4 ms·

I find the Monty Hall problem becomes intuitive when scaled. 100 doors, 1 car, 99 goats. Pick a door. All doors open (with goats except 1) ... Do you think you

I find the Monty Hall problem becomes intuitive when scaled. 100 doors, 1 car, 99 goats. Pick a door. All doors open (with goats except 1) ... Do you think you picked the 1/100 door with the car, or that it's the other only one left standing.

rho4 5y ago
Beautiful short explanation. I always struggle getting this point across in discussions.
- rho4 5y ago
  My hopes for this explanation were too high. I just tested it on my cooworkers. No luck. I reverted to suggesting to write a short simulation script to convince themselves by experimental data.
  - v-erne 5y ago
    I personally find the Generalized Monty Hall problem as unintuitive as the original one. But I belive that I found a batter way to make it intuitive (at least for people that have heard about classic and conditional probability). And it event worked on one of my friends :) If Your first choice was lucky and two goats remained in other gates than Monty does not change anything by showing You the goat in one of them (it't the same as if Monty selected random gate). Things becomes intresting if You was unlucky and selected Goat. Monty MUST show You the other goat (because there is only one available for him to select). And thus he introduces information to otherwise random selection (it stops beeing random). And by doing this he eliminates for You conditional possiblity of selecting second goat when You selected one in first round (and in this conditional scenariu swiching is sure bet which changes the overall odds to 2/3)
Mordisquitos 5y ago
Exactly. It also helps to emphasise that the eponymous host knows which door contains the prize and will never open the prize door before asking you whether you want to switch.
- lovecg 5y ago
  The variants where the host does not know are also worth considering. If they’re just guessing and happened to not open the prize door by chance, you’re back at 50/50. If they have a hangover and you think there’s a 5% chance they forgot where the prize is and still just happened to open the doors correctly, well that makes the math even more interesting.
  - rho4 5y ago
    As long as the host doesn't accidentally open the prize door, it doesn't matter whether he forgot or not. To test my statement you could write a simulation where the host randomly opens a door.
    - lmm 5y ago
      Completely false. Draw up a probability table for the case where the host picks a door at random (so 1/3 times they reveal the prize) and you'll see.
      - rho4 5y ago
        What happens if the host reveals the prize? Is the game repeated? I'm saying as long as he doesn't accidentally open the prize door, it doesn't matter. If you insta-loose in that case, we're talking a different game.
        lmm 5y ago
        > What happens if the host reveals the prize? Is the game repeated? Maybe. Maybe you insta-lose. Maybe you insta-win. It doesn't matter. (But by definition if the host picks a door at random, there is a possibility that they will pick the door with the prize behind it, so something must happen in that case). > I'm saying as long as he doesn't accidentally open the prize door, it doesn't matter. But that's false. If the host picked the door to reveal at random then your chance is 1/2 if you switch and 1/2 if you keep your original door. The 1/3-2/3 case only happens if the host deliberately picked a door that didn't have the prize.

rho4 5y ago

Beautiful short explanation. I always struggle getting this point across in discussions.

rho4 5y ago
My hopes for this explanation were too high. I just tested it on my cooworkers. No luck. I reverted to suggesting to write a short simulation script to convince themselves by experimental data.
- v-erne 5y ago
  I personally find the Generalized Monty Hall problem as unintuitive as the original one. But I belive that I found a batter way to make it intuitive (at least for people that have heard about classic and conditional probability). And it event worked on one of my friends :) If Your first choice was lucky and two goats remained in other gates than Monty does not change anything by showing You the goat in one of them (it't the same as if Monty selected random gate). Things becomes intresting if You was unlucky and selected Goat. Monty MUST show You the other goat (because there is only one available for him to select). And thus he introduces information to otherwise random selection (it stops beeing random). And by doing this he eliminates for You conditional possiblity of selecting second goat when You selected one in first round (and in this conditional scenariu swiching is sure bet which changes the overall odds to 2/3)

Mordisquitos 5y ago

Exactly. It also helps to emphasise that the eponymous host knows which door contains the prize and will never open the prize door before asking you whether you want to switch.

lovecg 5y ago
The variants where the host does not know are also worth considering. If they’re just guessing and happened to not open the prize door by chance, you’re back at 50/50. If they have a hangover and you think there’s a 5% chance they forgot where the prize is and still just happened to open the doors correctly, well that makes the math even more interesting.
- rho4 5y ago
  As long as the host doesn't accidentally open the prize door, it doesn't matter whether he forgot or not. To test my statement you could write a simulation where the host randomly opens a door.
  - lmm 5y ago
    Completely false. Draw up a probability table for the case where the host picks a door at random (so 1/3 times they reveal the prize) and you'll see.
    - rho4 5y ago
      What happens if the host reveals the prize? Is the game repeated? I'm saying as long as he doesn't accidentally open the prize door, it doesn't matter. If you insta-loose in that case, we're talking a different game.
      - lmm 5y ago
        > What happens if the host reveals the prize? Is the game repeated? Maybe. Maybe you insta-lose. Maybe you insta-win. It doesn't matter. (But by definition if the host picks a door at random, there is a possibility that they will pick the door with the prize behind it, so something must happen in that case). > I'm saying as long as he doesn't accidentally open the prize door, it doesn't matter. But that's false. If the host picked the door to reveal at random then your chance is 1/2 if you switch and 1/2 if you keep your original door. The 1/3-2/3 case only happens if the host deliberately picked a door that didn't have the prize.