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/
Search the Dr. Math Library:
Ask Dr. MathTM
© 1994-2015 The Math Forum