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One of the best examples of the idea of Bayes is the Monte Hall problem. It is a good example of how prior probability (3 unopened doors) can lead to a more cle

• daveFNbuck 8y ago
  Bayes' theorem also helps clear up some of the subtleties behind the Monty Hall problem. Switching is advantageous because Monty knows which door has the prize. If Monty doesn't know which door has the prize, seeing a non-prize door results in 50-50 odds that your door has the prize. Depending on how you model Monty Hall's prior probability of revealing the prize, seeing a non-prize door can result in the probability anywhere between 0 and 2/3 of switching being advantageous.
  • loup-vaillant 8y ago
    My take on the subtleties, in painstaking details, with source code so you can reproduce my experimental results: http://loup-vaillant.fr/tutorials/monty-hall http://loup-vaillant.fr/tutorials/monty-hall I consider several kind of Monties there, including an "enemy" that will try to open the prize door if he can.
    • daveFNbuck 8y ago
      I like your attention to detail on this! I'm a bit surprised you don't include the overall probability of winning against each Monty with an optimal strategy. I think it's very interesting that the helper doesn't increase your odds beyond 2/3.
      • loup-vaillant 8y ago
        Yeah, that and the spelling mistakes… I think I'll have another pass at this. I just re-read it, and I glossed over the numbers at times.
• 13415 8y ago
  It's better to switch doors according to any statistical method, whether you're a frequentist or Bayesian does not matter. You can also show that you should switch doors by making an exhaustive truth table, by writing a computer program, or by experimentation, if you prefer these kind of approaches.
  • daveFNbuck 8y ago
    A truth table doesn't necessarily get you to the right answer. There are 3 doors I could pick, 3 doors Monty could pick, and 3 doors the prize could be behind. If I make a truth table of all 27 possible combinations, there are 12 combinations where Monty doesn't choose the same door as the contestant or the prize. Of these 12 options, exactly 6 have the contestant choosing the right door and 6 have the contestant choosing the wrong door. You can certainly create a different truth table that arrives at the correct answer, but the truth table approach does not help ensure you get to the right answer like the Bayesian approach does.
    • deleted 8y ago
      [deleted]
    • 13415 8y ago
      No method on earth necessarily gives you the right answer to a problem. You first have to represent the problem correctly, of course. Check out Scenario 2 here https://medium.com/@ProfessorF/visualizing-the-solution-to-the-monty-hall-problem-bf4a08c8ed8f https://medium.com/@ProfessorF/visualizing-the-solution-to-t... for a correct tree.
      • daveFNbuck 8y ago
        Neither method guarantees the right answer, but it's a lot easier to get it wrong with a truth table.
        • 13415 8y ago
          Maybe you're right, I'm not fully convinced. I mentioned a full truth table/decision tree because a long time ago these gave me the insight why switching is the right solution in the standard formulation of the problem, and they also illustrate why the problem is a purely deductive/logical problem whose solution does not require any inductive inference. Then, to me it was a valuable lesson to learn that the Monty Hall problem does not reveal any perceived or real fundamental problem of probability theory.
• platz 8y ago
  This cheats frequentist because in the game the frequentist isn't allowed to change the distribution. With frequentist the trick is always in choosing a distribution. You can't update it according to a rule, but there is no reason why you simply can't pick a different populatuon distribution to operate under