Brain bender

derbski

derbski

Active member
This is called the Monty Hall problem.A contestant on a game show has to chose between door 1, 2, and 3, and there is a car behind 1 door and there is a goat behind the other 2 doors, and we assume that the placement of the car is random. A player first picks a door, and then the host opens a door with a goat. The player is then allowed to either stay with their original choice, or switch.

Here is the question, and if you cheat and look up the answer, at least let some people guess before you ruin it. Statistically speaking, is a player better off staying with their original choice, switching, or does it not matter at all?
 
i don't think i understand the question because of the way it is written.....why the fuck would you choose to stay with a door that has a goat? of course increasing the the odds from 1:3 to 1:2 is going to be beneficial to the player. but, if you mean the door contains a goat and the player is unaware of this fact, then choosing a different door does not change the players chances of finding the car...it is still 1:3. i'm still confused
 
Player chooses a door. The host opens a door with a goat. Out of the doors that are still closed, one of which the player is standing at, there is one door with a car, one door with the other goat. They can stay where they are at, or they can switch to the other unopened door.
 
you are not getting it. You pick a door, the host opens a door that you did not pick, which has a goat in it. if you picked a goat door, he opens the other goat door, if you picked the car door, he opens either one of the goat doors. After he opens one of the loosing doors, you are allowed to either switch to the other remaining door or stay with your original choice. In either case, all 3 doors will be opened
 
statistically speaking i will take a prize that is not a goat 99.9% of the time. i'm looking for prizes, not chores.
 
It doesnt make any difference. you have to look at each choice as a new situation, not a change. in the first situation, you have 33% chance to guess right, in the second, your previous choice makes no difference, there is 66% chance of getting it right, doesnt matter which you chose for the first situation.

its like flipping a balanced coin, 50% chance heads, 50% chance tails.

there is 1/4 chance of getting heads twice in a row, so on your second throw, you shouldnt bet on heads because its unlikely to bet on heads, its not often you get heads twice in a row right? doesnt matter, that specific throw is still 50% heads and 50% tails.

looking at the entire situation based on grade 12 statistics, you should change your guess. In reality you arent going to be any more likely to win the car.
 
Finally somebody had the same misconception as I initially did.

thats how I was thinking about it too, that it shouldnt matter, at that point it is a coin flip and your chances are 50/50 whether you stay with your original choice or you switch. But when they ran simulations, there actually was around a 16.6% difference, and when somebody wrote about it in a paper, people from all over the country were sending it letters saying how there is no reason it should make a difference, even professors with PHDs.

Think about it this way. When you first pick a door, there is a 1 in 3 chance that you get the car, a 2 in 3 chance that there is a goat. When the first door is opened, it would seem that you have a 50/50 chance 50% chance that your door has the goat, 50 percent chance that your door has the car. But really, opening the first door doesnt make a difference, you know that he will be opening a door when you choose the first one.

SO.... by opting to switch, you are betting that you were wrong the first time, the odds of that being 2/3. It seems counter intuitive that with 2 identical doors, one would have a 66 percent chance and the other would have 33, but that is what the statistics show.
 
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i explained that in my post, trust me we already did this problem in statistics, make a probability tree, and look at the second split. that is the probability of choosing the car or the goat after the first goat has been removed. reread through my post to get a better understanding of exactly what i am saying, i already said everything you did. then i explained why in that instance it doesnt matter.

let me explain this in a better way.

imagine i pick door 1 of 3, they then open door 3 which is reveals a goat, so now i am left with my original choice of door 1, or changing my choice to door 2.

then they say, "sorry you took to long to choose wether to switch or not, NEXT!"

they then take the revealed goat away, hide the 3rd door (the one they opened, now only door one and door 2 are visible from the contestants chair) and bring in a new guy who doesnt know that i ever existed, and they dont tell him that there used to be three doors. instead they tell him that there are 2 doors and that one holds a goat, and the other holds a new car, and he has to choose.

now look at the situation, the 2/3 chance of being wrong that seemed to guide me to change my guess does not apply to this man, he only knows that there are two doors, one with a car and one with a goat.

there is a 50/50 chance for either door to have the car, but that doesnt match up with the 2/3rds chance that door 2 had for me. thats because the two thirds chance only applied to the situation as a whole, my little theoretical explanation breaks the rules of the proposed question, but it shows how statistical analysis and reality don't always agree.

math says switch, the goat says baaaaa.
 
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