the second doomed one, and an original 1/3 chance of Lucky being the second doomed one. We normalize these so that we get the sum of probabilities to be 1, and we get the wanted 1/3 chance of Dusty being doomed and 2/3 chance of Lucky being doomed.

    1. First, survival has to be the car, it's the "prize" that Lucky is after. I think each person represents a door because they each have a fate attached to them, be it execution (a goat) or survival (the car). Grouping them might be a little more vague.

      Second, we (together with Lucky) knew all along that at least one of his Amigos is going to be executed.

      Because you have made Lucky the car (erroneously in my opinion), your possibilities are incorrect. But the numbers do work out the same anyway, if we use your approach:

      Lucky slated for survival:

      50% chance that Ned is identified
      0% chance the Dusty is identified

      Dusty slated for survival:

      100% chance that Ned is identified

      Ned slated for survival:

      0% chance that Ned is identified

      Therefore, using this approach, it is twice as likely that Ned would have been identified if Dusty was going to survive, so we would have a 2/3 likelihood of Lucky being doomed.

      I'm not convinced by this approach though (although that might just be exhaustion!)

    2. Woops, I added a new comment instead of replying to this. Oh well.

  2. Survival is indeed the prize, but the prize is not automatically the car. Somethimes you win a goat. And sometimes Dusty survives.

    Lucky has to be identified with the car, as they share the quality that they will never be revealed until the end. The probabilities should of course be the same even if you talk about the problem in a different way, as long as you're describing the same scenario.

    1. When I write "prize", I mean the intended prize. The goats (and execution) aren't really prizes, they are just humorous mechanisms for telling you that you didn't get the prize.

      Survival shares the characteristic of "never being revealed until the end", irrespective of whether it is assigned to Lucky or Dusty.

      The reason why I am not convinced by this approach is because while the guard and El Monty act as the host (El Monty somewhat more forcefully), Lucky actually initiates the encounter. When he does so, Lucky knows that at least one of his Amigos is scheduled for death, even if he does not know which one, and that each of them has a 2/3 of being executed (as is Lucky himself). Equally, he knows that each of the Three Amigos has a 1/3 likelihood of escaping execution.

      When Lucky cajoles the guard into revealing that one of his cell-mates is going to be executed tomorrow, he is not really getting any new information about his own chances. For this reason, I think that it is possible that this scenario might be more akin to Monty Faffs About rather than the Reverse Monty Hall Problem. I understand the argument about the random blurting out of Ned's name, but equally - the position that the guard is in as to whether he can safely blurt out either name or is obliged to blurt out Ned's is dependent on the status of Lucky, and the status of Lucky was randomly assigned.

    2. Lucky knows that at least one of his Amigos will die, the same way the contestant in the RMH problem knows that one of their doors contain a goat. The scenario is exactly the same.

      Imagine instead the scenario where you get NO information from the guard. You could still swap nametags if you wanted, but as you don't know which one of your amigos is destined to die, you might just end up swapping with him. As indeed you will 50% of the time. What you learn is which one of your companions you DEFINITELY SHOULD NOT swap nametags with. This is information that you cannot just ignore.

    3. I agree that we learn which Amigo we should not swap our nametag with. I went away from the computer after writing my last comment and thought about it some more, specifically with respect to the point that the guard is, to an extent, controlled in his response by the status of Lucky.

      There is a 2/3 likelihood that Lucky will be slated for execution and, if so, the guard will have only one other Amigo that he can identify. There is a 1/3 likelihood that Lucky will not be slated for execution and in this case the guard could choose either of the other Amigos to identify.

      I am coming around (rather swiftly now) to the idea that I am wrong, not only on this but also with respect to the Reverse Monty Hall Problem and the classic Monty Hall Problem.

      The two things that remain very interesting to me are: 1) what is shown in Monty Does Play Dice because while that article did get a few comments (10 that weren't me), there was no real attempt by anyone to engage with what I was getting at there (although I think that chrysics' comments are pertinent), and 2) the link between Two Balls One Urn and The Reverse Monty Hall Problem. I do have a couple more articles up my sleeve, but as my argument is now crumbling around me, I will have to re-write them and examine their arguments from the perspective of having thought they were convincing but now understanding that they were wrong.

      And, if I do settle on the "I was wrong" position, then I'll have to sit down with those people who I did convince and attempt to convince them that I was wrong and that they were wrong to have been convinced by me. But to do that, I would first have to get my head firmly around why the dice are wrong.

      In answer to ChalkboardCowboy's comment below, the good is that the process of going through different perspectives allows us to see the problem in different ways (although that is primarily "me", rather than "us"). This should not be an issue to anyone, if the answer is concrete and unassailable no matter what perspective you look at it from we should come up with the same answer. Also, you're not obliged to prove me wrong, and at the end of the day, if I am wrong, I really need to prove that to myself - even if others can help along the way. Finally, if I settle on being wrong, I will be very curious as to precisely what it was that convinced me that I was right when I was wrong (at this stage, my money is on the dice).

    4. I can't really see why the dice scenario is any different from any of the other scenarios you have cooked up. The important detail is always that while we have now whittled all the possibilities down to two, they are not equally likely. This is true independent on whether it's dice, goats, amigos, or swedish meatballs.

      I can't see really see any direct correlation between the urn question and the Monty Hall question, except that they both use conditional probability.

      I think it's important to note that the Monty Hall problem is well known BECAUSE it gives a non intuitive result, so that your intuition was wrong on this is kind of expected. But where I convinced a friend by repeatedly playing this game with him, so he noticed that he won more often by switching, you have created some kind of disconnect between the probability and the "real world", making it harder to find common ground.

    5. The thing is that there are circumstances in which 1/2 is the right answer - Monty Falls is an example. I think the dice might be equivalent to a Monty Falls, and it's possible that the same applies to Two Balls One Urn.

    6. I read through the comments on the urn problem. The difference, which (unless I've misunderstood) I think you agree with, is that the second ball is randomly revealed to be white, the same way Monty randomly reveals a door with a goat behind it in the Monty Falls problem. I feel the Monty Dice problem is not described with enough rigor for me to say which one it is equivalent to. Not sure what you want from it.

    7. There isn't a separate "Monty Dice problem". I was just using the dice to role play the Reverse Monty Hall Problem. As I said above, chrysics comments are pertinent, because when I have done the dice-based role play for people, I've not gone out of my way to hide the dice, because it was the principles I was after, not the actual fun of playing the game. However, the consequence of that was that I and the person I was talking with became as gods. We knew where the "goats" and "car" were. This changed the situation to the one that chrysics (correctly in my revised opinion) identified as resulting in a 1/2 likelihood of benefitting from a switch. What I was not correctly modelling for these people (if I am now correct, instead of just being wrong in a different way) was the situation in which the contestant found herself.

      What I would like to do now is to point out that I am crediting chrysics for what he actually did here. There was plenty that he wrote and said along the way that was very valuable.

  3. What good will it do, at this point, to introduce yet another scenario, abandoning all the terminology and notation that we've collectively developed over the last 10-odd days this has been going on? It's just sowing confusion, which I assume is at least ostensibly not your goal.

  4. > One of my interlocutors recently encouraged me to learn more of conditional probability. While I was pretty sure that I knew a bit about it, this didn’t stop me from trying to expand my knowledge. While I was fossicking around the internet I came across an interesting and related little scenario

    Wow, I did not think that I could be misunderstood on such a simple sentence.

    I did not want you to read about new scenarios, new games, new problems, so that you could make other mistakes ...

    I wanted you to focus on the formalism. To learn how a probability is modeled in math, to understand what independence of random variables mean, to understand that in math the same universe can be given different probability measures (like Monty Hall and Monty Fall. Same universe, different probability distribution). So that once you know the vocabulary, you could express yourself correctly and we could have an educated discussion.

    I didn't want you to lose time posting another blog post on another related problem where you can show once again that you do not know enough of probability theory to make a meaningful answer ... and spend countless hours answering every objection with the same argument.

    Because for me, hearing this sentence hurts :

    > I'm not convinced by this approach though

    What you are doing basically is spitting in the face of hundreds of mathematician who devoted their entire life to introduce a formalism that makes everything easier and unequivocal. Right now, you lose time doing hand-waves and saying "I'm not sure that this ... I'm not convinced by that ...", while you could have been convinced 2 weeks ago by my complete formal answer (that is still unanswered) if you just learned the formalism.

    1. I accept that it's likely that I am wrong, Mathematician, but while I remain unconvinced would you prefer that I just pretend to be convinced, to fake it until I make it? I have been forced to study statistics formally, albeit a few years ago now, and I do understand (if not perfectly) probability. As I have said elsewhere, until quite recently I was totally and absolutely convinced that 2/3 was the correct answer. I've been there already and it's entirely possible that after this short detour of doubt, I will end up there again.

      I'm sorry that my personal lack of utter conviction hurts you. It was not my intent.

  5. I have not read over previous posts because that would be too much effort. This post however has the same probability as the original Monty Hall problem only a different objective. You could have saved yourself a lot of typing by just changing the original problem to the contestant needs to pink a goat instead of a car because then he will get a million dollars.So the answer is don't change because in your original position you have 1/3 chance of picking the car, if you swap there is 2/3 chance you will pick the car, and you don't want to pick the car so you go for the lowest probability of getting it.`

  6. An important conditional in the Monty Hall problem is that the game show host knows which door the car is behind. This means that no matter which door he opens (even if there are a thousand doors and he opens 998) it never changes your original probability chance of picking the car.


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