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Introducing Students to Classic Problems in Probability

Introducing Students to Classic Problems in Probability. Allan Rossman Department of Statistics Cal Poly – San Luis Obispo arossman@calpoly.edu. What’s my point?. Classic problems in probability Provide great opportunities for exploring, understanding concepts of randomness

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Introducing Students to Classic Problems in Probability

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  1. Introducing Students to Classic Problems in Probability Allan Rossman Department of Statistics Cal Poly – San Luis Obispo arossman@calpoly.edu NZAMT 2013

  2. What’s my point? • Classic problems in probability • Provide great opportunities for exploring, understanding concepts of randomness • Lend themselves to interactive investigations • Can be presented at various mathematical levels • Illustrate mathematical as well as statistical ideas • Create pedagogical alternatives to rolling dice and flipping coins for their own sake • Are intellectually stimulating • Can be fun! NZAMT 2013

  3. What are my goals for this workshop? • Acquaint you with some classic problems in probability • Demonstrate solutions to these problems • Using simulation • Using enumeration • Provide ideas for activities that you can use with your students • In variety of courses, levels NZAMT 2013

  4. What concepts will be introduced? • Randomness • Probability • Simulation • Relative frequency • Permutations • Expected value • Decision making under uncertainty NZAMT 2013

  5. What are these classic problems? • Problem of the points • Matching problem • Collector problem • Secretary problem NZAMT 2013

  6. Problem of the points • Helped to motive the study of probability • Discussed in correspondence between Pascal and Fermat NZAMT 2013

  7. Problem of the points • A simplified version: • Suppose that Heather and Tom each put in $5 to play a game • They flip a fair coin repeatedly • If 4 Heads occur before 4 Tails, Heather wins • If 4 Tails occur before 4 Heads, Tom wins NZAMT 2013

  8. Problem of the points • Outcome: H H T H T (3 Heads, 2 Tails) • Then game is interrupted, cannot be completed! • How should the prize ($10) be divided? • What might you propose? NZAMT 2013

  9. Problem of the points • My students’ answers • Game was not finished, so each takes $5 back • Heather won 3/5 of the flips, so she gets $6, Tom gets $4 • More mathematical approach • Divide prize proportionally to each player’s probability of winning if the game were continued NZAMT 2013

  10. Problem of the points • How to calculate these probabilities? • Simulation • Enumeration • I’ll ask everyone to use a coin to simulate 5 repetitions of this random process • Remember that Heather wins with one more Head, but Tom needs two Tails to win • Estimate Heather’s probability of winning by proportion of times that she wins NZAMT 2013

  11. Problem of the points • Enumeration analysis H: Heather wins TH: Heather wins TT: Tom wins • Are these equally likely, so Heather has 2/3 probability of winning? • No: H (prob .5), TH (prob .25), TT (prob .25) • Heather’s probability of winning: ¾ =.75 • So, Heather should take $7.50, Tom $2.50 NZAMT 2013

  12. Problem of the points • Variation: What if they were playing to get 10 heads or tails and were interrupted after 9 heads and 8 tails? • Heather needs 1 more Head, Tom needs 2 more Tails • Same analysis, same answer! NZAMT 2013

  13. Problem of the points • New variation: What if they were playing to 5 and had stopped after H H T H T? • More complicated enumeration: Heather wins with • H H (prob .25) • H T H (.125) • H T T H (.0625) • T H H (.125) • T H T H (.0625) • T T H H (.0625) • Heather has .6875 probability of winning • Heather should take $6.875 NZAMT 2013

  14. Matching problem • Four mothers give birth to baby boys on the same night at the same hospital • As a very, Very, VERY sick joke (do not try this at home!), the hospital staff returns the babies to the mothers at random • How likely is it that everyone will get the right baby? Or nobody will? Or at least one will? • What is the average number that would get the right baby in the long run? NZAMT 2013

  15. Matching problem • Simulate • Put baby’s name on each of four index cards • Shuffle cards thoroughly • Random distribute cards to four “mothers” • Repeat … • Repeat … • Repeat … • Approximate probabilities by proportions NZAMT 2013

  16. Matching problem • Simulate • Turn to technology: Random Babies applet (www.rossmanchance.com/applets/) • Which is the most likely outcome? • How unlikely is getting 4 matches? • Is 3 very unlikely or impossible? • What’s the long-run average number of correct matches? • What happens as you conduct more repetitions? NZAMT 2013

  17. Matching problem NZAMT 2013

  18. Matching problem NZAMT 2013

  19. Matching problem NZAMT 2013

  20. Matching problem NZAMT 2013

  21. Matching problem • Enumeration 1234 1243 1324 1342 1423 1432 2134 2143 2314 2341 2413 2431 3124 3142 3214 3241 3412 3421 4123 4132 4213 4231 4312 4321 • Count # of matches for each outcome NZAMT 2013

  22. Matching problem • Enumeration 1234 1243 1324 1342 1423 1432 4 2 2 1 1 2 2134 2143 2314 2341 2413 2431 2 0 1 0 0 1 3124 3142 3214 3241 3412 3421 1 0 2 1 0 0 4123 4132 4213 4231 4312 4321 0 1 1 2 0 0 NZAMT 2013

  23. Matching problem • Exact probabilities • 0 matches: 9/24 = .375 • 1 match: 8/24 ≈ .333 • 2 matches: 6/24 = .250 • 3 matches: 0 • 4 matches: 1/24 ≈ .042 • Expected value • 0(9/24) + 1(8/24) + 2(6/24) + 4(1/24) = 24/24 = 1 • Long-run average value NZAMT 2013

  24. Collector problem • Each cereal box is equally likely to contain any one of 3 prizes • You want to collect the entire series of 3 prizes • Suppose that you buy boxes one at a time • What is the fewest that you might need to buy? • What is the most that you might need to buy? • How many boxes should you buy to have a 95% chance of success? • How many boxes do you “expect” to need (on average)? NZAMT 2013

  25. Collector problem • What if there are k prizes? • Same questions as above • How likely are you to succeed if your strategy is to buy twice as many boxes as prizes? NZAMT 2013

  26. Collector problem • Simulate! • Write each prize on an index card • Shuffle cards thoroughly • Select one card at random, note which prize • Repeat, until all prizes have been obtained • Note the number of selections needed • Repeat … • Repeat … • Estimate probability distribution (of number of boxes needed) by distribution of results NZAMT 2013

  27. Collector problem • 3 prizes • 100,000 repetitions • Average: 5.483 boxes • Probability of success with 6 boxes: 0.742 • Boxes needed for ≥.95 probability: 11 NZAMT 2013

  28. Collector problem • 4 prizes • 100,000 repetitions • Average: 8.326 boxes • Probability of success with 8 boxes: 0.625 • Boxes needed for ≥.95 probability: 16 NZAMT 2013

  29. Collector problem • 10 prizes • 100,000 repetitions • Average: 29.322 boxes • Probability of success with 20 boxes: 0.214 • Boxes needed for ≥.95 probability: 51 NZAMT 2013

  30. Secretary problem My all-time favorite probability/math problem! Your task is to hire a new employee, subject to these constraints: • You know how many candidates have applied (n) • The candidates arrive in random order • You interview candidates one at a time • You can rank candidates after interviewing them • But you have no prior sense of quality NZAMT 2013

  31. Secretary problem More constraints: • After you have interviewed a candidate, you must decide immediately whether to hire • No follow-up interviews • No going back later • Your task is to choose the best! • Any other result: you’ve failed! NZAMT 2013

  32. Secretary problem Predict the optimal probability of success when: • n = 3 • n = 12 • n = 500 • n = 4,484,451 (population of NZ) • n = 7,182,483,662 (world population) NZAMT 2013

  33. Secretary problem • Enumeration for n = 1: A: 1 Probability of success = 1! NZAMT 2013

  34. Secretary problem • Enumeration for n = 2: A: 12 B: 21 • Two strategies • Hire first candidate • Hire second candidate Probability of success = .5 NZAMT 2013

  35. Secretary problem • Enumeration for n = 3: A: 123 B: 132 C: 213 D: 231 E: 312 F: 321 • Seems like probability of success = 1/3 • Can we do better? NZAMT 2013

  36. Secretary problem • Key insight • We can learn from the first candidate • Optimal strategy • Let one go by • Then hire first candidate who is the best so far A: 123 B: 132 C: 213 D: 231E: 312 F: 321 NZAMT 2013

  37. Secretary problem • Enumeration for n = 4: A: 1234 B: 1243 C: 1324 D: 1342 E: 1423 F: 1432 G: 2134 H: 2143 I: 2314 J: 2341 K: 2413 L: 2431 M: 3124 N: 3142 O: 3214 P: 3241 Q: 3412 R: 3421 S: 4123 T: 4132 U: 4213 V: 4231 W: 4312 X: 4321 • Same form for optimal strategy • But should we let 1 go by or let 2 go by? NZAMT 2013

  38. Secretary problem Let 1 go by A: 1234 B: 1243 C: 1324 D: 1342 E: 1423 F: 1432 G: 2134 H: 2143 I: 2314 J: 2341 K: 2413 L: 2431 M: 3124 N: 3142 O: 3214 P: 3241 Q: 3412 R: 3421 S: 4123 T: 4132 U: 4213 V: 4231 W: 4312 X: 4321 • Probability of success = 11/24 ≈ .4583 NZAMT 2013

  39. Secretary problem Let 2 go by A: 1234 B: 1243 C: 1324 D: 1342 E: 1423 F: 1432 G: 2134 H: 2143 I: 2314 J: 2341 K: 2413 L: 2431 M: 3124 N: 3142 O: 3214 P: 3241 Q: 3412 R: 3421 S: 4123 T: 4132 U: 4213 V: 4231 W: 4312 X: 4321 • Probability of success = 10/24 ≈ .4167 NZAMT 2013

  40. Secretary problem • n = 4: • Optimal strategy • Let 1 go by, then hire first one who is best so far • Probability of success = 11/24 ≈ .4583 • Decreasing, but not as quickly as most expect NZAMT 2013

  41. Secretary problem NZAMT 2013

  42. Secretary problem NZAMT 2013

  43. Secretary problem • Recall: • For large values of n: • And so: NZAMT 2013

  44. Secretary problem • Recall: • Goal: choose r to maximize f(r) • Elementary calculus gives: • Optimal probability of success becomes: NZAMT 2013

  45. Secretary problem Remarkable result: • As n gets infinitely large • Optimal strategy is to let ≈ first 1/e (about 37%) go by • Then hire first who is best so far • Optimal probability of success 1/e, about 37% NZAMT 2013

  46. Secretary problem • Extensions • Hidden number game • Finding your soul-mate in life NZAMT 2013

  47. Thanks! • arossman@calpoly.edu NZAMT 2013

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