3 ms·

Right, good luck with that. 2 doors to choose from, 1 price, 50% chance of winning, period. And I don't care if there were initially a million doors to choose

• the_gipsy 5y ago
  The missing piece is: have the doors been allocated before the reveal and allowing the switch? If yes, then the chance is 2/3 for switching.
• ColinWright 5y ago
  If there are two doors, and one prize, and you can choose one door, then you're right, it's a 50% chance. That's not the situation, and the chance is not 50%. I don't care if you agree or not, you are mistaken if you think the chances are 50:50.
  • codr7 5y ago
    But that is the situation, there are exactly two doors and one price left.
    • ColinWright 5y ago
      We are either talking past each other, or you are missing the point. In the Monty Hall problem it's not 50:50. Really. It's not. * There are three door, one conceals a prize; * You choose one door; * The host, knowing where the prize is, opens one of the other doors to reveal a non-prize; * You have the option to stick with your choice, or to switch; * If you stick, your probability of having the prize is 1/3; * If you switch, your probability of having the prize is 2/3. There's enough discussion elsewhere, or you can write your own simulation (but make sure you simulate the right thing, and not something different). As others have pointed out, you need to know in advance that (a) the host will always open another door, (b) the host knows where the prize is located, and (c) the host will never reveal the prize. But given that, the odds really are not 50:50.
      • codr7 5y ago
        I am aware, and at one point I accepted the whole thing as is; but the older I get the more I feel like it doesn't make any sense to me, theories and proofs be damned. I just can't see the point in switching; opening the door provides new information, but that information applies equally to both remaining doors. Sorry if that makes people upset, but what am I supposed to do?
        • ColinWright 5y ago
          > opening the door provides new information, but that information applies equally to both remaining doors. This is intuitively obvious, but where it goes wrong. You've chosen door A. Doors B and C remain, at least one hides a goat, and the host knows where everything is. They can always open one of B and C, so the fact that they do so tells you nothing about A. But if they open B it tells you they didn't open C, and vice versa. So when they open a door they are telling you nothing about the door you chose, and something about the other unopened door. > ...what am I supposed to do? You can try writing a simulation and watch it work, but another way to think about it is this. Suppose you choose one door, put your hand on it, then close your eyes. Now you're given the choice of sticking, or switching to both the other doors. Clearly it's to your advantage to switch. Now with your eyes still closed the host opens one of the other doors. You can still stick, or switch. In effect, switching lets you choose both of the other doors, but you can ignore the one the host opened. Does that help?
• Biganon 5y ago
  How is this the top comment? Someone with the hubris to think that their intuition is better than dozens of proof of the Monty Hall paradox? Or maybe I missed the joke
  • Jensson 5y ago
    The comic didn't fully describe the problem so there is no correct answer. Like 99% of people who think they understand Monty Hall don't really understand it. If you can't spot when the problem isn't accurately described then you don't understand it.
  • zargon 5y ago
    It was likely because HN often puts new comments at the top of the thread for a few minutes to give them exposure. (I suspect it probably puts different new comments at the top for different visitors.)