Who's Going to Win the Playoff?

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  • Who's Going to Win the Playoff?Author(s): JACK A. OTTSource: The Mathematics Teacher, Vol. 78, No. 7 (OCTOBER 1985), pp. 559-563Published by: National Council of Teachers of MathematicsStable URL: http://www.jstor.org/stable/27964614 .Accessed: 13/07/2014 07:48

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  • Who's Going to Win the

    Playoff? By JACK A. OTT, West Hempstead Schools, West Hempstead, NY 11552

    Acontinuing challenge in teaching

    high school mathematics is finding interesting but accessible examples to illustrate the usefulness of the topic be

    ing studied. This dilemma seems espe cially acute in the area of probability, since the material taught stays quite elementary, but many applications use

    complex theories of statistical inference. As a result, exercises are often about colored balls in urns and tossed coins, not really the stuff of everyday situa tions.

    This article sets forth a realistic prob lem, computing the probability of win

    ning a sports playoff series, like the World Series or Stanley Cup, if the proba bility of winning a single game is known, which requires only simple permutation formulas and. some basic logic to solve.

    However, it is not a trivial problem and has a number of features that can be used to explore other areas.

    Attempting to solve this problem by a complete listing of all cases is possible but prone to error. However, a partial listing lays the framework for a more

    formal algebraic solution. In addition to

    discussing a listing and an algebraic ap

    proach, this article will also give a com

    puter solution using Monte Carlo meth ods. This technique provides an example where the computer makes solving a

    problem much simpler than the usual

    algebraic method. This Monte Carlo ap

    proach has been explored in a number of

    recent Mathematics Teacher articles (see the Bibliography). The program makes

    clear that a computer solution is more

    flexible and needs only simple adjust ments to solve variations on the original problem. The computer program is quite short, and its logic is very straightfor

    ward. It can be explained to a class by anyone who has learned some introduc

    tory programming concepts. The code used will run on any computer that al lows multiple statements on one line, though you might have to shorten some of the names if your machine has restric tions on the length of names for varia bles.

    The problem is this: Let W be the event that team Able

    wins a game and L be the event that team Able loses a game. If the proba bility of team Able beating team Baker is P(W) = a, then what is the probabili ty that team Able will win a best-of

    seven-game series with team Baker? Note that P(L) = 1

    - a.

    The statement of the problem shows a

    certain naivete. The probability of one team winning a game against another team would vary according to which team has the home advantage, whether

    key personnel are injured, and many other factors, but the solution to the

    problem as stated would be the first step in the solution of a more complex and realistic statement of the problem. The

    question of home advantage is covered

    by a second computer program given at the end of the article.

    The first step in the solution is to

    decide what might be the possible out comes of a seven-game series that team

    Able wins. These are as follows:

    1. four wins and no losses

    2. four wins and one loss

    3. four wins and two losses

    4. four wins and three losses

    Because of the unequal number of

    games played in each outcome, it is diffi

    October 1985 559

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  • cult to treat the problem as just one big permutation. Treating each of the four cases separately is simplest. Of course, how much of the structure of the solu tion to reveal to a class, and at what

    point in their struggle toward an answer to do so, is a judgment that must be left to the individual teacher.

    Four wins and no losses

    The probability of winning four straight games is

    P(WWWW) = (a)(a)(a)(a) = a\ since it can only happen in one way.

    Four wins and one loss

    The probability of winning four out of

    five games is computed by multiplying the probability of one such case times

    the number of ways this can happen. The

    standard formula for the number of per mutations when some of the objects be

    ing permuted are identical is

    where r

    ni = n i=l

    The last game must be a W, so we are

    really permuting L, W, W, W, or =

    4, n\ = 3, n2

    = 1. So

    4! -= 4

    3! 1!

    permutations are possible. Therefore, the probability of exactly four wins and one loss is

    (4)(1 - a)(a)(a)(a)(a) = 4(1 -

    a) a4.

    Four wins and two losses

    Similarly, the probability of four wins and two losses is

    [(1 -

    a)2 a4] = 10(1 -

    a)2 a\

    Four wins and three losses

    Finally, the probability of four wins and three losses is

    560 ?

    ^ [(1 - a)3 a4] = 20(1

    - a)3 a4.

    Total probability The total probability that team Able wins the series is denoted by P(Able) and is

    P(Able) = a4 + 4(1 -

    a) a4 + 10(1

    - a)2 a4 + 20(1

    - a)3 a4

    or

    (1) P(Able) = a4[l + 4(1 - a)

    + 10(1 -

    a)2 + 20(1 -

    a)3].

    Typical results are these:

    If a = P(W) = 0.75 (75%), then P(Able)

    = 0.929.

    If a = P(W) = 0.55 (55%), then P(Able)

    = 0.608.

    These results show what is intuitively known. A multigame series gives the

    stronger team a better chance to win the

    playoff when compared to this team's

    probability of winning a single playoff game.

    It follows from the method used that the general solution for finding the prob ability of winning a series of games is

    P(Able) = a'jl

    + ^ _^ (1

    - a)1

    (/-+ >-?>?

    UV-i>-*-'} where r = (n + l)/2, assuming is a

    positive odd integer. This approach is not the only way to

    solve the problem. It is possible to list all favorable outcomes for this finite sample space. Very often students show a pref erence for this approach. One way to

    organize such a list is shown in table 1. Students will likely make at least one

    mistake in trying to construct table 1.

    Nevertheless, such an attempt at listing can motivate students' learning of the

    permutation formulas used earlier. The

    attempt to make a complete list also leads them to organize the possible cases

    -Mathematics Teacher

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  • TABLE 1

    All Victorious Permutations

    4 wins, 0 losses

    4 wins, 1 loss

    4 wins, 2 losses

    4 wins, 3 losses

    WWWW LWWWW WLWWW WWLWW WWWLW

    LWWWLW LWWLWW LWLWWW LLWWWW WLWWLW WLWLWW WLLWWW WWLWLW WWLLWW WWWLLW

    LWWWLLW LWWLWLW LWLWWLW LLWWWLW LWWLLWW LWLWLWW LLWWLWW LWLLWWW LLWLWWW LLLWWWW WLWWLLW WLWLWLW WLLWWLW WLWLLWW WLLWLWW WLLLWWW WWLWLLW WWLLWLW WWLLLWW WWWLLLW

    in some way, and it can easily serve as a

    model from which to elicit the proper structure for an algebraic solution.

    Therefore, a teacher might even suggest that students start the problem by try

    ing to list the sample space.

    Program 1 attempts to solve this

    problem by Monte Carlo methods and some sample runs. With this approach the approximate probability of a certain outcome is determined by programming the machine to simulate the experiment. The computer's random-number genera tor is used to insure that the different outcomes of any single stage match the

    probabilistic specifications of the prob lem. Then the experiment is run repeat edly, and the computer counts the num

    ber of times the desired outcome is obtained. The approximate probability is

    the usual ratio of favorable events to

    total trials. If many trials are conducted, this approximate result will be quite close to the theoretical answer. The ad

    vantage of this method is that it is often very easy to find formulations for the

    individual stages of an experiment and,

    through the use of conditional state

    ments, call for the proper stages to follow one another. One does not have to ac

    count individually for all the possible outcomes.

    Another advantage of the computer method is the ease with which it can be

    changed to compute answers to varia tions on the original problem. Program 1 will permit any number of games in the

    playoff (line 250) and the assignment of any number to the probability of team Able winning a single game (line 300). This means the one program can solve

    many forms of the problem without the extra human effort required in recom

    puting the algebraic solution.

    Problem Extensions

    First, the probability that team Able wins is shown in equation (1). By similar

    logic it is easy to show that the probabili ty that team Baker wins the series is

    P(Baker) = (1 - a)4[l + 4a + 10a2 + 20a3].

    Since someone must win, then P(Able) +

    P(Baker) = 1, so

    a4 [1 + 4(1 -

    a) + 10(1 -

    a)2 + 20(1 -

    a)3] + (1

    - a)4 [1 + 4a + 10a2 + 20a3] = 1

    is an identity, that is, this equation must be true for all values of a. Independent of considerations of probability, can you prove this identity algebraically?

    Selecting the method of proof is most of the work. One keeps looking for bino mial expansion "hooks" for the problem but it is really best done by multiplying out all terms.

    Secondly, one can use the problem to discuss the whole concept of Monte Carlo

    probability. How many trials are

    enough? Can you use too many trials? How accurate is the answer? How repro ducible is the answer?

    Finally, one can improve the realism of the model by assigning two probabili ties. One is the probability of team Able winning a single game played at home.

    The other is the probability of team Able winning a single game when played away. These significantly complicate the

    algebraic solution. The computer pro gram can quite easily be modified to cope

    October 1985 561

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  • PROGRAM 1

    Program to Compute Playoff Probabilities by Monte Carlo Methods

    200 REM INITIALIZE COUNTERS FOR NUMBER OF SERIES WON

    210 LET AVICTORY = 0 220 LET BVICTORY = 0 230 REM INPUT PARAMETERS FOR

    THIS RUN 240 PRINT "HOW MANY GAMES IN

    THIS SERIES "; 250 INPUT NUMBER 260 PRINT 270 PRINT "WHAT IS THE PROBABILITY

    THAT TEAM A" 280 PRINT "WILL WIN A GAME ?

    PLEASE ENTER AS A" 290 PRINT "DECIMAL NUMBER BETWEEN 0

    AND 1 "; 300 INPUT PROBA 310 PRINT 320 PRINT "HOW MANY TRIALS SHOULD I

    MAKE IN" 330 PRINT "THIS RUN "; 340 INPUT TRIALS 350 FOR SERIES = 1 TO TRIALS 360 REM INITIALIZE COUNTERS FOR GAMES

    WON DURING ONE SERIES 370 LET AGAMESWON = 0 380 LET BGAMESWON = 0 390 FOR GAMES = 1 TO NUMBER 400 REM PICK A RANDOM NUMBER FROM

    0TO 1 410 LET X - RND(1) 420 REM SEE WHO WON THIS GAME BASED ON 430 REM THE PROBABILITY GIVEN BY THE

    USER 440 IF PROBA I X THEN LET AGAMESWON =

    AGAMESWON + 1 450 IF PROBA l= X THEN LET BGAMESWON ?

    BGAMESWON + 1 460 REM SEE IF THIS PARTICULAR SERIES

    IS FINISHED YET 470 IF AGAMESWON = (NUMBER + 1)/2 THEN

    LET AVICTORY - AVICTORY + 1 : LET GAMES = NUMBER : REM END LOOP TO AVOID STACK OVERFLOW

    480 IF BGAMESWON = (NUMBER + 1)/2 THEN LET BVICTORY = BVICTORY + 1 : LET GAMES = NUMBER

    490 NEXT GAME 500 NEXT SERIES 510 PRINT 520 PRINT "THE APPROXIMATE PROBABILITY" 530 PRINT "THAT TEAM A WILL WIN" 540 PRINT "A(AN) "NUMBER" GAME SERIES

    IF ITS" 550 PRINT "CHANCE TO WIN A SINGLE GAME

    IS" 560 PRINT 100

    * PROBA"% IS "

    AVICTORY/TRIALS"." 570 END

    HOW MANY GAMES IN THIS SERIES ?7

    WHAT IS THE PROBABILITY THAT TEAM A WILL WIN A GAME?PLEASE ENTER AS A DECIMAL NUMBER BETWEEN 0 AND 1 ?.6

    HOW MANY TRIALS SHOULD I MAKE IN THIS RUN ?100

    THE APPROXIMATE PROBABILITY THAT TEAM A WILL WIN A(AN) 7 GAME SERIES IF ITS CHANCE TO WIN A SINGLE GAME IS 60% is .73.

    APPENDIX

    Program to Compute Playoff Probabilities by Monte Carlo Methods Considering a Home and Away Series

    200 REM INITIALIZE COUNTERS FOR NUMBER OF SERIES WON

    210 LET AVICTORY = 0 220 LET BVICTORY = 0 230 PRINT "REMEMBER THAT TEAM A IS

    THE TEAM THAT" 240 PRINT "WILL BE THE HOME TEAM

    ONE MORE TIME" 250 PRINT "THAN TEAM WILL BE." 260 PRINT 270 REM INPUT PARAMETERS FOR

    THIS RUN 280 PRINT "HOW MANY GAMES IN

    THIS SERIES "; 290 INPUT NUMBER 300 PRINT 310 PRINT "WHAT IS THE PROBABILITY

    THAT TEAM A"

    320 PRINT "WILL WIN A GAME AT HOME "; 330 INPUT PHOME 340 PRINT 350 PRINT "WHAT IS THE PROBABILITY

    THAT TEAM A" 360 PRINT "WILL WIN A GAME PLAYED

    AWAY "; 370 INPUT PAWAY 380 PRINT 390 PRINT "HOW MANY TRIALS SHOULD I

    MAKE IN" 400 PRINT "THIS RUN "; 410 INPUT TRIALS 420 FOR SERIES = 1 TO TRIALS 430 REM INITIALIZE COUNTERS FOR GAMES

    WON DURING ONE SERIES 440 LET AGAMESWON = 0 450 LET BGAMESWON = 0

    562 Mathematics Teacher

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  • APPENDIX?Continued

    460 FOR GAMES = 1 TO NUMBER 470 REM PICK A RANDOM NUMBER FROM

    OTO 1 480 LET X = RND(1) 490 REM SEE WHO WON THIS GAME?THE

    RESULT IS BASED ON THE 500 REM PROBABILITIES INPUT BY THE

    USER WITH ODD NUMBERED 510 REM GAMES BEING USED AS HOME

    GAMES FOR TEAM A 520 IF PHOME I X AND (GAMES/2) I

    INT(GAMES/2) THEN LET AGAMESWON =

    AGAMESWON + 1 530 IF PHOME l= X AND (GAMES/2) I

    INT(GAMES/2) THEN LET BGAMESWON =

    BGAMESWON + 1 540 IF PAWAY I X AND (GAMES/2) =

    INT(GAMES/2) THEN LET AGAMESWON =

    AGAMESWON + i 550 IF PAWAY l= X AND (GAMES/2) =

    INT(GAMES/2) THEN LET BGAMESWON =

    BGAMESWON + 1 560 REM SEE IF THIS PARTICULAR SERIES

    IS FINISHED YET 570 IF AGAMESWON = (NUMBER + 1)/2 THEN

    LET AVICTORY = AVICTORY + 1 : LET GAMES - NUMBER : REM END LOOP TO AVOID STACK OVERFLOW

    580 IF BGAMESWON = (NUMBER + 1)/2 THEN LET BVICTORY = BVICTORY + 1 : LET GAMES = NUMBER

    590 NEXt GAME 600 NEXT SERIES 610 PRINT 620 PRINT 'THE APPROXIMATE PROBABILITY" 630 PRINT "THAT TEAM A WILL WIN" 640 PRINT "A(AN) "NUMBER" GAME SERIES

    IF ITS" 65? PRINT "CHANCE TO WIN A SINGLE HOME

    GAME IS" 660 PRINT 100 *...