@rjquillin well, in short, if there are 3 doors, initially the doors have a 1/3d chance of winning each. Also, door 1 has 1/3rd chance and doors 2 and 3 combined has 2/3rd chance.

Now that you know door 3 is not a winner, door 1 still has a 1/3rd chance and door two still has 2/3rds chance. It is better to switch to door 2 because it has a better chance of winning.

@chipgreen given that the choice of what door to reveal isn’t independent of the prize (i.e. they can’t open door #3 if it has the prize) they don’t reset to 50:50.

@chipgreen@radiolysis@rjquillin
How likely were you to be right the first time? 33%
Did the prize move after revealing the third door? No
The chance you guessed right the first time is still 33%.
Therefore the chance the prize is behind the other door is the remainder: 67%

When I tried to explain this to my dad the first time it took an entire weekend…

@chipgreen@klezman@radiolysis
Perhaps your dad and I would get along just fine.
For me, that 1/3 that just got eliminated just equally adds to the two remaining, making them both 50%.
Why would it only add to one? You now have two doors, equally likely to have the prize. The fact you initially picked 1 instead of 2 have no effect on 3 being reviled.

@chipgreen@radiolysis@rjquillin
OK, make it 100 doors, only one of which has a prize.
You choose door #1. (I doubt you’d argue that was anything other than 1% chance of being right.)
Monty opens doors 3 through 100, revealing nothing, and leaving only doors 1 and 2, one of which you know has the prize.
Did the likelihood of you picking the right door jump from 1% to 50% just because other doors got eliminated?
Or, put another way, the chance you guessed wrong was 99%. There’s nothing you can do to change that. So the 1% right/99% wrong guess you made the first time remains the same. That means that the other remaining doors - in their entirety - have a 99% probability of being correct. But now there’s only one door. So you’ll be correct 99% of the time if you always switch.

Your chances are better if you change your choice. But don’t tell that to the woman whose favorite number is 1! https://en.m.wikipedia.org/wiki/Monty_Hall_problem

@kshannon1

why?

@rjquillin A Wikipedia article was linked along with that statement. Did you read that article? All your questions would be answered.

@cengland0 tl:dr (all of it), attention span at 21:30 too short.

Guess I’m not in the 13% group.

And, way more questions than I had.

@rjquillin well, in short, if there are 3 doors, initially the doors have a 1/3d chance of winning each. Also, door 1 has 1/3rd chance and doors 2 and 3 combined has 2/3rd chance.

Now that you know door 3 is not a winner, door 1 still has a 1/3rd chance and door two still has 2/3rds chance. It is better to switch to door 2 because it has a better chance of winning.

@kshannon1 @rjquillin Ron, I figured you’d have known about these problems for ages!

@cengland0 @rjquillin

Seems to me that once door #3 is eliminated, doors #1 and #2 would each have a 1/2 chance of being the winning door.

@chipgreen given that the choice of what door to reveal isn’t independent of the prize (i.e. they can’t open door #3 if it has the prize) they don’t reset to 50:50.

@radiolysis

Thanks, I get it now. Interesting…

@chipgreen @radiolysis

But how does that skew door 2 to be more likely?

It’s still one of two…

@chipgreen @radiolysis @rjquillin

How likely were you to be right the first time? 33%

Did the prize move after revealing the third door? No

The chance you guessed right the first time is still 33%.

Therefore the chance the prize is behind the other door is the remainder: 67%

When I tried to explain this to my dad the first time it took an entire weekend…

@chipgreen @klezman @radiolysis

Perhaps your dad and I would get along just fine.

For me, that 1/3 that just got eliminated just equally adds to the two remaining, making them both 50%.

Why would it only add to one? You now have two doors, equally likely to have the prize. The fact you initially picked 1 instead of 2 have no effect on 3 being reviled.

@chipgreen @radiolysis @rjquillin

OK, make it 100 doors, only one of which has a prize.

You choose door #1. (I doubt you’d argue that was anything other than 1% chance of being right.)

Monty opens doors 3 through 100, revealing nothing, and leaving only doors 1 and 2, one of which you know has the prize.

Did the likelihood of you picking the right door jump from 1% to 50% just because other doors got eliminated?

Or, put another way, the chance you guessed

wrongwas 99%. There’s nothing you can do to change that. So the 1% right/99% wrong guess you made the first time remains the same. That means that the other remaining doors - in their entirety - have a 99% probability of being correct. But now there’s only one door. So you’ll be correct 99% of the time if you always switch.