Fun Probability problem
 

Imagine you have three doors....



Behind one of these three doors is a car and the other two are goats. Your goal is to win the car. After you have chosen your door, a goat on one of the other doors will be shown. Would it be in your interest to switch doors at this stage?

What is your chances of winning ?

If you stick with your original choice your chances remain 1 in 3.

If you reselect... 1 in 2.

Do I get the car?
:)

Nope...lol

:)

The solution....
 

The fact is that your chances of chosing the goat is 2 in 3 and the car 1 in 3.

If you choose either goat, the other goat will be eliminated and if you then choose to switch - you will get the car. There is 2 in 3 chance of this occuring

If you initially pick the car, switching will come out incorrect. There is a 1 in 3 chance of this occuring.

Therefore switching doubles your chances from 1 in 3 into 2 in 3.

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Still confused? This is far more apparent of a case of 100 doors. Your chance of getting the first one right is 1 in 100. If once you had selected your door, 98 were then opened and eliminated, it would be far more likely that the other door left would be the correct door. The correct strategy would then be to switch....

Wrong as far as i see it.

They are mutually exclusive events...the outcome of the first goat being revealed does not change the second result...you have a 50% chance of being correct either way once the other doors are revealed.

With your 100 door example, once the other 98 are revealed you are still left with a 50% chance of either door being right...the fact that incorrect doors have been removed from the equation means nothing...in choosing between the final 2 doors switch or stay means nothing, the door you initially chose becomes irrelevant.

The chances that you picked the correct door out of 100 is 1/100. They then eliminate 98 incorrect ones, leaving 2 doors. At this point, there is a great chance - 99/100 - that the car is behind the 'other' door. Your original guess still has a 1/100, or 1%, chance of being correct.

Same obviously works for 3 doors. Your 'guess' has a 1/3 chance of being correct, meaning there's a 2/3 chance it's incorrect. When they eliminate an incorrect door, your chances with the original door remain at 1/3. However, the 2/3 that before were assigned to the other two doors now are assigned to the only remaining door. So you have a 2/3 chance if you switch, and the same, original 1/3 chance if you stay.

I see and understand the logic you are trying to use, but I think it is wrong.

The fact one door was preselected has NO bearing on which door has the car behind it.

If you walk into a room with 3 doors, 1 is open showing a goat, and the other 2 are closed, you then have to pick a door knowing a car is behind 1, what is your chance of picking the car?

I say 50%...

This is exactly the same scenario, the fact that there was a previous action and door reveal does not change the probabilites related the this event...like I said look up Mutually exclusive...

It is like a coin toss, if you toss a coin and get a head 5 times in a row there is still a 50% chance the next will be a head, the previous events don't change the probabilities involved in this event.

This called the Monty Hall Problem or Dillema. It is an example what is logical is not always instinctive.

Think of it like a process tree

Goat 1 - Goat 2 is shown - switching gets car.
Goat 2 - Goat 1 is shown - switching gets car.
Car - Either goat is shown - switching loses car.

Therefore you get into a situation where you have a choice between the two doors and two times out of three it is the other door than the one you originally picked. Therefore you double your chances by switching.

If you would like to bet $4.50 (you give me when I win) to my $5.50 (I give you when I lose) then I would be willing to accept (repeated 100 times to counteract any variance). If your right and its a cointoss - you will be in positive expectation. If I'm right I will be...lol

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

http://www.youtube.com/watch?v=mhlc7peGlGg

Here you can try it out yourself

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

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)

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%

Correct...

Correct

Correct

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 ?

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

..opening the door is not a random choice..
 

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.

Again, look at it with 100 doors...99 goats and 1 car.

You get to pick one door. The chances of picking the car are 1/100.

They open the other 98. You can either stick with your original choice or switch to the one remaining door.

Your original choice still has a 1/100 chance of being the car. Agreed?

The chances of the one other remaining door having the car...99/100?

Just because you have two choices doesn't make it a 50/50 proposition.

Matthew,

It's been fun, but the snow is starting to melt and it's time to start thinking golf. I'll admit I was already familiar with this problem and its "correct" answer a long time ago (I think it was in a column by Marilyn vos Savant), but it's still fun to argue the "wrong" side of an argument. Sorry I couldn't challenge you more.

Cheers.

Thats cool :)

Maybe I should keep this going...

Heres another probability problem

What is the chances of being dealt the Ace of spades and the Ace of diamonds when playing texas hold'em?

Say after the flop comes with 3 spades showing. What is the chances of making a flush with the turn card, the river card and both?

Lets say there were $2500 in the pot preflop and your opponent moves all in after the flop and has you covered, and you have $2100 left. Would calling this bet be profitable over the long run?

Not that familiar with Hold 'em, so I'll sit this hand out. Just to finish up on Monty, here is what I consider to be the correct rundown of the probabilities: (My previous "flawed" argument conveniently excluded possibilities #4 and #6.)

1. The player originally picked the car. The door with goat 1 is revealed. The remaining door has goat 2. (1/6)
2. The player originally picked the car. The door with goat 2 is revealed. The remaining door has goat 1. (1/6)
3. The player originally picked goat 1. The door with goat 2 is revealed. The remaining door has the car. (2/6)
4. The player originally picked goat 1. The door with the car is revealed. The remaining door has goat 2. (0/6)
   (Rules don’t allow this option, so probability of #3 outcome is doubled.)
5. The player originally picked goat 2. The door with goat 1 is revealed. The remaining door has the car. (2/6)
6. The player originally picked goat 2. The door with the car is revealed. The remaining door has goat 1. (0/6)
   (Rules don’t allow this option, so probability of #5 outcome is doubled.)

But what if all you really want is a goat?