Three Curtains

You are a contestant on a game show. There are three curtains. Behind one of the curtains is a new car. You are asked to choose one of the curtains. Lets say that you choose curtain #1. The host of the show - who knows where the car is so as not to end the game prematurely - opens curtain #3 and there is no car behind it. The host now gives you a choice. You can stay with curtain #1 or you can change your choice to curtain #2. The question now is: would it be to your advantage to stay with curtain #1, or would it be to your advantage to change to curtain #2 or would there be no advantage either way?

It would be to my advantage to boycott this game show. I don't need a new car and I don't want to get stuck paying taxes and insurance on it.

But if you want the car, you have a 50/50 shot either way.

Ergo. No, the car behind the curtain is a Yugo. Your first impression about the boycott was the correct choice.

Wasn't this a real game show once upon a time? 🤔

What would you do if after your pick of curtain #1, and before any curtains were opened, the host told you that you could change your pick to both curtains #2 and #3?

Without knowing what's behind any of the curtains, that would change your odds to two out of three that you get the car. It would be better to pick two and three than stay with one.

I'm still trying to guess where you are going with this stuff. Can you give us a hint?

Deleted

Lynx,
re: "It would be better to pick two and three than stay with one."

What do you know about curtains #2 and #3? You know that at least one of them doesn't have a car behind it. But you say you would switch anyway. Which means that you get to open both curtains #2 and #3. So what is the difference if you open both of them or the host opens one of them for you?

re: "I'm still trying to guess where you are going with this stuff."

It's just a logic problem. I'm not going anywhere with it.

If all three are unknown, the probability of any curtain having the car is one third. If you stay with one, you have a one third chance. If you switch to two AND three, you have a two thirds chance.

If the host already opened three, each of the two remaining curtains has a one in two chance.

[QUOTE="Lynx, post: 4598148, member: 203470"

If the host already opened three, each of the two remaining curtains has a one in two chance.[/QUOTE]


You can look at it this way: Think of it as two areas. Area "A" contains curtain #1 and area "B" contains curtains #2 and #3. There is a 1/3rd chance that the car is in area "A" and a 2/3rds chance that it is in area "B". Before opening any curtains, you KNOW that at least one of the curtains in area "B" doesn't have the car behind it. So by the host knowingly opening a curtain in area "B" that doesn't have a car behind it doesn't change the 2/3rds odds that area "B" still has a car in it. So it is to your advantage to switch to area "B".

Not at all. When the host opens #3, it changes the probability of the remaining curtains. Your area B now has a 50/50 chance, not a two thirds chance.

That's right. Thanks! I knew there was something but I couldn't remember.

Maybe you'll see why you're wrong if you go to the below link.

https://www.mathwarehouse.com/monty-hall-simulation-online/

I ran it three times... and got a car three times. I think that program is rigged.

Still doesn't change simple math though. If one door has been opened, there are two doors left. If there is a car behind one door and there are two doors, the probability of the car being behind a given door is 50% no matter what.

[QUOTE="Lynx, post: 4598161, member: 203470 When the host opens #3, it changes the probability of the remaining curtains. Your area B now has a 50/50 chance, not a two thirds chance.[/QUOTE]

Area "B" contains curtain 2 and curtain 3 which means that there is a 2/3rds chance that the car is in area "B".
You know that at least one of the curtains in area "B" doesn't have the car behind it.
So how is knowing that it isn't behind curtain 3 change the odds that the car is still in area "B"? The only thing that changes is the odds that it is behind curtain 3 which goes from 1/3rd to Zero.

Those odds go somewhere though. They don't just disappear. Specifically, half the odds go to each of the remaining doors.

That is correct. They go to curtain #2, the remaining curtain in area "B" which has a 2/3rds chance of containing the car.

If you want to play it that way, go ahead and have fun. Hope you like goat milk.

Actually I'll take the goat you win. (Do they ever actually give people the goat? Or is it just the "you lose" symbol that they keep and reuse?) I like goat milk. I make homemade boiled custard out of it.

I'd rather a goat than a car, too.