Math or Science related quiz?

[reddevil0126]

So you watched movie "21".  Kevin spacey asks "There are 3 doors, behind one of them is a car, and behind the others goat.  First you were asked to pick a door, and then Monty Hall (host of the quiz show) opens one door, which has goat behind it.  (Host never opens the door with the prize.)

Again, you are given the choice either  to stay with the first choice or to switch to the other door (at this moment there are only two doors since one with goat behind was open).  Do you have to stay or switch to raise the probability of winning the car?  Explain why?  What is the ratio of winning the prize? 

[Buddy]

I think I would stick with the goat. Goats are cute and I love goat milk products. 

[Tank]

You switch. I saw a program on this once and there is some sound statistics behind it but I can't remember them 

[xSilverPhinx]

You switch. I saw a program on this once and there is some sound statistics behind it but I can't remember them 

The Monty Hall problem? Something about removing the door the presenter opened from the whole, I think, but this is one of the most counter-intuitive problems in maths, and as it says under my name, maths is just not for me.

[Tank]

You switch. I saw a program on this once and there is some sound statistics behind it but I can't remember them 

The Monty Hall problem? Something about removing the door the presenter opened from the whole, I think, but this is one of the most counter-intuitive problems in maths, and as it says under my name, maths is just not for me.

That's the one.

[Guardian85]

The Mythbusters did a number on that (pardon the pun). For reasons that I don't know it seemed that changing door is better then standing put. I wonder why...

[The Magic Pudding]

I heard this recently, I haven't heard the official explanation, I thought it must be the person choosing where the prize was put wasn't doing so in a random manner.  They aren't choosing 1, 2, or 3 randomly, perhaps with use of a dice.  They are choosing to move it or not and then making a second decision as to where.  But I could be wrong, quite likely I am.

[xSilverPhinx]

Wiki helps explain the probability problem. IMO the decision tree makes it a bit easier.

Every possible outcome if the player picks door 1:

Sources of confusion

When first presented with the Monty Hall problem an overwhelming majority of people assume that each door has an equal probability and conclude that switching does not matter (Mueser and Granberg, 1999). Out of 228 subjects in one study, only 13% chose to switch (Granberg and Brown, 1995:713). In her book The Power of Logical Thinking, vos Savant (1996:15) quotes cognitive psychologist Massimo Piattelli-Palmarini as saying "... no other statistical puzzle comes so close to fooling all the people all the time" and "that even Nobel physicists systematically give the wrong answer, and that they insist on it, and they are ready to berate in print those who propose the right answer." Interestingly, pigeons make mistakes and learn from mistakes, and experiments, Herbranson and Schroeder, 2010, show that they rapidly learn to always switch, unlike humans.

Most statements of the problem, notably the one in Parade Magazine, do not match the rules of the actual game show (Krauss and Wang, 2003:9), and do not fully specify the host's behavior or that the car's location is randomly selected (Granberg and Brown, 1995:712). Krauss and Wang (2003:10) conjecture that people make the standard assumptions even if they are not explicitly stated. Although these issues are mathematically significant, even when controlling for these factors nearly all people still think each of the two unopened doors has an equal probability and conclude switching does not matter (Mueser and Granberg, 1999). This "equal probability" assumption is a deeply rooted intuition (Falk 1992:202). People strongly tend to think probability is evenly distributed across as many unknowns as are present, whether it is or not (Fox and Levav, 2004:637). However, if a player believes that sticking and switching are equally successful and therefore randomizes their strategy, they should, in fact, win 50% of the time, reinforcing their original belief.

In addition to the "equal probability" intuition, a competing and deeply rooted intuition is that revealing information that is already known does not affect probabilities. Although this is a true statement, it is not true that just knowing the host can open one of the two unchosen doors to show a goat necessarily means that opening a specific door cannot affect the probability that the car is behind the initially-chosen door. If the car is initially placed behind the doors with equal probability and the host chooses uniformly at random between doors hiding a goat (as is the case in the standard interpretation) this probability indeed remains unchanged, but if the host can choose non-randomly between such doors then the specific door that the host opens reveals additional information. The host can always open a door revealing a goat and (in the standard interpretation of the problem) the probability that the car is behind the initially-chosen door does not change, but it is not because of the former that the latter is true. Solutions based on the assertion that the host's actions cannot affect the probability that the car is behind the initially-chosen door are very persuasive, but lead to the correct answer only if the problem is completely symmetrical with respect to both the initial car placement and how the host chooses between two goats (Falk 1992:207,213).

The problem continues to attract the attention of cognitive psychologists. The typical behaviour of the majority, i.e., not switching, may be explained by phenomena known in the psychological literature as: 1) the endowment effect (Kahneman et al., 1991); people tend to overvalue the winning probability of the already chosen—already "owned"—door; 2) the status quo bias (Samuelson and Zeckhauser, 1988); people prefer to stick with the choice of door they have already made. Experimental evidence confirms that these are plausible explanations which do not depend on probability intuition (Morone and Fiore, 2007).

Is there a good way to explain this without using a formula or a decision tree to show every possible outcome?  ???

[reddevil0126]

I guess the decision tree is the easiest one to understand and the way I was taught.

Your answer is more than I expected and I really appreciate the attachment.  Good reading.
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Another counter-intuitive statistics-related question: 

First, don't try to come up with the mathematical solution.  Let your instinct or expectation do the work.

====Question:
You are having a party for your daughter; total of 50 friends and parents showed up.  What is the probability of finding AT LEAST two people with the same birthday (NOT your daughter's birthday - this may reduce the probability in real situation)?

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(Stop here if you did not give an answer!!!)



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Then, this is one way to think:

Let's assume we have only two people.  What's the probability of the two having different birthday?
The way to think is you pick one, and the probability of her having different birthday is 365/365, because she is the only one so far.

The probability of the second person having different birthday is 364/365.  (2nd person can't have the same b-day as 1st) 

So, two people having different B-day is (365/365)*(364/365) = close to 1.

Then let's add 3rd person.  The probability of the 3rd person having different B-day is 363/365 (363 = 365 minus previous two people's B-days).
So, for all three having different B-days = (365/365)*(364/365)*(363/365). (It's still close to 100 %) 

But, you see the pattern?  The chance of all 50 people having different B-days will be
(365/365)*(364/365)*(363/365)*........................*(316/365) . 
Guess the chance of all having different B-days?

Then, at least two of them having the same B-day is {1 - (the answer above)}


The final answer is? almost 100%.  If you ask at least 45 - 50 people randomly, you may find at least one person with the same B-day.

(But why you never met anyone with the same B-day in your workplace?) 

LEt me know if you find any error.  Need beer?

J