Simple probability problem

[slayer202]

Originally posted by: BigJ

Originally posted by: slayer202

Originally posted by: kranky
I never would have believed you should switch!

I still don't. if you picked a door other than the open one with the goat, that leaves 2 doors left. 1 has a goat, the other has a car. either one you pick gives you a 50% chance at a car. switching does not change jack ******

At the point when the player is asked whether to switch, there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):

* The player originally picked the door hiding goat number 1. The game host has shown the other goat.
* The player originally picked the door hiding goat number 2. The game host has shown the other goat.
* The player originally picked the door hiding the car. The game host has shown either of the two goats.

If the player chooses to switch, the car is won in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the odds of winning by switching are 2/3. In other words, players who switch will win the car on average two times out of three.

thanks for summing up way too much to read on the wiki page, although i'm sure the detailed explanation is quite interesting, and mind blowing

 

[BigJ]

Originally posted by: Special K

Originally posted by: slayer202

Originally posted by: kranky
I never would have believed you should switch!

I still don't. if you picked a door other than the open one with the goat, that leaves 2 doors left. 1 has a goat, the other has a car. either one you pick gives you a 50% chance at a car. switching does not change jack ******

Let's change the scenario a bit. Suppose there are 10,000 doors, and only one of them has the prize. You intially pick one of the doors. The game show host then opens 9998 of the doors, leaving the door you choose, and the other one. The door you chose has a 1/10000 chance of having the prize. The door the host left has a 1/2 chance. Clearly you would be better off changing your guess.

No. The door that the host left has a 9,999/10,000 chance of being the correct one.

 

[Cawchy87]

Originally posted by: slayer202

Originally posted by: BigJ

Originally posted by: slayer202

Originally posted by: kranky
I never would have believed you should switch!

I still don't. if you picked a door other than the open one with the goat, that leaves 2 doors left. 1 has a goat, the other has a car. either one you pick gives you a 50% chance at a car. switching does not change jack ******

At the point when the player is asked whether to switch, there are three possible situations corresponding to the player's initial choice, each with equal probability (1/3):

* The player originally picked the door hiding goat number 1. The game host has shown the other goat.
* The player originally picked the door hiding goat number 2. The game host has shown the other goat.
* The player originally picked the door hiding the car. The game host has shown either of the two goats.

If the player chooses to switch, the car is won in the first two cases. A player choosing to stay with the initial choice wins in only the third case. Since in two out of three equally likely cases switching wins, the odds of winning by switching are 2/3. In other words, players who switch will win the car on average two times out of three.

thanks for summing up way too much to read on the wiki page, although i'm sure the detailed explanation is quite interesting, and mind blowing

Conditional Probablity isn't that mind blowing, interesting mabye.

 

[JulesMaximus]

Originally posted by: slayer202

Originally posted by: kranky
I never would have believed you should switch!

I still don't. if you picked a door other than the open one with the goat, that leaves 2 doors left. 1 has a goat, the other has a car. either one you pick gives you a 50% chance at a car. switching does not change jack ******

READ THIS:

Increasing the number of doors

It may be easier to appreciate the result by considering a hundred doors instead of just three. In this case there are 99 doors with goats behind them and 1 door with a prize. The player picks a door; 99% of the time, the player will pick a door with a goat. Thus, the chances of picking the winning door at first are very small: only 1%. The game host then opens 98 of the other doors revealing 98 goats and offers the player the chance to switch to the only other unopened door. On 99 out of 100 occasions the other door will contain the prize, as 99 out of 100 times the player first picked a door with a goat. At this point a rational player should always switch.

To drive the point home, one only has to imagine the host having to start with the first door and go down the line, opening the doors but skipping over only the player's door and one other door. The player can then more easily appreciate the randomness of his first choice and the large amount of information he has gained since he made that choice and then see the wisdom in switching.

 

[FoBoT]

the odds don't change just because you open the door, there are still three doors
whether the door is open or shut, one of the three is the winner

read the wiki part about how it isn't like a coin toss

 

[thesurge]

Originally posted by: FoBoT
the odds don't change just because you open the door, there are still three doors
whether the door is open or shut, one of the three is the winner

read the wiki part about how it isn't like a coin toss

Wrong. Because he isn't allowed to open your door to show you whether it is a donkey or a car but rather he shows you a door with a donkey from the other two, the door that he did not open (not your door) gains the probability of the door that he did show you. So it becomes 66%.

Basically, the 33% of the other door merges with the 33% of the unopened and unpicked door (since the host automatically eliminates one of those two doors from the problem).

 

[DrPizza]

What's kinda funny, at least to me, is that I already have 3 cars.
I raise goats; I'd be just as happy winning the goat.

 

[JujuFish]

Originally posted by: nycxandy
Yes... and did you get it from Numb3rs?

The Monty Hall problem is older than dirt.