PARADOXES OF PROBABILITY
A treatise on probability theory wouldn't be complete without a number
of paradoxes to thoroughly confuse you. Here are a few classic ones:
Monty Hall hosted a game show called Let's Make a Deal. He presented
his game show contestant with three doors numbered 1, 2, and 3. Behind one
of the doors is the grand prize. The contestant chooses a door. Then Monty,
who knows what's behind each door, opens up one of the two remaining doors
which doesn't have the grand prize behind it. Monty then asks the contestant, "Do
you want to stay with your original choice, or would you like to switch to the
other remaining door?"
Should the contestant stay with her original choice, should she change
to the other door, or does it not make any difference?
A doctor has two drugs, A and B, which he can prescribe to patients
with a certain illness. The drugs have been rated in terms of their
effectiveness on a scale of 1 to 6, with 1 being the least effective and
6 being the most effective. Studies show that drug A is uniformly
effective at a value of 3. Drug B varies in its effectiveness. 54% of
the time it scores a value of 1, and 46% of the time it scores a value
of 5.
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.
(The problem of the three chests)
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?
THE CHILD PARADOX
(Assume equal probability of a child being a boy or girl)
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? |
WHICH IS SAFER? FLYING VS. AUTOMOBILE?
In college I came across some statistics which said the accident rate
per mile was less for airplanes than it was for automobiles. Thus as
we all know traveling by air is safer than traveling by car. On the
same page however were statistics indicating the accident rate per
HOUR was MORE for airplanes than it was for automobiles. Thus it
would appear automobiles are actually safer by the hour while air
travel is safer by the mile. The reader is left to ponder the
interpretation of statistics.
HOW TO MAKE 44 A MAJORITY OF 140
“The persons who composed the Assembly of the Notables were all nominated by the
King, and consisted of one hundred and forty members. But as M. Calonne could not
depend upon a majority of this Assembly in his favor, he very ingeniously arranged them
in such a manner as to make forty-four a majority of one hundred and forty; to
effect this he disposed of them into seven separate committees, of twenty members each.
Every general question was to be decided, not by a majority of persons, but by a majority
of committees; and as eleven votes would make a majority in a committee, and four
committees a majority of seven, M. Calonne had good reason to conclude that as forty-four
would determine any general question he could not be outvoted.”
—Thomas Paine The Rights of Man Part the First. (1791)
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.
ANSWERS:
The Monty Hall 3 door problem:
The contestant should always switch doors. Without switching doors she
has a 1/3 chance of winning. Switching doors gives her a 2/3 chance of
winning.
Her chance of winning was 1 in three to begin with—and after Monty opens a door, the
chance of her winning is still just one in three.
Here's another way of thinking about it. Let's say you pick door #1. You've divided the
doors into two sets:
{1}{2,3}
Set {1} has a 1/3 chance of holding the winning door. Set {2,3} has a 2/3 chance of
holding the winning door. You now decide to switch from set {1} to set {2,3}. You now
have a 2/3 chance of holding the winning door. Monty opens one of the doors in set {2,3}.
Set {2,3} still has a 2/3 chance of holding the winning door, but now you know which of
{2,3} is not the winning door. You choose the other door in set {2,3} and you have a 2/3
chance of having the winning door.
A computer simulation of the game using a random number
generator should convince any skeptics.
(See http://www.macchiato.com/humor/monty_hall_skeptics.htm
for Java code that executes the Monty Hall problem.)
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 not always intuitive. (See http://astro.uchicago.edu/rranch/vkashyap/Misc/mh.html)
The a priori probability that the prize is behind door X, P(X) = 1/3
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!
Simpson's Paradox:
Proof:
|
|
| 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).
Three-Card Game
The chance that the underside is red is 2 to 1.
THE CHILD PARADOX
| 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
|
