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A treatise on probability theory wouldn't be complete without a number of paradoxes to thoroughly confuse you. Here are a few classic ones:

Should the contestant stay with her original choice, should she change to the other door, or does it not make any difference?

The doctor, wishing to provide his patients with the best possible care, asks his statistician friend which drug has the highest probability of being the most effective. The statistician says, "It is clear that drug A is the most effective drug 54% of the time. Thus drug A is your best bet."

Later a new drug C becomes available. Studies show that on the scale of 1 to 6, 22% of the time this drug scores a 6, 22% of the time it scores a 4, and 56% of the time it scores a 2.

The doctor, again wishing to provide his patients with the best possible care, goes back to his statistician friend and asks him which drug has the highest probability of being the most effective. The statistician says, "Well, seeing as there's this new drug C on the market, your best bet is now drug B, and drug A is your worst bet."

Show that the statician is right.

Three cards are in a hat. One card is white on both sides; the second is white on one side and red on the other; the third is red on both sides. The dealer shuffles the cards, takes one out and places it flat on the table. The side showing is red. The dealer now says, "Obviously this is not the white-white card. It must be either the red-white card or the red-red card. I will bet even money that the other side is red." Is this a fair bet?

Question 1:

A mother has two children. The younger one is a daughter named Mary. What is the probability that the other child is a girl? |

Question 2:

A mother has two children. The older one is a daughter named Mary. What is the probability that the other child is a girl? |

Question 3:

A mother has two children. One of them is a daughter. What is the probability that the other child is a girl? |

Question 4:

A mother has two children. One of them is a daughter named Mary. What is the probability that the other child is a girl? |

—Thomas Paine

I leave the reader with the following book recommendation:

How to Lie With Statistics by Darrell Huff, Irving Geis (Illustrator) (1954)

A classic — still in print.

(From the book
*Ten Questions About Human Error: A New View of Human Factors and System
Safety (Human Factors in Transportation)*, (2004) by Sidney W.A. Dekker.)

There are about 700,000 physicians in the United
States. The U.S. Institute of Medicine estimates that each year between
44,000 and 98,000 people die as a result of medical errors.
^{1} This makes
for a yearly accidental death rate per doctor of between 0.063 and
0.14. In other words, up to one in seven doctors will kill a patient
each year by mistake. Take gun owners in contrast. There are 80,000,000
gun owners in the United States. Yet their errors lead to "only" 1,500
accidental gun deaths per year. This means that the accidental death
rate, caused by gun-owner error, is 0.000019 per gun owner per year.
Only about 1 in 53,000 gun owners will kill somebody by mistake.
Doctors then, are 7,500 times more likely to kill somebody by mistake.
While not everybody has a gun, almost everybody has a doctor (or
several doctors), and is thus severely exposed to the human error
problem.^{2}

It is also true statistically that one is more likely to be killed by a family member than by a serial killer. Should we abandon our family and move in with a serial killer? I leave the reader to once again ponder the meaning and interpretation of statistics.

References:

- Kohn, L.T., Corrigan, J. M., & Donaldson, M. (Eds.). (1999).
*To err is human: Building a safer health system.*Washington, DC: Institute of Medicine.

- Sidney W.A. Dekker,
*Ten Questions About Human Error: A New View of Human Factors and System Safety (Human Factors in Transportation)*, (2004)

[A supurb book on modern day philosophy. See my review at Amazon.]

A proper use of Bayes' Theorem is required to mathematically solve this problem. Bayes' Theorem tells us how to properly update the odds after we are given new information. The result is often counterintuitive. However, before we jump to the math, let's first just reason our way through this. (Reasoning, though, usually works only

Let's say you pick door #3. Think of the doors as now being divided into two sets:

{1 & 2} {3}

Your door {3} has a 1/3 chance of being the grand prize. The remaining doors, {1 & 2}, taken together as a group, collectively have a 2/3 chance of having the grand prize behind one of them.

Monty, who knows what's behind each door, now opens up one of the first two doors which doesn't have the grand prize behind it.

Nothing has changed. Door {3} still has a 1/3 chance of being the grand prize, and the first two doors, {1 & 2}, taken together as a group, still have a 2/3 chance of having the grand prize behind one of them. Except now there's only one door left unopened in that group.

A computer simulation of the game using a random number generator is good way to cross check the result. (See http://macchiato.com/humor/monty_hall_skeptics.htm for Java code that executes the Monty Hall problem.)

And here's the math, in case anyone cares. Again, a proper use of Bayes' Theorem is required to mathematically solve this problem. Bayes' Theorem tells us how to properly update the odds after we are given new information, the new information being:(Note: for the following M.o. = Monty opens)The grand prize is not behind door number <whatever door Monty opens>For this analysis let's label the doors {A,B,C}. The a priori probability that the prize is behind any door X, P(X) = 1/3 Let's examine the case where the contestant selects door A, and Monty then opens door B. The probability that Monty Hall opens door B if the prize were behind A, P(Monty opens B|A) = 1/2 The probability that Monty Hall opens door B if the prize were behind B, P(Monty opens B|B) = 0 The probability that Monty Hall opens door B if the prize were behind C, P(Monty opens B|C) = 1

The probability that Monty Hall opens door B is then p(Monty opens B) = p(A)*p(M.o. B|A) + p(B)*p(M.o. B|B) + p(C)*p(M.o. B|C) = 1/6 + 0 + 1/3 = 1/2 Then, by Bayes' Theorem, P(A|Monty opens B) = p(A)*p(Monty opens B|A)/p(Monty opens B) = (1/6)/(1/2) = 1/3 and P(C|Monty opens B) = p(C)*p(Monty opens B|C)/p(Monty opens B) = (1/3)/(1/2) = 2/3

In other words, the probability that the prize is behind door C is higher
when Monty opens door B, and you SHOULD switch!

Pr(A>B and A>C) | = | Pr(A>B | and | A>C | | | A=3 | ) | * | PR(A=3) | |||

= | .54 | * | .56 | * | 1 | |||||||

= | .3024 | |||||||||||

Thus A is the most effective drug ~ 30% of the time. |

Pr(B>A and B>C) | = | Pr(B>A | and | B>C | | | B=1 | ) | * | Pr(B=1) | |

+ | Pr(B>A | and | B>C | | | B=5 | ) | * | Pr(B=5) | ||

= | 0 | * | .56 | * | .54 | |||||

+ | 1 | * | .78 | * | .46 | |||||

= | .3588 | |||||||||

Thus B is the most effective drug ~ 36% of the time. |

Pr(C>A | and | C>B | ) | = | Pr(C>A | * | C>B | | | C=2 | ) | * | Pr(C=2) | |

+ | Pr(C>A | * | C>B | | | C=4 | ) | * | Pr(C=4) | |||||

+ | Pr(C>A | * | C>B | | | C=6 | ) | * | Pr(C=6) | |||||

= | .3388 | ||||||||||||

Thus C is the most effective drug ~ 34% of the time. |

Comparing the results:

A is the most effective drug | ~ 30% | of the time. |

B is the most effective drug | ~ 36% | of the time. Most effective most of the time. |

C is the most effective drug | ~ 34% | of the time. |

= 100% |

Granted this is a silly way of rating medications, but Simpson's Paradox actually turned up in a (now not so) recent investigation to see if there was sex bias in the admissions of men and women to graduate studies at the University of California at Berkley. Independent studies of admissions of men and women in the fall of 1973 showed a positive sex bias against female applicants. Then when the data for men and women were combined, there was a small but statistically significant bias in FAVOR of women. (See "Sex Bias in Graduate Admissions: Data from Berkeley," by P. J. Bickel, E. A. Hammel and J. W. O'Connell in 'Science', Vol. 187, February 7, 1975, pages 398-404).

Answer 1: | 1/2 | The possibilities are Daughter-Girl, Daughter-Boy. | ||

Answer 2: | 1/2 | The possibilities are Girl-Daughter, Boy-Daughter. | ||

Answer 3: | 1/3 | The possibilities are Girl-Boy, Boy-Girl, Girl-Girl.
For a further analysis of Question 3 click here |
||

Answer 4: | I'm not sure. Here's my guess: | |||

It depends on the probability of the mother naming both
her children Mary. If she names all her children Mary
then knowing one of them is named Mary doesn't help us
and the answer as you know from question 3 is 1/3. If
she names only one child Mary, then this uniquely
identifies the child and the probability is 1/2. That
is, there are two mutually exclusive possibilities: Mary
is the older child or Mary is the younger child. In
either case the probability of the other child being a
girl is 1/2. For a further analysis of Question 4 click here |

Conclusions | Generating random integers within a desired range | |||

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