The Probability of Winning a Series. Gregory Quenell

Transcription

1 The Probability of Winning a Series Gregory Quenell

2 Exercise: Team A and Team B play a series of n + games. The first team to win n + games wins the series. All games are independent, and Team A wins any single game with some fixed probability p. What is the probability that Team A wins the series? p p A p p p B AA AB BA BB p p AAA AAB ABA ABB BAA BAB BBA BBB p + p ( p) + p ( p) = p p

3 Exercise: Team A and Team B play a series of n + games. The first team to win n + games wins the series. All games are independent, and Team A wins any single game with some fixed probability p. What is the probability that Team A wins the series? p p A p p p B AA AB BA BB p p p p AAA AAB ABA ABB BAA BAB BBA BBB p + p ( p) + p ( p) + p ( p) = p p

4 Vertical solution for n + games: P (A wins the series) = = = n+ k=n+ n+ k=n+ n+ k=n+ n+ = p n+ P (A wins the series on the k th game) P (A wins n of the first k games) P (A wins the k th game) ( ) k p n ( p) k n p n k=n+ ( ) k ( p) k (n+) n

5 Horizontal solution for n + games: P (A wins the series) = = n+ k=n+ n+ k=n+ P (A wins k of the n + games) ( ) n + p k ( p) n+ k k A B AA AB BA BB AAA AAB ABA ABB BAA BAB BBA BBB

6 The probability that Team A wins a (n+)-game series n+ p n+ k=n+ ( k n ) n+ ( p) k (n+) k=n+ ( n + p (n = 0) p p p (n = ) p p k ) p k ( p) n+ k 0p p + p (n = ) 0p p + p p 8p + 0p 0p (n = ) p 8p + 0p 0p.0 P (A wins) p

7 What are the coefficients? n+ p n+ k=n+ ( k n ) ( p) k (n+) = n+ r=n+ After some messy reindexing, we get [ ( ) r (n+) n+ k=r ( k n ) ( ) ] k (n + ) p r r (n + ) n+ k=n+ ( n + k ) p k ( p) n+ k = n+ r=n+ [ r k=n+ ( n + ( ) r k k ) ( n + k r k ) ] p r

8 A Pascal s triangle relation Equating the coefficients of p r in the vertical and horizontal polynomials, we get a two-parameter family of relations: [ ( ) r (n+) n+ [ r k=n+ ( k n k=r = ( n + ( ) r k k ) ( ) ] k (n + ) r (n + ) ) ( n + k r k for each n and for each r in {n +, n +,..., n + }. ) ] 8

9 ( ) r (n+) n+ k=r ( k n ) ( ) k (n + ) r (n + ) = r k=n+ ( n + ( ) r k k ) ( n + k r k ) Example: n =, r = ( )( ) ( )( ) + 0 ( )( ) + 0 ( )( ) 0 = = ( )( ) 0 9

10 ( ) r (n+) n+ k=r ( k n ) ( ) k (n + ) r (n + ) = r k=n+ ( n + ( ) r k k ) ( n + k r k ) Example: n =, r =. ( )( ) 0 0 ( 0 )( ) ( )( ) = 8 = ( 0 0 )( ) + 0 ( )( ) 0 0

11 ( ) r (n+) n+ k=r ( k n ) ( ) k (n + ) r (n + ) = r k=n+ ( n + ( ) r k k ) ( n + k r k ) Example: n =, r = ( )( ) + ( )( ) = 0 = ( )( ) ( )( ) + ( )( ) 0

12 ( ) r (n+) n+ k=r ( k n ) ( ) k (n + ) r (n + ) = r k=n+ ( n + ( ) r k k ) ( n + k r k ) Example: n =, r =. ( )( ) = 0 = ( )( ) + ( )( ) ( 0 0 )( ) + 0 ( )( ) 0 0

13 Remark: In the boundary case where r = n +, we get the hockey stick relation =

14 Remark: In the other boundary case where r = n +, we get an alternating hockey stick = +

15 Back to the polynomials We translate and scale the functions to get a sequence of odd polynomials on [, ] s 0 (x) = x x x s (x) = s (x) = x 0x + x 8..0

16 Observation This sequence converges pointwise and monotonically to the signum function s 0 (x) = x x x s (x) = s (x) = x 0x + x 8..0

17 Question Can we make this sequence converge to sgn(x) in some vector space of functions on [, ]? That is, can we find a system {φ 0, φ, φ,...} of polynomials on [, ] so that each φ k is a polynomial of degree k + there is an inner product, with φ m, φ n = δ m,n sgn(x) = sgn, φ k φ k (x) k=0 our s n polynomials are the partial sums of this series: s n (x) =? n sgn, φ k φ k (x) k=0

18 First attempt: Use the normalized Legendre polynomials The partial sums converge to sgn(x), but they are not our s n polynomials. 8

19 Second attempt: Use the normalized Chebyshev T polynomials Again, we do not get the monotonic convergence. 9

20 Inconvenient Truth: Our polynomials s n cannot be the partial sums of a series sgn, φk φ k Reason: In such a series, the difference between two partial sums, s n+ s n.0 is a scalar multiple of φ n+, so it must be orthogonal to s n. But in our sequence, s n (x) and s n+ (x) s n (x) always have the same sign, so their inner product s n, s n+ s n = cannot be zero. s n (x)[s n+ (x) s n (x)]w(x) dx 0

21 Question: Why does f, g have the form f(x)g(x)w(x) dx where w is some non-negative weight function? In R n, we can take x, y to be [ x T ] [ ] [ ] A y for any symmetric, positive-definite matrix A (not just a diagonal A). So why not say f, g = where A is symmetric and positive definite? f(x)a(x, y)g(y) dy dx

22 Try this on s 0 (x) = x s (x) s 0 (x) = x( x ) s (x) s (x) = 8 x( x ) s (x) s (x) = x( x ) s (x) s (x) = 8 x( x ) These functions are orthonormal with respect to the inner product f, g = where A is the function graphed above. f(x)a(x, y)g(y) dx dy,

23 Better news: The polynomial s n has the following properties: s n is odd of degree n s n () = s (r) n () = 0 for r =,,..., n

24 Better news: The polynomial s n has the following properties: s n is odd of degree n s n () = s (r) n () = 0 for r =,,..., n Better still, these properties determine s n..0 Example: Write s (x) = a x + a x + a x. Then = s () = a + a + a 0 = s () = a + a + a 0 = s () = a + a a = /8 a = 0/8 a = /8

25 Observation: This makes our s n polynomials look a lot like power-series convergents to the signum function Sort of:.0 An n-coefficient polynomial approximation to a function f is determined by values of f at 0 and ; the right number of derivatives of f at ; and being an odd polynomial.

26 Something else to play with: Here are some even polynomials v n (x) determined by v n (0) = 0; v n () = ; v n() = for n > ; v (r) n = 0 for n > and r v (x) = x v (x) = (x x ) v (x) = 8 (x 0x + x ) v (x) = (0x 0x + x 0x 8 ).

27 Something else to play with: Here are some even polynomials v n (x) determined by v n (0) = 0; v n () = ; v n() = for n > ; v (r) n = 0 for n > and r. And here is an even function with all those same properties v (x) = x v (x) = (x x ) v (x) = 8 (x 0x + x ) v (x) =. (0x 0x + x 0x 8 )