Probability Riddle

If you have 3 doors, 1 which contains a BMW and the other two contain goats. You are told to pick one of the doors, and you do so. For arguements sake, lets say you picked 1, but it really doesn't matter. Next, I open door number 3 and show you a goat. Now, I ask you, do you want to keep door number 1 or switch to door number 2? Why (Prove/Justify your answer)?

And of course the whole point of this riddle is because you get to win the BMW if you guess the door with the BMW, and you get nothing if you get the door with the goat. No, you do not get to keep the goat....

Keep door number one, because u never said that door number one DIDNT contain teh bwm, maybe it did.

Not good enough of an answer. More elaboration is needed.

I would switch to 2, because there was a probabillity that you would get the goat, now it is even, and on the sat tests the answer is usually the middle answers , c and b, f and g...


2 is in the middle... right..?


BTW i say this because the middle is common for the answer to be there... Im probablly totalllly wrong though arent i?




I gave it my best shot :P

If the guy showed you the goat, he probably did it so you'd switch your answer and not get the car that is behind door number 1. If you had picked door number 1 and it had a goat behind it, he wouldn't even bother telling you about the goat behind door number 3.

What if the guy asked you if you want to go to door #2 for the specific reason of making you think he's trying to get you to stray away from door #1?

Anyways... Goats are fuckin loud ... lol so you can tell what door their coming from!

This is a pretty standard question in probability, often referred to as the "Monty Hall question" There are 3 standard arguments:

Argument 1: It does not matter. The probability of finding the car in the remaining two doors was equal in the beginning, and they are still equal now. The fact that you put your hand on one of them cannot increase or decrease its probability of having the car under it.

Argument 2: If we repeated this experiment a million times, you would get the the car only one third of the time by sticking to your first door. People who consistently switch would win the other two thirds. Therefore you should switch.

Argument 3: Think about what you would do if there were a thousand doors, rather than three, and you opened 998 doors with goats behind them.

BAHHHHHHH!!!!!!

Right, very good Doofus, although I'm not surprised someone with such worldly knowledge as yourself would have heard this riddle before.

Right, and in lamens terms:

You pick door 1. That is a 1 out 3 chance. 1/3.
The other two doors combined are 2 out of 3 chance. 2/3.
When door 3 is revealed, in essence, door number 2 now has that same 2/3 ratio chance.
Hence, stick with door 1, you have 1/3, switch with door 2, you have 2/3.

Answer: Switch.

I don't get it.

Door A, Door B, Door C.

1/3 chance of getting the car at this point.

Door C is opened, it's not the car.

Door A, Door B.

It could be in either one. 1/2 chance. Where do you get 2/3's?

Door A, Door B, Door C.
1/3 - A, 1/3 - B, 1/3 - C.
You choose A. 1/3 chance.
That means, B and C have a combined 2/3 chance.
You find that C is definitely wrong.
Therefore, in essence, B now has a 2/3 chance.
Stay with A - 1/3.
Switch to B - 2/3.
Answer: Switch.

The classical answer to this problem is yes, because the chances of winning the prize are twice as high when the player switches to another door than they are when the player sticks with their original choice. This is because upon the original choice, the player has only a 1/3 chance of choosing the door with the prize; this probability does not change when a door opens with a goat. Hence the chances of winning the prize are 1/3 if the player sticks to their original choice, and thus 2/3 if the player switches.

Instead of one door being opened and thus eliminated from the game, it may equivalently be regarded as combining two doors into one, as a door containing a goat is essentially the same as a door with nothing behind it. In essence, this means the player has the choice of either sticking with their original choice of door, or choosing the sum of the contents of the two other doors. Clearly, the chances of the prize being in the other two doors is twice as high.

It may also be easier to appreciate the result by considering a hundred doors instead of just three, with one prize behind only one of the doors. After the player picks a door, 98 doors are opened with goats behind them. Clearly, there's now a very high chance (precisely 99/100) that the prize is in the other door not opened.

Edit:

Yeah, double post... I'm forewarned...

You find that C is definitely wrong. Therefore, in essence, B now has a 2/3 chance.

......What?

It may also be easier to appreciate the result by considering a hundred doors instead of just three, with one prize behind only one of the doors. After the player picks a door, 98 doors are opened with goats behind them. Clearly, there's now a very high chance (precisely 99/100) that the prize is in the other door not opened.

Um.. theres nothing clear about this, it's still 50/50 to me. I'd like to see them do it 1 million times.

There's no reason for the other door to now worth 99 out of 100 while the one you picked is 1 out of 100. The one you picked could be worth the 99/100 and the other one worth 1/100 with that kind of logic.


Also, the obvious answer to this riddle is to get a heartbeat monitor.

Yea, it sounds stupid...

What if the combination was:

Door 1: BMW
Door 2: Goat
Door 3: Goat

?

No, IMO, when a door is opened, the problem is restated anew. You start with 3 doors, one of which is the correct door there is a 33% possiblity of any given door being the right one. You then open 1 of the doors showing that nothing is behind it. You now have 1 door with a 0% possibility of being right and 2 doors each with a 50% chance of being right. The fact that you picked in the first place is inconcequential if they don't open the door you picked.

Possibilities:
1 (no car when they open door 3)
2 (no car when they open door 3)
3 (find car when they open door 3)

If you always pick door 1 and they always open door 3, there is a 1/3 possibility that the car is behind that door. Since you said that that case never happens, the only real possibilities are that the car is behind door one or door two, thus your odds of either being correct is 50%

Yeah, I'll have to go against you on this one titus, it's 50% chance if it's not behind door 3.

I am gonna go with K_A, because this example doesnt work. Probability becomes a bit ineffective with a crooked game show host looking behind the doors. However, shows want you to win, because they don't have to give you the car, the sponsor does, but when people win more people watch the show. I would switch.

WW is my hero. Since he posted, my opinion is further strengthened, because no one reads my posts anyway.

I think the 2/3 possibility stems from the original chances, they are just stating it from that point in the hypothesis.