

A108885


Let T(m,p) be the value of the following game: there are m "minus" balls and p "plus" balls in an urn, for a total of n=m+p balls. You may draw balls from the urn one at a time at random and without replacement until you decide to stop drawing. Each minus ball drawn costs you $1 and each plus ball drawn gets you $1. Sequence gives triangle of numerators of T(np,p), 0 <= p <= n, read by rows.


4



0, 0, 1, 0, 1, 2, 0, 0, 4, 3, 0, 0, 2, 9, 4, 0, 0, 1, 3, 16, 5, 0, 0, 0, 17, 12, 25, 6, 0, 0, 0, 12, 58, 10, 36, 7, 0, 0, 0, 0, 1, 71, 30, 49, 8, 0, 0, 0, 0, 4, 113, 145, 21, 64, 9, 0, 0, 0, 0, 1, 47, 93, 527, 56, 81, 10, 0, 0, 0, 0, 0, 127, 21, 235, 294, 36, 100, 11, 0, 0, 0, 0, 0, 61, 284, 199, 202
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OFFSET

0,6


REFERENCES

L. A. Shepp, Stochastic Processes [Course], Statistics Dept., Rutgers University, 2004.


LINKS



FORMULA

T(m, 0)=0, T(0, p)=p; T(m, p) = max{0, (m/(m+p))*(1+T(m1, p))+(p/(m+p))*(1+T(m, p1))}.


EXAMPLE

Triangle of values T(np,p), 0 <= p <= n, begins:
[0]
[0, 1]
[0, 1/2, 2]
[0, 0, 4/3, 3]
[0, 0, 2/3, 9/4, 4]
[0, 0, 1/5, 3/2, 16/5, 5]
[0, 0, 0, 17/20, 12/5, 25/6, 6]
[0, 0, 0, 12/35, 58/35, 10/3, 36/7, 7]
[0, 0, 0, 0, 1, 71/28, 30/7, 49/8, 8]


MAPLE

M:=60; for m from 0 to M do T(m, 0):=0; od: for p from 0 to M do T(0, p):=p; od: for n from 1 to M do for m from 1 to n1 do p:=nm; t1:=(m/(m+p))*(1+T(m1, p))+(p/(m+p))*(1+T(m, p1)); T(m, p):=max(0, t1); od: od:


MATHEMATICA

M = 60; Clear[T]; For[m = 0, m <= M, m++, T[m, 0] = 0]; For[p = 0, p <= M, p++, T[0, p] = p]; For[n = 1, n <= M, n++, For[m = 1, m <= n1, m++, p = nm; t1 = (m/(m+p))*(1+T[m1, p]) + (p/(m+p))*(1+T[m, p1]); T[m, p] = Max[0, t1]]]; Table[T[np, p] // Numerator, {n, 0, 12}, {p, 0, n}] // Flatten (* JeanFrançois Alcover, Jan 09 2014, translated from Maple *)


CROSSREFS



KEYWORD



AUTHOR



STATUS

approved



