An Optimization Problem With a Surprisingly Simple Solution
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- Maude Ray
- 6 years ago
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1 An Optimization Problem With a Surprisinly Simple Solution D. Drinen, K. G. Kennedy, and W. M. Priestley 1 INTRODUCTION. Suppose you and n of your friends play the followin ame. A random number from the uniform distribution on [0, 1] will be enerated. This number is called the taret. Each of you will independently uess what the taret number will be and the person whose uess is closest will be declared the winner. In order to investiate an optimal stratey for this ame, we need to assume somethin about your friends uesses. Consider first the case where you have complete knowlede of all your friends uesses before you make yours. In that case the optimal stratey is trivial: simply order their uesses and then find the larest ap between successive uesses. If that ap is at least twice as lare as that between 0 and the smallest uess and also that between the larest uess and 1, then position yourself halfway between those two uesses. If not, then position yourself as close as the rules allow to the left of the smallest uess or the riht of the larest uess, as appropriate. Your friends would likely not find this ame to be worth playin and even you would probably not find it interestin. Let us now assume that you do not have exact knowlede of your friends uesses. Most of the papers on similar ames (see, for example, [1], [7], [], and [3]) make the assumption that your friends will behave ame-theoretically. That is, they will make uesses based on what they think your uess will be, knowin that your uess also depends on what they are likely to uess. With this approach it seems necessary also to assume the number of players is small, for if n is lare such strateical thinkin is probably not as feasible and the computations may become much more difficult. Here we are primarily interested in lare n, so we take a different viewpoint: we assume that you have probabilistic knowlede about your friends uesses. That is, you do not know what the uesses will be, but you know that they will all come from a particular probability distribution. We will consider several distributions in the sections ahead, but for now let us focus on the mathematics of a relatively simple case. If your friends are all computers prorammed to uess a uniform random number between 0 and 1, then the optimal stratey can be found. Before we do it, thouh, take a moment to uess what the answer is. If you 1
2 were playin this ame for a lare cash reward, what number would be your choice? Most people we have asked both mathematicians and others have said that they would choose 1/, and this seems like a reasonable choice. Indeed, you miht define your score to be the distance from your uess to the taret number, and since the player with the lowest score is the winner, it would make sense to try to minimize your expected score. A uess of 1/ accomplishes that, as the reader may easily verify. But the crux of the matter is that minimizin your expected score is not equivalent to maximizin your probability of attainin the minimum score. As it turns out, for n < 5 the best choice would indeed be 1/. But when n = 5 the optimal choice is not 1/, but is approximately.3 (or, if you prefer it,.7). A uess of.3 ives you rouhly an 18.39% chance of winnin, compared to a uess of 1/, which ives you a 17.97% chance. When n = 10, the best choice turns out to be almost exactly 1/6. In eneral, if n is lare, the optimal uess is approximately /(n + ). We will prove this in Section 4. Now let s examine what happens to those of us who would have uessed 1/. When n = 10, the probability of winnin with a uess of 1/ turns out to be about If n = 0, the win probability associated with a uess of 1/ is about If n = 99, the probability of winnin if you uess 1/ is This number, of course, is extremely close to 1/(n + 1), which is what your win probability would be if you made a completely random (uniform) uess. To see this, note that if you made a random uess, you would be indistinuishable from each of your friends and would therefore have the same win probability as each of them. Thus your win probability would be 1/(n + 1). It therefore turns out to be the case that, for lare n, a uess of 1/ is no better than a completely random uess. Later we will show that, for lare n, what is true for a uess of 1/ is also true for any other number in the open interval (0, 1). That is, if W n () denotes the win probability associated with a uess of, then for any in the open interval, the ratio of W n () to 1/(n + 1) approaches 1 as n ets lare. Officially, the theorem can be stated as follows: Theorem 1. Let X 1, X,..., X n, and T be independent uniform random variables on the unit interval. For any in the unit interval, define W n () = P ( T < min i X i T ). Then for any in the open interval (0, 1), as n approaches infinity. W n () 1/(n + 1) 1
3 In Section 4 we will establish in formula (6) an explicit expression for W n. For now, we can et a feel for the behavior of W n by lookin at its raph for various values of n. Fiure 1 shows raphs of W n () for n = 4, 9, and 49. Also shown on each plot is a horizontal line at 1/(n + 1). Fiure 1: Win probability as a function of your uess, with taret and uesses distributed uniformly on [0, 1]. 0. n = 4 n = 9 n = Two symmetric small peaks materialize in the raph of W n when n = 5 and evolve thereafter in an intriuin way. One can see that the peak on the raph of W n (for the sake of definiteness, we will always focus our attention on the left peak) moves toward zero as n ets larer. Also, the rane of choices for that yield win probabilities nontrivially reater than 1/(n + 1) becomes narrower. Hence, if you know n exactly you can make an optimal choice for, but if you only have the vaue knowlede that n is lare, then you miht not wish to try for the optimal value. If you think n is 500 for example, then the uess you make will be disappointin if n turns out to be only 00. So if a few thousand of your friends are makin uniform random uesses at a uniform random number, then it does not much matter what your uess is. Perhaps this is not surprisin. It may be surprisin, however, that it does not matter that the distribution is uniform. If the taret number is exponentially distributed, for example, and your friends uesses are also exponentially distributed (with the same parameter), is an analoue of Theorem 1 still true? It is. A main result of this paper is the followin eneralization of Theorem 1. Theorem. Let X 1, X,..., X n, and T be independent random variables from the same continuous distribution havin density function f. Let be any number for which both lim t f(t) and lim t + f(t) exist and are nonzero. Define W n () = P ( T < min i X i T ). Then as n approaches infinity. W n () 1/(n + 1) 1 By a continuous distribution we mean one with a continuous cumulative distribution function (cdf), thouh its associated density function (pdf) need 3
4 not be continuous. We restrict our attention to continuous random variables (those associated with continuous distributions), where the probability of a tie in the ame described at the beinnin of the paper will be zero. To et a feel for the content of this theorem, consider the case where the taret random number T is a standard normal random variable (mean 0, standard deviation 1), and your friends uesses are also distributed as independent standard normal random variables. The taret number is likely to be near zero, so you miht be tempted to uess zero or somethin close to it. But the uesses are likely to be clustered around zero as well, so a uess near zero will win only if the taret number ends up in a very narrow rane around your uess. At the other extreme, if you make a uess that is far away from zero, you have carved out a wider rane of winnin possibilities for the taret number. But because you are far from zero, the probability of the taret s landin in that wide rane may (or may not) be less than the probability of landin in some narrow rane near zero. Rouhly speakin, your choice is between claimin a tall skinny piece of the distribution (a uess near zero), a short fat piece (a uess far from zero), or somethin in between. The restriction on in the statement of the theorem simply means that your uess is reasonable in some sense. So Theorem says it doesn t matter what you choose, as lon as your uess is reasonable. Guess whatever you want. As lon as n is lare enouh, the crowdin out of uesses near the tall part of the distribution will almost exactly counterbalance the unlikelihood of uesses in the short part, makin your uess no better or worse than a random uess from the same distribution as the taret number. In Fiure we see Mathematica-enerated raphs of the function W n () for n = 9, 99, and 999 in the case where the X i s and T come from a standard normal distribution. Note the asymptotic converence to a horizontal line at 1/(n+1), and also note the small peaks movin outward. We shall see in Section 5 that this behavior, which we also saw in the uniform case, is typical. Fiure : Win probability as a function of your uess, with taret and uesses distributed as standard normals. n = 9 n = 99 n = In the next section we will state and prove a eneral formula for W n (), and the followin section contains the proof of Theorem. In Section 4 we specialize once aain to the settin considered in Theorem 1 and explore a few interestin aspects of the uniform case. Section 5 will compare raphs of W n () 4
5 for several different probability distributions, whose behavior is clarified by a eneralization of Theorem. We will close in Section 6 with remarks about related questions. A GENERAL FORMULA FOR W n (). Theorem 3. Suppose X 1, X,..., X n are independent identically distributed continuous random variables with pdf f X and cdf F X. Suppose that T is a continuous random variable with pdf f T, and that X 1, X,..., X n, T are independent. For any real number, let W n () = P ( T < min i X i T ). Then W n () = as (1 F X ()+F X (t )) n f T (t) dt+ (1 F X (t )+F X ()) n f T (t) dt. Before beinnin the proof, we should note that W n () can also be written (1) h (t) n f T (t) dt, () where h (t) is the probability that X 1 does not lie between and t. Formula (1) is handy in eneratin Mathematica plots as in Fiure, while the conceptually simpler formula () will be seen to suest a proof of Theorem. To prove Theorem 3, we start by notin that W n () = P (you win) = P (T, you win) + P (T >, you win). We will show that these two terms correspond to the two interals in formula (1). If you uess, you win if and only if each of X 1,..., X n lies outside the interval of lenth T centered at T, as seen in Fiure 3. We will consider in detail the case when T. P (T, you win) = P (T, and all X i are outside of [T, ]). If you are to win, then k of the X i s must be reater than and n k must be less than T, for some k between 0 and n. So your probability of winnin is n k=0 ( ) n P (T, X 1,..., X k >, and X k+1,..., X n < T ). k We will now rewrite the above as a sum of interals of the joint pdf f X1,...,X n,t : n k=0 ( ) n k t } {{ } k times t } {{ } n k times f X1,...,X n,t (x 1,..., x n, t)dx n dx 1 dt. 5
6 Fiure 3: T T T T By our independence assumption this becomes n ( ) n t f X1 (x 1 )dx 1 f Xk (x k )dx k f Xk+1 (x k+1 )dx k+1 k k=0 Each of the latter n k interals is equal to F X (t ) and each of the former k interals is equal to 1 F X (). So we have n ( ) n (1 F X ()) k F X (t ) n k f T (t)dt. k k=0 Passin the sum inside the interal and applyin the binomial theorem yields P (T, you win) = (1 F X () + F X (t )) n f T (t)dt. A similar computation produces the analoous result in the T > case: P (T >, you win) = (1 F X (t ) + F X ()) n f T (t)dt. These two equalities imply formula (1), which completes the proof of Theorem 3. The rewritin of formula (1) as () is straihtforward. 3 PROOF OF THEOREM. The representation of W n () in terms of h (t) as iven in equation () motivates our proof. Fix as in the hypotheses of Theorem and fix ɛ > 0. Focussin t f Xn (x n )dx n f T (t)dt. 6
7 for now on the positivity of the riht-hand limit at, we first bracket this limit close enouh to obtain positive numbers p 1, p, and δ so that 1 ɛ < p 1 /p < p /p 1 < 1 + ɛ and p 1 < f(x) < p (3) for x in (, + δ). If necessary, we shrink δ so that p δ < 1. Note that if t, h (t) = P (X 1 [, t ]) = 1 t f(x) dx. To use inequality (3) in estimatin the interand we must have x in (, +δ), for which it is sufficient to restrict t to the interval (, + δ/). Thus, the riht inequality in (3) implies h (t) 1 p (t ) for all t in (, + δ/). (4) Note also that p δ < 1 implies that the riht side of inequality (4) is strictly positive. Now, the left inequality in (3), toether with (4) and a simple antiderivative, can be used to conclude that +δ/ (n + 1)h (t) n f(t) dt p 1 p > p 1 p ( 1 1 ) (1 p δ) n+1 ( 1 ɛ ) if n is sufficiently lare. Because p 1 /p > 1 ɛ it follows that for such n, 1 ɛ < +δ/ (n + 1)h (t) n f(t) dt. Similar estimates usin inequality (3) in the other direction show that +δ/ and hence for sufficiently lare n we have 1 ɛ < (n + 1)h (t) n f(t) dt < 1 + ɛ, +δ/ (n + 1)h (t) n f(t) dt < 1 + ɛ. (5) Now we consider the interal +δ/ (n + 1)h (t) n f(t) dt. The function h is decreasin on (, ), so h (t) h ( + δ/) 1 p 1 δ if t > + δ/. Thus, for sufficiently lare n we have 0 +δ/ (n + 1)h (t) n f(t) dt (n + 1)(1 p 1 δ) n < (n + 1)(1 p 1 δ) n < ɛ. +δ/ f(t) dt 7
8 Combinin the above with inequality (5), we see that for lare n, so that 1 ɛ < lim n (n + 1)h (t) n f(t) dt < 1 + ɛ, (n + 1)h (t) n f(t) dt = 1. Similar reasonin applies to the interal on (, ), so and the result follows. lim n (n + 1)h (t) n f(t) dt = 1, 4 THE UNIFORM CASE. We now specialize to the ame described at the beinnin of Section 1, where the other players uesses and the taret are uniformly distributed with cdf F. In this case, { 0 if x / [0, 1], f(x) = f X1 (x) = f X (x) =... = f Xn (x) = f T (x) = 1 if x [0, 1]. Since F (t ) is 1 for t 1, 0 for t 0, and equal to t when t (0, 1), we have: 0 if t, F (t ) = t if < t < 1+, 1 if t 1+. Assume 0 1. To calculate W n () here, we will break up the first interal in equation (1) as 0 (1 F () + F (t )) n f(t)dt + and the second interal as 1+ 1 (1 + F () F (t )) n f(t)dt + 1+ (1 F () + F (t )) n f(t)dt, (1 + F () F (t )) n f(t)dt. Applyin what we know about the functions F and f, we deduce that W n () is iven by 0 (1 ) n dt + (1 + t) n dt (1 + t) n dt + n dt. 1+ 8
9 Routine interation yields W n () = 1 n Rearranin terms, we have ( (1 ) n + n (1 ) 1 n + 1 (n+1 + (1 ) n+1 ) ). (6) (n+1)w n () = n+1 +(n+1) n (1 )+(n+1)(1 ) n (1 ) n+1, (7) showin that the raph of W n has the same shape as the polynomial in on the riht-hand side but shifted and scaled. For lare n, this polynomial is easily seen to take very small positive values on a lare portion of the center of the interval [0, 1], but its behavior near the endpoints requires more scrutiny. Despite takin its minimum value of 1 at the endpoints 0 and 1, the polynomial takes a maximum value approachin e close to the points 1 = /(n+) and = n/(n+), thereby producin throuh formula (6) the small peaks in the raph of W n when n 5. To see this more clearly, let us first restrict our attention to in the interval [0, 1/]. Here the first two terms of the riht-hand side of equation (7) approach 0 uniformly, so the two sides of the approximation (n + 1)W n () (n + 1)(1 ) n (1 ) n+1 become uniformly close on [0, 1/] as n increases. It is a routine calculus exercise to verify that the riht-hand side attains at = 1 = /(n + ) its maximum value of (1 /(n + )) n, which approaches e as n. From this it follows that (n + 1)W n ( 1 ) 1 + e, (8) a result that, with a little more effort, can be derived directly from formula (6) via l Hôpital s Rule. By symmetry of W n the same limit is approached near the riht endpoint when = = n/(n + ). Note that (8) holds even thouh Theorem uarantees that for any fixed in the open interval (0, 1) we have (n + 1)W n () 1. The overshoot revealed by (8) may remind us of the Gibbs phenomenon encountered in approximations by Fourier series near a jump discontinuity of the limit function [4, pp. 6 66]. Here, this behavior may be related to constraints encountered in approximations by Bernstein polynomials, as we shall note below. Let us first summarize, however, the results of the foreoin discussion of the uniform case. Remark 4. If n 5, the maximum of W n () iven by formula (6) occurs at approximately 1 = /(n + ) and = n/(n + ), and max W n ()/(1/(n + 1)) approaches 1 + 1/(e ) as n approaches infinity. The implications for anyone who is playin the uessin ame described at the beinnin of the paper are as follows: 9
10 If you know precisely how many players n are in the ame, and if n is lare, you should uess /(n + ) or, equivalently, uess n/(n + ). This stratey will provide you with a win probability approximately 1/(e ) 6.767% better than a uniform random uess. Readers familiar with approximation properties expected of Bernstein polynomials [5] may be unsurprised by (8), once it is reconized that the polynomial on the riht-hand side of formula (7) is the (n+1)th Bernstein approximation of a function f specified to be identically zero on [0, 1] except near the endpoints, where f(0) = 1 = f(1) and f(1/(n + 1)) = 1 = f(n/(n + 1)). The polynomial must then aree with f at the endpoints and must try to approximate f uniformly on [0, 1], while of course encounterin difficulty because of the quick fluctuations of f near the endpoints. The resultin trade-off produces the shape of the raph of this Bernstein polynomial and, throuh formula (6), produces overshoots in W n () near the endpoints. The uniform case discussed in this section is one of the few situations where we have been able to simplify equation (1) to obtain an explicit formula for W n (). As we shall observe in the next section, thouh, the overshoots in the uniform case appear to manifest themselves in more eneral settins at points naturally correspondin to 1 and. In fact, the existence, for lare n, of overshoots near 0 and 1 in the uniform case may be seen to be due not so much to the fact that these points are endpoints, but that they are unreasonable uesses that lie in the support of the pdf under consideration. 5 NON-UNIFORM DISTRIBUTIONS. For non-uniform distributions, the interals in equation (1) are very difficult or impossible to evaluate symbolically, so we asked Mathematica to estimate them numerically. Shown here are the raphs of W n when n = 9, 99, and 999 for several different distributions. In accordance with Theorem, the raphs of W n approach 1/(n + 1) asymptotically at all reasonable values of. We will comment shortly on some of the unreasonable uesses. Fiure 4: Win probability as a function of your uess, with taret and uesses distributed accordin to the pdf f(x) = x on [0, 1] n = 9 n = 99 n =
11 Fiure 5: Win probability as a function of your uess, with taret and uesses distributed accordin to the pdf f(x) = e x on [0, ) n = 9 n = 99 n = In Section 4 we found that the uniform distribution of uesses produces the function W n iven by equation (6), whose raph exhibits small peaks near 1 = /(n + ) and = n/(n + ) if n 5. For other distributions we miht be led on intuitive rounds (see [6, Proposition 9.3.1]) to expect the resultin W n to approximate the composition of the function of equation (6) with the cdf F of the distribution in question, provided we are at a point where F has a positive derivative. We should then enerally expect Mathematica to plot new peaks near F 1 ( 1 ) and F 1 ( ) when n 5, and this seems to be the case in Fiures, 4, and 5. In Fiure 4, for example, the peaks seem to be located, as we should expect, near 1 and, for n = 9, 99, and 999, respectively. Surprises may occur, however, in more complicated settins where the cdf fails to be differentiable or where its derivative vanishes, as seen in Fiures 6, 7, and 8. Also, at the end of this section we note what can happen if the raph of the cdf has a vertical tanent. Fiure 6: Win probability as a function of your uess, with taret and uesses distributed accordin to the piecewise uniform distribution with pdf f(x) = 1/ on [0, 1/] and f(x) = 3/ on (1/, 1] n = 9 n = 99 n = Recall that the pdf of the distribution need not be continuous in order for Theorem to hold, but the example in Fiure 6 shows that some intriuin behavior may occur near a point of discontinuity. In particular, we did not expect to see a local minimum just to the left of 1/. The two distributions used in Fiures 7 and 8 are very similar and in each 11
12 Fiure 7: Win probability as a function of your uess, with taret and uesses distributed accordin to the pdf f(x) = 4 x 1/ on the unit interval n = 9 n = 99 n = Fiure 8: Win probability as a function of your uess, with taret and uesses distributed accordin to the pdf f(x) = 1(x 1/) on the unit interval n = 9 n = 99 n = case the hypotheses of Theorem fail to be satisfied at = 1/. We should therefore not be too surprised that W n (1/) does not approach 1/(n+1) asymptotically. That the limitin value of (n+1)w n (1/) differs between the two cases is attributable to the followin eneralization of Theorem. Theorem 5. Let X 1, X,..., X n, and T be independent random variables from the same continuous distribution and let f be their common pdf. For a iven, suppose there exist real numbers k and k, both reater than 1, such that lim t + f(t) and lim t k t f(t) t k both exist and are nonzero. Then lim (n + 1)W n() = 1 n k k +1. (9) Theorem is Theorem 5 with k = k = 0. Fortunately, the proof of Theorem eneralizes to this larer settin, althouh we encounter slihtly more complicated expressions. The requirement p δ < 1 made at the outset of the proof of Theorem must be modified to ensure that p δ k+1 < k + 1 in eneral. Other modifications suest themselves. We ask the reader to fill in the details of the proof of Theorem 5. What does Theorem 5 tell us? At = 0 in Fiure 4, we have essentially a one-sided case where k = 1, so (n + 1)W n (0) 1/4. In Fiure 7, we have 1
13 k = k = 1 at = 1/, so (n + 1)W n (1/) 1/; while in Fiure 8 we have k = k =, so (n + 1)W n (1/) 1/4. We rarely deal with unbounded pdf s, but they can produce interestin effects, such as makin the riht-hand side of equation (9) exceed 1 and forcin the raph of the associated cdf to have a vertical tanent. Consider, for example, the distribution on [ 1, 1] iven by the pdf f(x) = (1/4) x 1/. Here, k = k = 1/ at = 0, so (n + 1)W n (0). Moreover, despite what our foreoin discussions miht have led us to expect, W n seems to attain its maximum here for all n at = 0. 6 RELATED QUESTIONS. Here we mention some issues that, for lack of space and/or insiht, seem to deserve more attention than we can provide. 6.1 UNREASONABLE GUESSES AND INDUCED OVER- SHOOTS? More precise definitions may help us better study the intriuin relationship suested in the Mathematica plots we have seen. To that end, we define an unreasonable uess to be a point lyin in the support of the common pdf f of the X i s and T such that lim sup(n + 1)W n () = L < 1. Theorem 5 often enables us to calculate L easily, and shows that unreasonable uesses include not only endpoints but also points like 1/ in Fiures 7 and 8 where the pdf tends to zero. We will say that we have an overshoot at if for each neihborhood N of, lim inf M n = M > 1, where M n is the supremum of (n + 1)W n on N. Remark 4 confirms that we have overshoots (M = 1 + e /) at 0 and at 1 in the uniform case. The heuristic arument in Section 5 suests that we should expect the same value of M associated with all unreasonable uesses in Fiures 1,, 4, and 5 where in Fiure we should consider and to be unreasonable uesses. We do not know whether there must necessarily be an overshoot at each unreasonable uess. Note, however, that if you decide to uess accordin to the same pdf f as your n friends then you would be indistinuishable from them, so all of you would have the same win probability of 1/(n + 1). Therefore, W n ()f() d = 1 n + 1. (10) Hence, for each n, the continuous function (n + 1)W n interates to 1 aainst the pdf f. This remark, toether with Theorem, seems to make it at least plausible that if is a point in the support of f where the lim sup L is less than 1, then the lim inf M should exceed 1, but we don t see a proof. 13
14 6. WHEN THE GUESSES DIFFER FROM THE TAR- GET. While most of this paper has focussed on the case where the taret and your friends uesses come from the same distribution, equation (1) is eneral enouh to handle situations where the taret random variable T differs from the uesses X 1,..., X n. Suppose for example that the taret is standard normal. Theorem says that, if your friends uesses are also standard normal, then for lare enouh n, any uess you make is essentially as ood as any other. Suppose instead, however, that your friends uesses cluster around zero more tihtly than the taret. Then uesses near zero will be crowded out, hence becomin less profitable for you, and uesses far from zero (thouh not too far) will become more profitable. If, on the other hand, your friends uesses are slihtly more spread out than the taret, then uessin near zero becomes a better stratey for you. Shown in Fiure 9 are pictures of W n for n = 999 if the taret T is standard normal while your friends uesses are normally distributed with mean zero and standard deviations.98 and 1.0, respectively. Compare these pictures with Fiure. Fiure 9: Win probability as a function of your uess, where the distribution of uesses differs from distribution of the taret. n = 999 n = THE PRICE IS RIGHT. One variation on the ame is to play it Price is Riht style, where the winner is the player whose uess is closest to the taret amon all uesses less than the taret. Here, a player is disqualified if he or she overbids. Followin the stratey outlined in the proof of Theorem 3, the reader may verify that in this case the probability of winnin with a uess of is iven by W n () = (1 F X (t) + F X ()) n f T (t) dt, where F X represents the cdf of the uesses and f T represents the pdf of the taret. 14
15 If the uesses are assumed to come from the same distribution as the taret, and if the pdf of the taret is a continuous function, then this interal is surprisinly simple to evaluate: W n () = 1 n + 1 (1 F X() n+1 ). Note that W n is decreasin (no overshoots here), so it follows that the optimal stratey is to make the minimum allowable uess. If the least uess is an allowable option, it ives you the optimal win probability of 1/(n + 1). Also note that, in effect, there are n + players here, since the house wins if you and your n friends all overbid. 6.4 AN APPLICATION? We close with a few remarks about the motivation for this problem. This paper arose out of an attempt by the first author to build an abstract model for the optimal selection of an entry in an NCAA basketball tournament prediction contest with a lare number of entrants, such as those run each year by major sports websites like ESPN.com and yahoo.com. There are some vaue similarities between a basketball pool and the ame described in this paper. In both cases, somethin will happen, lots of people will make a uess at what that somethin will be, and the one who is closest will be declared the winner. In the abstract ame of this paper, the participant must choose between makin a uess in the tall part of the taret distribution and potentially bein crowded out by many other participants makin similar uesses, and makin a uess in the short part of the taret distribution, which has less likelihood of bein close to the taret but a better chance of winnin if it does happen to be close. In the basketball pool, the former choice corresponds to pickin the teams who are favored to win while the latter corresponds to choosin an entry with many lon shot teams. If you believe that the entries in a lare pool satisfy somethin approximatin the conditions of Theorem, then our results miht be loosely interpreted to mean that it doesn t really matter what kind of entry you fill out. Serious basketball fans miht be disappointed to learn that all their basketball knowlede is for nauht in these lare contests. Casual basketball fans, on the other hand, miht find this result liberatin. They can fill in their brackets based on the eoraphy of the teams, their uniform color, or the relative fierceness of their mascots, and their chances will be as ood as anyone else s. While we were pleased with the mathematics that arose from the investiation, even the first author must concede that the connection between our abstract ame and its real-life inspiration is probably too loose to offer much practical advice for those who wish to enter such contests. Dedication. This paper is dedicated to Sherwood F. Ebey, whose enthusiasm for the study of probability has proven to be infectious. 15
16 References [1] S. Even, The price is riht ame, this Monthly 73 (1966) [] T. Feder, Toetjes, this Monthly 97 (1990) [3] T. Feruson and C. Genest, Toetjes na, Mathematical Statistics and Applications: Festschrift for Constance van Eeden, IMS Lecture Notes/Monoraph Series 4 (003) [4] T. Körner, Fourier Analysis, Cambride University Press, Cambride, [5] G. Lorentz, Bernstein Polynomials, Chelsea Publishin, [6] S. Ross, Probability Models for Computer Science, Harcourt/Academic Press, San Dieo, 00. [7] J. Steele and J. Zidek, Optimal strateies for second uessers, Journal of the American Statistical Association 75 (1980) Dou Drinen is an associate professor at the University of the South. He received a B.A. from Trinity University (TX) in 1993 and a Ph.D. from Arizona State in He only very occasionally thinks about anythin other than sports or math, so he is always excited when he can save some time by thinkin about both simultaneously. Department of Mathematics and Computer Science, Sewanee: The University of the South, Sewanee, TN ddrinen@sewanee.edu K. Grace Kennedy raduated from the University of the South in 006 majorin in mathematics and French literature. She is currently pursuin a Masters in roup theory at the Université de Picardie in Amiens, France. In the spirit of the liberal arts tradition, Grace will try to see every work of art in the Louvre that she studied in Art History 103 at Sewanee. Soon, she hopes to return to the US for raduate study on roup theory topics related to eneralizations of the braid roup. kracekennedy@alumni.sewanee.edu Mac Priestley has tauht at the University of the South since 1967, havin received a B.A. there in 196. His Ph.D. deree is from Princeton. He wrote an elementary text, Calculus: A Liberal Art (Spriner-Verla, 1998), to support a one-semester eneral education course for liberal arts majors who may lack a stron backround in mathematics. In 006 he received the Distinuished Teachin Award from the Southeastern Section of the MAA. Department of Mathematics and Computer Science, Sewanee: The University of the South, Sewanee, TN wpriestl@sewanee.edu 16
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