*Tricky Probability Problem 2* 'Lets Make A Deal'

[BigShow]

Let's Make a Deal! Consider the following variation on this popular problem. There exists 4 seperate doors. Behind 3 of the doors is a goat, and behind the remaining door is a shiny, brand new SUV. You select a door at random, the door you believe the SUV is concealed behind. Then the "host" selects a door with a goat behind it at random, and opens it. (Here the host has perfect information, since he knows which door the SUV is hidden behind). You then have an option: to stick with your initial choice or switch to one of the 2 other doors, picking at random.

What is the probability of winning the car if you decide
(i) not to switch?
(ii) to switch?

[szw]

This is called the monty hall problem. (a little differnet, it is usually only 3 doors total). Too lazy to write out the explaination but it is a very common problem, so it will be easy to google/adapt to 4 doors.

[skandalouz_08]

if you decide not to switch isn't it 1/4 or 0.25 and then if you do switch you're still choosing between 3 doors so its 1/3 or 0.33

I dunno makes sense to me to do it that way

[szw]

Originally posted by skandalouz_08
if you decide not to switch isn't it 1/4 or 0.25 and then if you do switch you're still choosing between 3 doors so its 1/3 or 0.33

I dunno makes sense to me to do it that way

No if you switch you are choosing between 2 doors.

[calgarygts]

Doesn't matter if you switch or not (unless you picked the door the host opened). You're left with three doors and you have an equal opportunity to win the SUV with any of them. So I would say you have 33.3333333(infinity)% chance of winning whether you switch or not.

I'm also not known as a math wiz so maybe someone can tell me if I'm right or wrong.

[szw]

You're wrong sorry. Just google monty hall, it will explain.

[calgarygts]

Shit, I just read up on the monty hall problem. At least I'm joined by many a mathematician in thinking that I was right. I guess if I'm ever given the choice, I'll switch.

[Audio_Rookie]

http://www.comedia.com/hot/monty-answer.html

[szw]

Bigshow, make sure u change ur answer from the ones posted if you are handing this in!

[BigShow]

Originally posted by szw
Bigshow, make sure u change ur answer from the ones posted if you are handing this in!

yea i know lol...

if you make the switch your chances of winning are 37.5% or 3/8 which is greater than if you dont make the switch (1/4 chance)

when you make the switch its something like 1/2 x 1/4 x 3...

[Charon]

Of course you switch. 75% chance of you picking the wrong door the first time.
If he removes one. Then you should switch as the chance you have the wrong door is still 75%

[sexualbanana]

yes switch.

You have 1/4 chance of picking the right door on your first try.

You have 3/4 chance of getting it wrong. If the host knows the the door where the SUV is and wants to pick a goat, he take away one of the goat doors. Then you are left with a 1/2 chance of picking the SUV door.

3/4 x 1/2=37.5% like BigShow said.

[pinoyhero]

I love this one, its a timeless math question, just love it!

[Tech2]

Sure, you can say your chances went up. But that has nothing to do with you changing your door. It does have to do with one of the false options being removed.

4 doors, you are 1/4.

host removes 1

3 doors, you have 1/3 (no matter if you switch or not)

host removes another false one

you have 1/2 chance, (regardless of switching)

host removes final false one

you are 1/1


If switching actually helped your odds off winning, then you always switch. But your decision to swtich has nothing to do with where the prize actually was in the first place. The fact there is one less door is what increases your odds, not any decision you can make.

The problem has nothing to do with any combination of your first and second decision.

[szw]

Originally posted by Tech2
Sure, you can say your chances went up. But that has nothing to do with you changing your door. It does have to do with one of the false options being removed.

4 doors, you are 1/4.

host removes 1

3 doors, you have 1/3 (no matter if you switch or not)

host removes another false one

you have 1/2 chance, (regardless of switching)

host removes final false one

you are 1/1


If switching actually helped your odds off winning, then you always switch. But your decision to swtich has nothing to do with where the prize actually was in the first place. The fact there is one less door is what increases your odds, not any decision you can make.

The problem has nothing to do with any combination of your first and second decision.

This is incorrect.

[Tech2]

If there is a 1 in 4 chance at the beginning, and the host takes away one wrong door, switching or not doesn't increase your odds.

Its like you are starting all over, and you have a 1 in 3 chance. Changing your decision doesn't make your chances better than 1 in 3.

It's like rolling dice, getting one 6 doesn't give you a lower or higher chance of getting another 6. It's always 1 in 6.

Rolling a 6 once, then making the die 5 sided without a 3 does increase your chances of getting your number, but you don't have to change your choice from being 6. It's now 1 in 5, not any better than that no matter what came before.

[HyperZell]

Oh man, I had this question covered in Human Resources and Organizational Dynamics last year, and it was a great lesson.

The correct answer is that yes, switching does improve your chances. Look at it this way - if it's 3 doors, and you picked one, you MOST LIKELY picked a losing door. Can't argue with that, right?

So the host reveals a losing door to you. Now there are 2 doors remaining. The common thing people do is assume that the events are independent, and that you nopw have a 1/2 chance of winning. However, if we go back and assume once again that you MOST LIKELY picked a losing door, then switching would leave you with...the winning door! Don't make it any more complicated than that if you don't have to.

I went around asking my friends this question, and most of the picked staying, but 'got it' when I explained. All except this one engineer (natch) who argued with me for the longest time and ended up getting pissed off until everyone around him told him he was wrong. Don't be that engineer - problem solved.

[Tech2]

Originally posted by HyperZell

The common thing people do is assume that the events are independent, and that you nopw have a 1/2 chance of winning.

Exactly. Like 2 rolls of a die, the events don't influence each other. Where the winning door is has nothing to do with what you chose in the first place, or with switching. You dont' control the odds, the number of doors does. 1/4 and 1/3 are the odds not because you make a choice, but because there is only one way to win with a limited number of choices. The likelihood of not winning on the prior choice doesn't change the odds that you could be right.

[HyperZell]

Originally posted by Tech2


Exactly. Like 2 rolls of a die, the events don't influence each other. Where the winning door is has nothing to do with what you chose in the first place, or with switching. You dont' control the odds, the number of doors does. 1/4 and 1/3 are the odds not because you make a choice, but because there is only one way to win with a limited number of choices. The likelihood of not winning on the prior choice doesn't change the odds that you could be right.

Wait, so...are you agreeing with me?

[Tech2]

Yeah, i wasn't very clear about agreeing or not. What I meant was that the 2 events are independent of each other. Going back to assume you were wrong in one decision is unnessisary. I ran this by some people at school today because it's been bugging me for a long time. A couple people said they heard about this trick before but also seemed to agree with my explanation.

What I did was take a paper with 4 circles and ask for the odds of choosing a correct one. 1/4 was the unanimous answer. Then I ripped off one of the circles. What are the odds now if you change or if you stay? There 3 cirlcles left, so 1/3.

I also used the example of dice. I roll a 1d6 and ask you to guess what it is. Girl says 3. She has a 1/6 chance of being right. Then I say, actually it was a 1d5, but you didn't guess 6 so you don't lose. What are the odds of still being right? 1/5.

I am totally open to being wrong about this, but so far no one has been able to explain to me otherwise...