The trick is in SEEING the problem correctly. With only 2 doors left, one tends to reason, the prize is behind one or the other; THEREFORE the odds are 50/50. But this is an illusion, because you are NOT picking between the door you first chose and the other door.
You are NOT picking between the door you first chose and the other door.
You are picking between
1 THE SET OF THE DOOR YOU FIRST CHOSE, and
2 THE SET OF *ALL* THE DOORS THAT AREN’T THE DOOR YOU FIRST CHOSE
This can be illustrated thus:
You are picking between
1 THE SET OF THE DOOR YOU FIRST CHOSE, and
2 THE SET OF *ALL* THE DOORS THAT AREN’T THE DOOR YOU FIRST CHOSE
This can be illustrated thus:
This can be made clearer by using a SET of 100 doors. You pick 1. It is 1% you picked right. The host opens 98 other doors, and ask whether you want to switch. It is STILL 1% you got it right. It is 99% likely you got it wrong, so you you should switch to the SET YOU DIDN’T PICK:
If you STILL don’t believe me about this, try it empirically. Get a friend to play “door opener" and use a set of 3 and run it about 60 trials where you DON’T SWITCH and then about 60 tries where you DO SWITCH.
After you keep losing 2/3rds of the time by not switching and winning 2/3rds of the time by switching, you’ll be convinced that it is NOT 50/50.
It converges on
not switch = 1/3 win
switch = 2/3 win
It converges on
not switch = 1/3 win
switch = 2/3 win
Or you’ll be convinced I’m wrong about it the problem and you just have REALLY REALLY BAD LUCK since you lose a 50/50 odds thing 66.666666% of the time! 😆
Final footnote:
Is it weird that the breakdown of “not switch” to “switch” was 1/3 to 2/3, exactly what the odds on the doors are?
Is it weird that the breakdown of “not switch” to “switch” was 1/3 to 2/3, exactly what the odds on the doors are?
