Fun Probability problem

[EdZ]

Mathew, I think you are not seeing the difference between a pattern and an event. The probability of a particular pattern can depend on previous events, the probability of a single event does not.

Any single coin toss is 50/50, all day, every day.

A choice between 2 doors is 50/50, all day, every day.

To answer your initial post - is it in your interest to switch - what are your odds of picking correctly?

50/50.

You assume a pattern (event A may/does alter event B), when in this example, there is no relationship between event A and B that justifies viewing it as such, absent any other data.

The 'feel vs real' of statistics.

[Mathew]

Mathew, I think you are not seeing the difference between a pattern and an event. The probability of a particular pattern can depend on previous events, the probability of a single event does not.

Any single coin toss is 50/50, all day, every day.

A choice between 2 doors is 50/50, all day, every day.

To answer your initial post - is it in your interest to switch - what are your odds of picking correctly?

50/50.

You assume a pattern (event A may/does alter event B), when in this example, there is no relationship between event A and B that justifies viewing it as such, absent any other data.

The 'feel vs real' of statistics.

You would be right that you will always end up with two doors - but to say it was 1/2 would be wrong because you fail to look at the process which got you there. One door is really 1/3 chance of being the car and the other door 2/3 chance being the car. Its really that simple....

Just because there is two doors does not mean its 1/2. If I had two boxes and one contained a prize and I put it in one box 1/3 of the time and the other box 2/3 of the time and you knew this, it would no longer be a 1/2 chance.

However you would be correct if lets say after you picked your door and a goat was revealed, you ignored the pass actions and then tossed a coin to determine whether you switched or not. Then it would be a fifty fifty chance.... this would also be true of 100 doors....

Here are some webpages which explain it in further detail.

This is a good page which describes it in detail....

http://en.wikipedia.org/wiki/Monty_Hall_problem

Some stuff on youtube

Here you can try it out yourself

http://math.ucsd.edu/~crypto/cgi-bin/monty2?1+13588

[armourall]

What would you think if I stated it like this...

Original Selection:

1. The player originally picked goat 1.
2. The player originally picked goat 2.
3. The player originally picked the car.

Chances of originally selecting the car: 1 in 3.



Second Selection:

1. The player originally picked goat 1. The door with goat 2 is revealed. The remaining door has the car.
2. The player originally picked goat 2. The door with goat 1 is revealed. The remaining door has the car.
3. The player originally picked the car. The door with goat 1 is revealed. The remaining door has goat 2.
4. The player originally picked the car. The door with goat 2 is revealed. The remaining door has goat 1.

Chances of switching and selecting the car: 2 in 4 (1 in 2)

[Mathew]

The statistics ....

From one of the webpages I just listed

http://math.ucsd.edu/~crypto/cgi-bin/monty2?0+4338

Switched - 319 players - 210 Winners - 65.8%
Didn't switch - 201 players - 72 Winners - 35.8%

[Mathew]

What would you think if I stated it like this...

Original Selection:

1. The player originally picked goat 1.
2. The player originally picked goat 2.
3. The player originally picked the car.

Chances of originally selecting the car: 1 in 3.

Correct...

Second Selection:[list=1][*]The player originally picked goat 1. The door with goat 2 is revealed. The remaining door has the car.

Correct

[*]The player originally picked goat 2. The door with goat 1 is revealed. The remaining door has the car.

Correct

• The player originally picked the car. The door with goat 1 is revealed. The remaining door has goat 2.
• The player originally picked the car. The door with goat 2 is revealed. The remaining door has goat 1.

Chances of switching and selecting the car: 2 in 4 (1 in 2)

Nope - there is only 1 car behind the doors not two....

Just because there are four permutations of what can occur doesn't mean you have four choices. It really doesn't matter if you eliminate goat 1 or goat 2 when you have picked the car. If you switch, you will lose either way....



Make this any easier to understand ?

[armourall]

Just because there are four permutations of what can occur doesn't mean you have four choices. It really doesn't matter if you eliminate goat 1 or goat 2 when you have picked the car. If you switch, you will lose either way....

Then what are the odds of switching and getting goat 1?

[Mathew]

Then what are the odds of switching and getting goat 1?

your chances of picking the car is 1/3

Your chances that goat 2 is shown after picking the car is 1/2

Half of 1/3 is 1/6

Your chances of switching and getting goat 1 is 1/6

[flogger]

When the door is opened to reveal a goat, that choice is not random. There's always a goat behind one of those doors for them to show you and they only choose after you have chosen. So, that has no bearing on the initial conditions. The probability that you did not pick the car with your first pick is 2/3 and that does not change.