Speaking of giving away cars. Back in the day on The Price is Right there was a game where the contestant had to guess the price of a car.
The first two numbers were always 34. As in $3400.
I don't think the game show host ever cared if someone won the car or not. They got the cars for free didn't they?
Monty Hall problem
Re: Monty Hall problem
ArturoBandini wrote:... In the latter two cases, there is a 50% probability of the game resolving to either of the two possible cases (assuming the host chooses from the two doors randomly), so that makes me think that these cases should be weighted less than the fully determinate starting choices.pjbogart wrote:Pick A, B is revealed, switch to C. WIN.
Pick B, A is revealed, switch to C. WIN.
Pick C, B is revealed, switch. LOSE.
Pick C, A is revealed, switch. LOSE.
It's still 50/50
...
This is exactly right. And the exact way you would reflect this weighting: you have a 1/3 chance of picking the car with your first choice, and a 2/3 chance of picking a goat with your first choice.
[nerdgasm]Excellent topic  probability problems are fascinating and a hobby of mine. I actually spent an afternoon writing a computer simulation of this problem, to prove to myself that the contestant should always switch.[\nerdgasm]
Re: Monty Hall problem
I guess I'm assuming that the host knows which door has a car behind it. Since one of the two remaining doors must have a goat, the host can always pick a goat to reveal.
Another way to look at this is to consider that if you picked the goat on the first try, the host could simply say, "too bad, you got a goat!" So you have to kind of remove the host's behavior in order to look at it statistically. After he opens one door with a goat, you're left with two doors and your chances of winning or losing becomes 50%, whether you choose to switch or not.
Another way to look at this is to consider that if you picked the goat on the first try, the host could simply say, "too bad, you got a goat!" So you have to kind of remove the host's behavior in order to look at it statistically. After he opens one door with a goat, you're left with two doors and your chances of winning or losing becomes 50%, whether you choose to switch or not.

 Forum God/Goddess
 Posts: 2256
 Joined: Sun Jan 27, 2008 11:54 pm
 Location: near west
Re: Monty Hall problem
Yes, this is critical to the setup of the problem.pjbogart wrote:Since one of the two remaining doors must have a goat, the host can always pick a goat to reveal.
If you remove the host's behavior, you have an entirely different problem (e.g. Not the Monty Hall Problem). But yes, if the host didn't know which door to open, the probability of switching and winning the car would change, because sometimes the host would unknowingly reveal the car. The probability at that point is indeterminate, because we don't know what the rules of the game would be in that situation (e.g. can you choose the door the host already opened?).pjbogart wrote:So you have to kind of remove the host's behavior in order to look at it statistically.

 Forum God/Goddess
 Posts: 3679
 Joined: Wed Aug 12, 2009 12:43 pm
 Location: Dane County
 Contact:
Re: Monty Hall problem
I think the confusion lies in the point in the process at which one determines the odds. The math supports the increased odds of getting the car if choosing to switch, but these odds are dependent upon the process as a whole (Pick a door, a different door is revealed, choose to switch.)
As mr titor alluded to with the coin flip comparison, if the decision to switch is treated as a truly independent event (choosing 1 of two doors), the odds are 50:50. In this case, unlike repeatedly flipping a coin, that decision is not independent of the previous events.
As mr titor alluded to with the coin flip comparison, if the decision to switch is treated as a truly independent event (choosing 1 of two doors), the odds are 50:50. In this case, unlike repeatedly flipping a coin, that decision is not independent of the previous events.

 Forum God/Goddess
 Posts: 515
 Joined: Wed Aug 04, 2010 12:43 pm
 Location: ???
Re: Monty Hall problem
Rowe's Rule: the odds are five to six that the light at the end of the tunnel is the headlight of an oncoming train [or Trane]."
Paul Dickson
I know it’s not The Full Monty, thus I must say so sorry for being slightly off topic. And Mr. Ball, I am very sorry about your assault.
(I hope you are feeling better:)
Edit: I finally watched The Blindside on Sunday night. I know,
I'm a lil' behind. What a phenomenal movie, based on a true story!
P.S. to “The Dumb Ass:” Dummy Amy does not mean I want to be related. I did NOT mean for it to be read like that. You two were so mean to me; I never, ever want to see you again. And what your sister said about my tobe exhusband: FU3!!! To believe I actually felt sorry for you 3:( And FYI: My Mommy told me not to be ashamed of having a happy childhood. And I will tell any damn story about being a lil' Amy that I want to! And Bookie (hint) that was to one specific person (not you).
Last edited by city2countrygal on Thu Nov 01, 2012 1:10 am, edited 1 time in total.
Re: Monty Hall problem
I was thinking about this problem a little more last night, and I thought of another way to suggest why you would always switch: You are, in effect, being given a choice between the one door you selected initially, and all other doors. That one of those doors is revealed to you doesn't change that you are effectively choosing two doors when you switch.
To take this line of reasoning further  imagine that there are 100 doors, instead of three, still with only one car. Would you rather keep your one initial door, or change to the 98 that are left after the host reveals one of the goats?
To take this line of reasoning further  imagine that there are 100 doors, instead of three, still with only one car. Would you rather keep your one initial door, or change to the 98 that are left after the host reveals one of the goats?
Who is online
Users browsing this forum: No registered users and 1 guest