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Title Author Date
Three doors and thousands of boxes Garcia , Alberto Feb 03, 2009
Dear Mr.Perakh,

I have been reading your articles for a while, not neccesarilly in chronological order, and i have just found an excuse to send an e-mail to
you, showing my most sincere appreciation. Thanks a lot for all of them.

The three door problem (AND, the compere knowing where the prize is) became a classic discussion between my wife and me some years ago.

Not in a formal way, i´ve found that people starts grasping the idea, when you tell them to consider that:

-ALWAYS changing your choice, what you are really doing, is DISCARDING the door of your choice. It´s like being able to choose all of the others.
-As someone else said, considering a lot more doors, makes the argument clearer. ¿Does it?. Just for fun:

I don´t know if you remember the last frames of the movie "Riders of the lost ark", where you can see a guy, "hiding" the wooden box which
contains the ark in between thousands of other wooden boxes.

I asked my wife, let´s imagine you are in front of this sea of wooden boxes and you have to guess where the ark is. Go ahead and make your
choice. Now the guy, who knows where the ark is, opens up all the boxes but one. ¿Do you think you were so lucky to have chosen the right one? Remember when you did it!

I know it doesn´t throw much light but, i liked to read about this after so long.

Best regards,

Alberto.

 

Title Author Date
Monty Hall problem Tremblay, Francois Jun 15, 2004
To Mark Perakh:

I appreciated your letter about the Monty Hall problem with the three doors. Before yesterday, I thought the solution was absurd and illogical. Your post seemed ridiculous. After making the calculations again, I realized that everything you wrote was completely correct.

Thanks!

Yours in Reason,
Francois Tremblay

 

Title Author Date
question on example Spencer , Steven Feb 08, 2004
Hi TalkReason,


Improbable Probabilities


Assorted comments on some uses and misuses of probability theory

First posted on June 22, 1999; last update September 2001.

By Mark Perakh

http://www.talkreason.org/articles/probabilities.cfm
The first problem is as follows. Imagine that you watch buses arriving at a certain stop. After watching them for a long time, you have determined that the interval between the arrivals of any two sequential buses is, on the average, one minute. The question you ask is: How long should you expect to wait for the next bus if you start waiting at an arbitrary moment of time? Many people asked to answer that question would confidently assert that the average time of waiting is 30 sec. This answer is wrong. It would be correct if all the buses arrived at exactly the same interval of 1 min. However, the situation is different in that 1 min is just the average interval between any two consecutive bus arrivals. This number 1 min, is a mean of a distribution wherein the interval varies between zero and a certain maximum which is larger than 1 min. Therefore, the average waiting time, regardless of when you start waiting, is 1 min rather than 30 sec.

========================================================
Ok, I have a problem with this.
I am going to switch to a daily watch rather than each minute, for thinking simplicity.

Take a train that leaves daily.. 7AM time is planned, but it varies a bit, leaving an average interval of 24 hours even though sometimes it is 22 other times 23, 24, 25.... but it generally leaves around 7AM.

You observe if for a few weeks and say, look the average interval is 24 hours.

Now I can mosey over at anytime of day..
Let's take 24 points, from midnight to 12PM to midnight.

At Midnight we have approximately a 7hr wait, then 6-5-4-3-2-1-0- then 24-23-22- etc.

Leading to the 12 hours, 1/2 day average.

(moving around a couple of hours won't make a big difference.. 6AM 7AM 8AM.

How do you get that the average time of waiting would be 24 hours ???????

Thanks.

Shalom,
Steven Spencer
Queens, NY


Schmuel@bigfoot.com
http://groups.yahoo.com/group/Messianic_Apologetic/

 

Title Author Date
Reply to 3 Doors (the Monty Hall show) Nord, Harald May 19, 2003
I was first reading the article "Improbable Probabilities" on professor Perakh's home-pages. Just like Nesa Simon David I also could not understand why the contestant should change his choice. I kept thinking
P(A)=P(C) even after door B had been opened. What if the contestant had chosen door C in the first place? was my thinking.

So I sent an e-mail to professor Perakh and he was kind enough to refer to the reply he had given on these pages.

Of course, the opening of door B is not a random event and the mathematical proof leaves no doubt.

I am thankful to professor Perakh for referring me to this web-site, as it contains much very interesting reading!

Kind regards Harald Nord ++

 

Title Author Date
3 Doors (the Monty Hall show) Simon David, Nesa Nov 02, 2002
In the chapter titled "Probability estimate is often tricky" of Mark Perakh's article Improbable Probabilities the author gives the example of the Monty Hall show, where participants are asked to choose from 3 doors, of which one has a prize behind it.

The author states that the participants double their chances of winning if they change their first choice after the compere opens one of the doors. My mind was quite frankly boggled by this statement. To me it is obvious that the door opened by the compere is irrelevant (after that door has been opened).

The game only really starts after the compere has opened one of the doors, thus taking that door out of the game... and leaving just 2 doors to choose from. One of the doors is a winning door, and the other is a losing door. The probability of winning at this game is thus clearly 50%.

I felt that i couldn't accept your explanation (because i had a nagging feeling that it somehow couldn't be right), so i decided to run a simulation (in my computer). After 8692 iterations, the number of wins to the total number of games is 49.459% (actually it's been fluctuating between 51% & 49%.

What is surprising however, is that the "percentage of games won where the participant changed his/her choice" is 32.881%, but the "percentage of games won where the participant kept his/her choice is 16.578%.

However, it must be made clear that the chance of winning a game is ultimately 50%. The 32% and 16% chances are "after the fact" percentages. A person who keeps first choice has a 50% chance of winning. A person who changes choice also has a 50% chance of winning. It just so happens that 32.881% of persons who win change choice, and 16.678% of persons who win keep first choice (I still don't understand why). But the fact remains, whatever you do, change choice or keep first choice, your chance of winning is 50%.