PARADOXES OF PROBABILITY

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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:

  1. The Monty Hall 3 Door Problem
  2. Simpson's Paradox
  3. The Three-Card Game
  4. The Child Paradox
  5. Which is safer? Flying vs. Automobile?
  6. How to make 44 a majority of 140
  7. ANSWERS



THE MONTY HALL 3 DOOR PROBLEM

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?

Answer



SIMPSON'S PARADOX

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.
Answer



THE THREE-CARD GAME

(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?


Answer


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?

Answer



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:
cover How to Lie With Statistics by Darrell Huff, Irving Geis (Illustrator) (1954)
A classic — still in print.




ANSWERS:

 Probability Theory is never intuitive. Our brains did not evolve to intuitively understand probability theory correctly. The results are often surprising and counterintuitive.

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.

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 after one has determined the correct answer by doing the math.)

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: 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
(Note: for the following M.o. = Monty opens) 
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 2:1/2The possibilities are Girl-Daughter, Boy-Daughter.

Answer 3:1/3The 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




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