3 ms·

I’ve been convinced since ‘91 that they Monty Hall problem will eventually get added to the list.

• Dylan16807 3y ago
  Monty hall is really easy to work out if you consider the 100 door version.
  • wruza 3y ago
    It isn’t trivial if the host has to open 98 or still 1 door though.
    • Dylan16807 3y ago
      I mean the one where you're left with 2 doors.
      • wruza 3y ago
        Yes, but my intuition doesn’t find obvious the jump from open-one to leave-two. These options feel at least equally noteworthy and nothing hints which one you should use.
        • Dylan16807 3y ago
          Well you can also look at opening 95 of the 100. The important takeaway is that the probability is not 1 divided by the number of doors left. And that is true no matter how many doors he opens, if it's a fixed nonzero number. If you ask yourself "at what point did the odds stop being 1 in 100", the answer is obviously not "when he opened the last door out of many" or "midway through opening the many doors", so it has to be when he opened the very first door. And if opening the first door out of 90+ changes the odds, it's hard to see a reason why opening the first door out of 1 wouldn't change the odds.