Comments on: The Movie 21, Variable Change, and Monty Hall

By: Shahu

Shahu — Sat, 13 Dec 2008 05:50:42 +0000

To the player, it is unknown whether the host knows whether which door has the car as not stated in the problem statement.. In this scenario, the probability that the host knows which door has the car, to the player is 1/2. Now if the host does not know which door has the car behind, the probabilities of the player’s winning or losing by keeping the 1st door is the same i.e. 1/2 when the host reveals that the 3rd door has the ghost. In case the host knows which door has the car, behind, the player indeed has a winning probability of 2/3 on switching according to simple Bayesian analysis.. Hence the cumulative probability of winnin will be..
Probability of host knowing the door having car X 2/3 + probability of the host not knowing of it X 1/2 = 1/2*2/3 + 1/2*1/2 = 1/3 + 1/4 = 7/12

Probabilty of losing hence would be 1-7/12 = 5/12 which can also be proved in the manner stated above.

By: Drew

Drew — Tue, 09 Dec 2008 22:20:33 +0000

Andrew, you have made an error in your analysis. You are counting the scenario where you picked the correct door initially twice. If you look at the row of numbers in your table denoted by “Your initial pick is door number:”, you will see that you are giving a 50/50 probability for having selected the correct door initially, which is wrong. It is a 1/3 probability for having selected the correct door initially.

You are essentially asking the question “Given that I have a 50/50 chance of having selected the correct door right off the bat, what is the likelihood that switching doors will make me a winner?”.

By: Ward

Ward — Sat, 06 Dec 2008 10:17:48 +0000

You would still feel the fool if on game night you swapped doors but the car was behind door #1. You could discuss probability with your mates at the bar and explain to them that if you repeated this process 2 more times you would have won the car twice!

By: Joseph

Joseph — Wed, 03 Dec 2008 11:16:53 +0000

First of all, I agree with the switch, but only if the host opening one door is constant. If the host has a choice of whether to open a door or not then you don’t switch. I’ll be the host and we’ll play 1000 times. I’ll open 1 door after you choose and give you a choice to switch 100 times out of the 1000. You switch every time you have a choice. You won’t win once out of the 100 becuase I can’t afford to give away a car. You must know whether the host intended to open a door no matter what door you picked in order to gain an advantage by switching.

If the host opening a door is a constant, if you switch every time then you win if you were wrong on your first choice and lose if you were right. 33 percent you were right and 67 you were wrong, so 67 percent you win and 33 you lose.

By: Andrew

Andrew — Sat, 29 Nov 2008 20:05:34 +0000

When in doubt, use brute force.

I created a table where I mapped out every possible permutation of the problem. There are only 24 possible outcomes, and in 12 of them staying the correct choice, and in 12 of them switching is the correct choice. 50/50.

http://www.andrewcmurphy.com/monty_hall.html

By: Drew

Drew — Mon, 27 Oct 2008 15:32:47 +0000

Hi there Rob, I see what you are saying, but you’re forgetting one thing. If your girlfriend had picked door #2, it probably would’ve been your door #1 that would’ve been opened with a goat, so you can’t think about it that way. It goes ONE person at a time, not two. The reason is because the door that the host opens depends upon where the car is. The host is tricking you. He’s gonna change his door selection so that it will always be 66.6% chance of it being the door you didn’t pick. That’s the catch.

Let me try to explain a different way.

Say you pick door #1 initially. We all agree that you have a 33% chance of winning the car. Then you ask the host to go behind the stage and find a goat and tell you which door has that goat. Do you think the act of the host looking behind the stage suddenly changes the 33% probability that you picked the car? Of course not, you said so yourself in your message. You have 1/3 chance that you picked the car, ALWAYS, no matter what happens. You can NEVER change the past and that 33% is FIXED IN STONE. So stop thinking of it like that. The 33% does not change to 66%, it does not change to 50%. It STAYS 33% and has to.

What does change is that the host was kind enough to let you pick the TWO doors (door #2 AND door #3). So when he asks if you want to change, he’s not asking, do you want to switch from door #1 to door #2. He is ACTUALLY asking, do you want to change your selection from door #1 to BOTH door #2 AND door #3. And of course you’d say yes, because picking two doors is always better than picking one.

It becomes super easy to visualize when you think about 1000 doors. You pick door #1, and then the host goes and removes goats from 998 of the other doors, leaving only door #1 and door #237. Knowing that there’s a car behind one of them, do you want to change your selection to door #237? Or do you want to stick with your original choice of door #1 and hope to heck that somehow the probability jumped from 1/1000 to 50% magically just cause somebody went behind the stage for a minute???

By: Rob Minich

Rob Minich — Mon, 27 Oct 2008 07:08:16 +0000

Ok People, I graduated from high shcool, started college and realized building houses was my gift. Needless to say I work with quit a few non graduating high school students. So sometimes I need to explaing myself clear in all languages, even when I only understand English so so. Lets go guys and girls stay with me.
When the game show host first gave us three doors to choose from I feel not one of us had a problem feeling we had a 33.3% chance of picking the right door. why did you and I say 33.3% chance of winning because 3doors X 33.3% = 100% right.
Now when door 3 is opened and is a goat that 33.3% needs to be eliminated from the starting point and needs to be divided equally in half which is 16.65% and added to door one and two evenly, which now brings doors one and two up to 50% each. Cut and dry. If I picked door one and my girlfriend picked door two and we both switched to the other door that would mean I would have a 66.6% chance and she would have a 66.6% chance of winning. 66.6% + 66.6% adds up to 133.2% now can we all see that this is not possible. Can we see that 133.2% would sound like a politician speaking. I hope you all can see this now in what ever language you read or what ever politician you are hearing. 133.2% is not what it can be. It can only be 100%. If I change my door and get 66.6% you can be rest a sure my girl friend will want her 66.6% as well. This is the problem we all want more than 100% and no one can give us more than 100%. Therefore door two can not get the 33.3% added to it and door one get nothing right?

By: Dan

Dan — Mon, 20 Oct 2008 09:29:38 +0000

Drew, I am an enthusiastic but rather mediocre first year maths student who very much appreciated your explanation, which makes perfect sense. I think if you combine the idea of sets with an increase in the number of goats (doors) involved, it does become quite intuitive. Thanks a lot. Also gunoieru simulation always helps too!

By: rich

rich — Thu, 09 Oct 2008 14:20:53 +0000

Svenn and patti got it correct from a math point of view. the problem was not stated correctly in the movie. Those that have taken probability knows that you must examine if the outcome of one trial affects the outcome of the next. For instance when you roll a dice the chance of rolling a 5 is always 1/6. Each role is independent of the other. So in this case it seems it would be 50% because you are picking 1 out of 2 doors.

I think this is what Imani and Dr. Jekyl are thinking. It is what I thought after watching the movie. But as stated by Svenn and patti, the rules of the game means that the previous outcome has an effect on the next outcome because the game show host knows where the car and will not be picking at random, therefore the trials are not independent of one another.

finally there is this site
http://www.shodor.org/interactivate/activities/SimpleMontyHall/?version=1.6.0_05&browser=MSIE&vendor=Sun_Microsystems_Inc.

play and you will see the probability of 66% when you switch doors. Notice how the game never picks your door, it always picks one of the other 2 doors which has a goat. That is the key to understanding the problem

By: Drew

Drew — Thu, 25 Sep 2008 13:10:40 +0000

Martin, actually Jack is right. You ARE given that choice in the beginning. When the host removes the goat from the equation, he essentially turns back time and makes the probability of your next choice change - hence variable change. The ONLY way what you are saying would be true is if the host picked a random door to remove, which means he might remove the car. Since this is not an option, the host’s choice is NOT random, and hence changes the variables in the equation.

The easiest way to think of it is imagine 1000 doors. You pick one of the doors, the host removes 998 other doors with goats. Do you think you now have a 50/50 chance that you picked the car in the first place when there were 1000 other doors? I think not.