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Darts and Wilensky's Paradox


Date: 03/17/99 at 14:03:34
From: Paul Morin
Subject: Probability

Problem: Find the probability of scoring 25 points with 2 darts if 
the dartboard is made up of 4 concentric circles with the inner circle 
having a radius of 2 inches and each successive ring 2 inches larger 
in radius. Possible points are 20 (inner circle), 15, 10, and 5 points 
(moving out from the center).  

I want to know why I got two different answers using what I thought 
were two equally valid approaches:

Approach 1: Probability of scoring a 25, P(25), is P(20)times P (5|20) 
plus P(5)times P(20|5) plus P(15) times P(10|15) plus P(10) times 
P(15|10). The answer is P(25)=11/64.

Approach 2: Divide up the target into equal area portions. The center 
is one portion, the next ring has 3 portions equal to the center area, 
the next has 5, and the outer ring has 7. The total number of possible 
combinations of two darts is C(16,2) = 120 plus the 16 possibilities 
where both darts hit the same square = 136 total possibilities. There 
are 7 possible ways to get 25 by hitting the 20 and 5, and 3X5 = 15 
ways to score 25 points hitting the 15 and 10. So the probability 
P(25) = 22/136 or P(25) = 11/63.

Both answers are close. The first is correct, but why don't I get the 
same answer with each method?

Thank you.


Date: 03/18/99 at 07:27:25
From: Doctor Mitteldorf
Subject: Re: Probability

Dear Paul,

The problem with the second approach is that when you count 
"combinations" of different regions you lump together AB and BA. In 
other words, the probability of hitting 5 and then 20 is counted along 
with the probability of hitting 20 and then 5. Compared to the 
probability of hitting 10 and then 10, this deserves to be weighted by 
an extra factor of 2, because there's only one way to hit 10-10, 
whereas 20-5 and 5-20 are separate possibilities.

If you double the C(16,2) count, you get 240; add the 16 possibilities 
for single regions twice in a row, and you are accounting for all 256 
possible combinations equally, which is what you should be doing.

The larger moral of this story is that knowing the formulas in 
probability theory is no assurance that you're going to get the right 
answer. Each and every problem has to be analyzed. When you apply a 
formula to a problem, you're really constructing a mathematical 
model, and it's your responsibility to be sure that the model is an 
accurate reflection of the situation in the problem.

Here's a fun probability problem to ponder, called Wilensky's paradox.  
You are presented with 2 envelopes, each with some money in it. All 
you know about the amounts is that one envelope contains twice as much 
money as the other. You can pick either envelope.  

You pick envelope A and it has $100. But the Master of Mischief says, 
"Wait - you can change your mind if you want." Let's see: envelope B 
either contains $200 or $50, with equal probability. You have a 50% 
chance of getting $200, and a 50% chance of getting $50. Your expected 
gain if you choose envelope B is 1/2 of $200 plus 1/2 of $50, or $125.  
For envelope A your expected gain is exactly $100. So you choose 
envelope B.

But wait a minute, here. How can envelope B always be better? After 
all, you didn't really change your mind because of anything you 
learned from opening envelope A. No matter how much money was in 
envelope A, you would have said the same thing, and chosen envelope B.  
There can't be any difference between the expected gain for envelope A 
and envelope B.

What do you think?

- Doctor Mitteldorf, The Math Forum
  http://mathforum.org/dr.math/   
    
Associated Topics:
High School Probability

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