3 ms·

the "trick" Monty Hall problems are another good one here: https://twitter.com/colin_fraser/status/1628461980645462016 https://twitter.com/colin_fraser/status/1

by rmorey 4y ago

the "trick" Monty Hall problems are another good one here: https://twitter.com/colin_fraser/status/1628461980645462016 https://twitter.com/colin_fraser/status/1628461980645462016 Apparently GPT-4 gets this one right!

ionwake 4y ago
Tbh I still can barely get my head round it even after coding a working solution.
- function_seven 4y ago
  I played role of Monty with another person. They chose a door, I revealed a different door, and they switched (or not). That got my head around it. Most of the time the door I "chose" to reveal was no choice at all. There was only one door I could open.
- dwaltrip 4y ago
  The best way to understand it: Think of a version with 100 doors, 99 with goats and 1 with a car. You choose a door, and the host opens 98 doors that have goats. Do you keep your randomly chosen door or switch to the single door that the host didn’t open?
  - 8note 4y ago
    The big insight is that at each attempt, the remaining door could have been chosen if it had a goat, but wasn't It would be extremely lucky for that door to have not been selected 98 times in a row and still have a goat
- jpoprocks 4y ago
  I had it explained to me like this: You pick one of three options, giving you a 1/3 chance of being correct, 2/3 odds you picked incorrectly. The host removes an option and give you the option to switch. Your options then are - Keep the same door: you win 1/3 of the time (your first guess was right) Switch doors: you win 2/3 of the time (your first guess was wrong) It really just comes down to, do I think I was right the first time, which was 1/3 odds, or wrong the first time, 2/3 odds.
- pynappo 4y ago
  Here's how I've explained it: Choose randomly between 3 doors. 1/3 of the time you end up with the door with the car, and switching loses. The other 2/3, you pick a door with a goat, the other door with the goat is eliminated, and switching wins. Basically, P(lose when switching) = P(choosing correct door at first), and P(win when switching) = P(choosing any incorrect door at first).
- sireat 4y ago
  Another way to verify the solution is to code a Monte Carlo simulation of Monty Hall This can be done with ChatGPT. No code corrections needed.
tomp 4y ago
GPT-4 gets it. https://twitter.com/tomprimozic/status/1635720278578692152 https://twitter.com/tomprimozic/status/1635720278578692152

by rmorey 4y ago

ionwake 4y ago

Tbh I still can barely get my head round it even after coding a working solution.

function_seven 4y ago
I played role of Monty with another person. They chose a door, I revealed a different door, and they switched (or not). That got my head around it. Most of the time the door I "chose" to reveal was no choice at all. There was only one door I could open.
dwaltrip 4y ago
The best way to understand it: Think of a version with 100 doors, 99 with goats and 1 with a car. You choose a door, and the host opens 98 doors that have goats. Do you keep your randomly chosen door or switch to the single door that the host didn’t open?
- 8note 4y ago
  The big insight is that at each attempt, the remaining door could have been chosen if it had a goat, but wasn't It would be extremely lucky for that door to have not been selected 98 times in a row and still have a goat
jpoprocks 4y ago
I had it explained to me like this: You pick one of three options, giving you a 1/3 chance of being correct, 2/3 odds you picked incorrectly. The host removes an option and give you the option to switch. Your options then are - Keep the same door: you win 1/3 of the time (your first guess was right) Switch doors: you win 2/3 of the time (your first guess was wrong) It really just comes down to, do I think I was right the first time, which was 1/3 odds, or wrong the first time, 2/3 odds.
pynappo 4y ago
Here's how I've explained it: Choose randomly between 3 doors. 1/3 of the time you end up with the door with the car, and switching loses. The other 2/3, you pick a door with a goat, the other door with the goat is eliminated, and switching wins. Basically, P(lose when switching) = P(choosing correct door at first), and P(win when switching) = P(choosing any incorrect door at first).
sireat 4y ago
Another way to verify the solution is to code a Monte Carlo simulation of Monty Hall This can be done with ChatGPT. No code corrections needed.

tomp 4y ago

GPT-4 gets it. https://twitter.com/tomprimozic/status/1635720278578692152 https://twitter.com/tomprimozic/status/1635720278578692152