Drug testing

7328 /www/nrich/html/content/id/7328/ 1 2011-06-15T15:20:20 <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <ul id="stemLinks"> <li><a href="http://nrich.maths.org/4311">Warm-up</a></li> <li><a href="http://nrich.maths.org/471">Try this next</a></li> <li><a href="http://nrich.maths.org/2381">Think higher</a></li> <li><a href="http://plus.maths.org/content/revelation-game">Read: mathematics</a></li> <li><a href="http://plus.maths.org/content/anyone-tennis">Read: science</a></li> <li><a href="http://plus.maths.org/content/do-you-know-whats-good-you-0">Explore further</a></li> </ul> <div> </div> <p><span style="font-style: italic;">This problem involves looking at drug testing and the payoff this might give to athletes.</span></p> <p>Imagine a drug test that is 99% accurate.</p> <p>That is, if you are drug-free, there's a 99% chance you'll pass the test, and if you have taken the drug, there's a 99% chance you'll fail the test.</p> <p>In addition, imagine we know that 99% of athletes DO NOT take the drug.</p> <p><strong>If an athlete is tested and fails the test, what is the probability that they have taken the drug?</strong></p> <p><br></br> Dave and Joe are athletes at approximately the same skill level - each has an equal chance of winning in a race between the two.</p> <p>If Dave takes the drug but Joe doesn't, Dave's chance of winning increases to 75%.</p> <p>If Joe takes the drug but Dave doesn't, Joe's chance of winning increases to 75%.</p> <p>If they both take the drug, then each has an equal chance of winning again.<br></br> <br></br> Here is a payoff matrix, showing the chances of winning:<br></br> </p> <br></br> <table style="" border="1"> <tbody> <tr> <td>Dave/Joe (%)</td> <td>Drug</td> <td>No Drug</td> </tr> <tr> <td>Drug</td> <td>50/50</td> <td>75/25</td> </tr> <tr> <td>No Drug</td> <td>25/75</td> <td>50/50</td> </tr> </tbody> </table> <br></br> <p>The payoff of taking the drug is always better than not taking the drug, so the best strategy for both athletes is to use the drug!</p> <p>The race officials decide to use drug testing, so that athletes who take drugs can be disqualified.<br></br> <br></br> <strong>How does the payoff matrix change if they drug test both Dave and Joe?<br></br> <br></br> How does the payoff matrix change if they only drug test Dave? </strong></p> <div> </div> <div><strong>How does the payoff matrix change if they randomly drug test either Dave or Joe with a 50% chance?</strong></div> <div> </div> <div> </div> <div>You could also explore the effects of the different testing regimes when three athletes are competing for Gold, Silver and Bronze.</div> <p>What drug testing regime do you think would be the fairest? Are there any practical issues arising from your suggestion?</p> <p> </p> <div class="framework"><strong>Notes and Background</strong><br></br> <br></br> Read more about screening on <a href="http://understandinguncertainty.org/dishonesty">Understanding Uncertainty</a>.<br></br> </div> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><p><strong>For the first part:</strong></p> <p>If test is positive, we are 99% sure of correctness<br></br> If test is negative, we are 99% sure of correctness</p> <p>99% of athletes DO NOT take the drug. Which means 1% do.<br></br> <br></br> 1% of 99% who have not actually taken the drug will test positive (false positive).</p> <p>99% of 1% who have taken the drug will test positive (true positive).</p> <p>Total positive tests = false positives + true positives = $0.01 \times 0.99 + 0.99 \times 0.01 = 0.0198 = 1.98%$ So 1.98% of tests are positive.</p> <p>Of these, 0.99% are false positives and 0.99% are true positives, so the probability an athlete has taken the drug given that they tested positive is 50%.</p> <br></br> <p><strong>If both athletes are drug tested:</strong></p> <p><br></br> If both athletes take the drug, then each has 1%*(100%-1%*50%)=0.95% chance of winning if he luckily passes the test, unless the other athlete also passes and beats him. </p> <p>If neither takes the drug, then each has 99%*(99%*50% + 1%)= 49.995% chance of winning by passing the drug test and beating the opponent, or if the opponent fails the drug test.</p> <p>If one athlete takes the drug and the other doesn't, then the drug-taking athlete has 1%*(99%*75% + 1%) = 0.7525% chance of winning by passing the drug test and beating the opponent, or if the opponent fails the drug test. Similarly, the undoped athlete has 99%*(1%*25%+99%) = 98.2575% chance of winning. So the payoff matrix is now<br></br> </p> <br></br> <table border="1"> <tbody> <tr> <td>Dave/Joe (%)</td> <td>Drug</td> <td>No Drug</td> </tr> <tr> <td>Drug</td> <td>0.95/0.95</td> <td> <div>0.7525/98.2575</div> </td> </tr> <tr> <td>No Drug</td> <td> <div>98.2575/0.7525</div> </td> <td> <div> </div> <div>49.995/49.995</div> <div> </div> </td> </tr> </tbody> </table> <div> </div> <div><strong>If only Dave is tested:</strong></div> <div>If both athletes take the drug, then Dave has 1%*50% = 0.5% chance of winning. Since Joe cannot be disqualified, he will have 1-0.5%=99.5% chance of winning. If only Dave takes the drug, then he has 1%*75%=0.75% chance of winning. If only Joe takes the drug, then Dave has 99%*25% = 24.75% chance of winning. If neither dopes, then Dave has 99%*50% chance of winning.</div> <div> </div> <table border="1"> <tbody> <tr> <td>Dave/Joe (%)</td> <td>Drug</td> <td>No Drug</td> </tr> <tr> <td>Drug</td> <td>0.5/99.5</td> <td>0.75/99.25</td> </tr> <tr> <td>No Drug</td> <td>24.75/75.25</td> <td>49.5/50.5</td> </tr> </tbody> </table> <div> </div> <div>Dave is much worse off as a result of drug testing. What's worse, Joe is better off taking the drug because he can't be found out! The winning strategy is for Dave not to take the drug, but for Joe to take it. If we had tested just Joe instead, then the entries of the payoff matrix will simply flip.</div> <div> </div> <div> </div> <div><strong>If Dave or Joe are tested randomly, with probability 50%</strong></div> <div>The answer is simply the average of the above payoff matrix (for testing Dave) and its flipped version (for testing Joe). We get the following table</div> <div> </div> <table border="1"> <tbody> <tr> <td>A/B</td> <td>Drug</td> <td>No Drug</td> </tr> <tr> <td>Drug</td> <td>50/50</td> <td>50/50</td> </tr> <tr> <td>No Drug</td> <td>50/50</td> <td>50/50</td> </tr> </tbody> </table> <div> </div> <div>Now it makes no difference whether an athlete takes the drug or not. In reality being caught drug-taking will severly damage the reputation of an athlete (and possibly lead to being banned from competing). So both athletes will choose not to take the drug.</div> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><br></br> <h3>Why do this problem?</h3> This problem invites students to work with some tricky conditional probabilities in the context of drug testing in sport. It also introduces the payoff matrix, an important representation in Game Theory.<br></br> <h3>Possible approach</h3> <div>Introduce the first part of the problem:</div> <div>"Imagine a drug test that is 99% accurate. That is, if you are drug-free, there's a 99% chance you'll pass the test, and if you have taken the drug, there's a 99% chance you'll fail the test. In addition, imagine we know that 99% of athletes DO NOT take the drug. If an athlete tested and fails the test, what is the probability that they have taken the drug?"</div> <div> </div> <div>You may wish to show students <a href="http://understandinguncertainty.org/screening">this animation</a> to introduce Bayes Theorem together with a visual way of thinking about conditional probability in circumstances such as this. <a href="http://plus.maths.org/content/logic-drug-testing">This Plus article</a> also examines conditional probability in the context of sports doping, including a tailored version of the animation.</div> <br></br> Once students have had a chance to engage with and work on this first question, introduce the second part of the problem. Give students time to make sense of the payoff matrix, and then split the class into two groups, with half the students working on the question "How does the payoff matrix change if they drug test both Dave and Joe?" and the other half working on "How does the payoff matrix change if they only drug test Dave?"<br></br> <br></br> After students have had time to tackle their question, bring the class back together and invite each group to present their working out and the payoff matrix.<br></br> <br></br> Finally, challenge students to answer the question "How does the payoff matrix change if they randomly drug test either Dave or Joe with a 50% chance?" and then take some time to discuss the pros and cons of different drug-testing regimes.<br></br> <h3>Key questions</h3> <div>Imagine Dave takes the drug and Joe doesn't, and they are both tested. What are the different possible outcomes?</div> <div>What is the probability of each outcome?</div> <br></br> <h3>Possible extension</h3> Invite students to consider the possible outcomes when three athletes compete for Gold, Silver and Bronze, with different drug testing regimes.<br></br> <br></br> <h3>Possible support</h3> <div>It may be helpful to consider an ordering of the events, with the associated probabilities at each stage.</div> <div>For example, first the athletes decide whether to take the drug or not. Then the race takes place. Then the drug testing takes place.</div> <br></br> <br></br> <br></br> <br></br> <br></br></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><p>It may be helpful to consider an ordering of the events, with the associated probabilities at each stage.<br></br> For example, first the athletes decide whether to take the drug or not. Then the race takes place. Then the drug testing takes place.</p></mdoxml> <?xml version="1.0" encoding="UTF-8"?> <mdoxml version="1.0"><span style="font-weight: bold;">Solution part 1</span>: false positive: 99%*1%=0.99%, false negative: 1%*1%=0.01%.) <br></br> <br></br> <span style="font-weight: bold;">Solution part 2</span>: true positive: 1%*99%=0.99%, false positive: = 99%*1% = 0.99%, positive = true positive + false positive = 1.98%, P(drug | +) = 0.99%/1.98% = 50%.)<br></br> <br></br> <span style="font-weight: bold;">Solution part 3</span>: If both athletes dopes, then each has 1%*(100%-1%*50%)=0.95% chance of winning if s/he luckily passes the test, unless the other athlete also passes and beats him/her. If neither dopes, then each has 99%*(99%*50% + 1%)= 49.995% chance of winning by passing the drug test and beating the opponent, or if the opponent fails the drug test. If one athlete dopes and the other doesn't, then the doped athlete has 1%*(99%*75% + 1%) = 0.7525% chance of winning by passing the drug test and beating the opponent, or if the opponent fails the drug test. Similarly, the undoped athlete has 99%*(1%*25%+99%) = 98.2575% chance of winning. So the payoff matrix is now<br></br> <br></br> <table border="1"> <tbody> <tr> <td>A/B (%)</td> <td>Drug</td> <td>No Drug</td> </tr> <tr> <td>Drug</td> <td>0.95/0.95</td> <td> <div>0.7525/98.2575</div> </td> </tr> <tr> <td>No Drug</td> <td> <div>98.2575/0.7525</div> </td> <td> <div> </div> <div>49.995/49.995</div> <div> </div> </td> </tr> </tbody> </table> <div> </div> <div> </div> <div><span style="font-weight: bold;">Solution part 4</span>: If both athletes dope, then A has 1%*50% = 0.5% chance of winning. Since B cannot be disqualified, s/he will have 1-0.5%=99.5% chance of winning. If only A dopes, then s/he has 1%*75%=0.75% chance of winning. If only B dopes, then A has 99%*25% = 24.75% chance of winning. If neither dopes, then A has 99%*50% chance of winning.</div> <div> </div> <table border="1"> <tbody> <tr> <td>A/B (%)</td> <td>Drug</td> <td>No Drug</td> </tr> <tr> <td>Drug</td> <td>0.5/99.5</td> <td>0.75/99.25</td> </tr> <tr> <td>No Drug</td> <td>24.75/75.25</td> <td>49.5/50.5</td> </tr> </tbody> </table> <div> </div> <div>As we expected, A is much worse off as a result of drug testing. What's worse, B is better off doping because s/he can't be found out! The winning strategy is for A not to dope, but for B to dope. If we had tested just B instead, then the entries of the payoff matrix will simply flip.</div> <div> </div> <div> </div> <div><span style="font-weight: bold;">Solution part 5</span>: The answer is simply the average of the above payoff matrix (for testing and its flipped version (for testing B). We get the following table</div> <div> </div> <table border="1"> <tbody> <tr> <td>A/B</td> <td>Drug</td> <td>No Drug</td> </tr> <tr> <td>Drug</td> <td>50/50</td> <td>50/50</td> </tr> <tr> <td>No Drug</td> <td>50/50</td> <td>50/50</td> </tr> </tbody> </table> <div> </div> <div>Now probablistically it makes no difference whether an athletes dopes or not. In reality being discovered of doping will severly damage the reputation of an athlete. So both athletes will choose not to dope. This scheme only requires one testing per pair. We've accomplished the same goal while saving the cost of drug testing!</div> <div> </div> <br></br> <br></br></mdoxml> 2 3 0 0 0 0 1 Drug testing How do different drug-testing regimes affect the risks and payoffs for an athlete who chooses to take drugs? Admin Individual Probability Conditional probability Decision Mathematics and Combinatorics Game theory Admin Stage 5 Decision mapping