How smart is Yas Forums?

100%

Optimal is defined as "best or most favorable."

The most favorable action for each player getting the correct door. 100%. You're wrong.

No. I'm OP. I'm right. It's my game.

It isn’t though

I love them. The Holy Grail film was the best.

It's not your game. Reword it properly if you don't want people to easily solve it.

it is
it really obviously is

>winning probability if all players act optimally
>optimally

102% then

I guess you are simply an obvious ignorant fool

Just because there are 3 doors you automatically think Monty Hall problem? This is not the same, you are a moron

no, he's right
this is monty hall
op just tried to hide it

(2/3)^3 = 8/27

He tried to hide it in such a way that he completely broke the premise, and therefore also what made it the Monty Hall problem

I see what you are trying to do, but you fucked up OP, you are stupid. The way you wrote it, it’s 100%

no, read the question
green guy has to find green car and can open two doors
if the first door has it then he's done and everyone wins
if not then he has to decide which door to try next
this is the equivalent of the presenter showing him a door without a donkey

Do they have to open both in the same turn? Or can they open 1. And open the second in a later turn?

This riddle boring. As long as they don't leave a door unpicked, the answer is 33% no matter which combination of doors they decide to open. If they all ignore one of the doors it's 0%. There isn't any optimal start besides make sure every door is picked at least once.

In the original Monty Hall problem the contestant chooses a door, he doesn’t open it. Then presenter opens just one door

>he may open two of the three doors. If he finds the red car, the doors are closed again.

Does this mean that he can open 1 door first, and if his car is there, the door is closed again and he can open his second and leave it open, or must he open them at the same time?

Every contestant has to win in order for everyone to win. They have no info on any door. If one wins, the next guy also gets three doors with no info. 2/3 for one guy to win. If won, the next has a 2/3 chance to win, and also the last. So 2/3 * 2/3 * 2/3, or (2/3)^3, which equals to 8/27, or roughly 29,6%.

He can open the doors one at a time.

You're wrong, the answer is 50%.
Essentially, you say that what you do is based on what is behind the first door.
Everyone opens d1 at the start of their turn.
If d1 is blue > r go to d2, g go to d3
If d1 is red > g go to d2, b go to d3
If d1 is green > b go to d2, r go to d3.
This way it is only one decision being made, a binary choice between door 2 and 3, made by one person. In the case of d1, the person with that color finishes correct immediately. With d2, if you are incorrect, then the other guy is incorrect, too. If you are correct, then he will be, too. Thus 50/50 shot of winning or losing based on conditional probability.

This.
OP wanted to be smart, but could not hide he is in fact a fag.
It is not Monty hall. And here is no strategy. You just randomly chose with probability 2/3 3 times

See

Shit, you’re right. I fucked up