Repost: Cookie Problem?

Question

A huge jar of cookies has N whole cookies. A cookie is drawn at random, broken into halves, of which one is eaten, the other tossed back into the jar. Subsequently, if any half cookie is drawn, it is eaten, if any whole cookie is drawn, it is broken into halves, etc. Eventually all the cookies are eaten. For a given large N, what is the maximum number of half cookies will the jar have, before all the cookies are eaten? "Half cookies" don't include whole cookies, but those halves broken from whole cookies.

Dr D, read the original posting of this question:

http://answers.yahoo.com/question/index;_ylt=AljLVo8RBtg.K8Pg4OSLuXHsy6IX?qid=20070527010425AA8gCwq

The odds are that the total number of half cookies will rise from 0 to a maximum and fall back to 0, since a random drawing could pull either a whole cookie or a half cookie, except for the first draw.

Answer 1

t - time
n(t) - number of whole cookies in the jar
h(t) - number of halves in the jar

probailty to draw whole cookie is n/(n+h)

dn/dt = -n/(n+h)
dh/dt = -h/(n+h) + n/(n+h) = (n-h)/(n+h)

dh/dn = (h-n)/n = h/n - 1

let h(t) = u(t)n(t)

d(un)/dn = u - 1
u + n du/dn = u - 1
n du/dn = - 1
du = - dn/n
u = -ln(n) + C

h = -n(ln(n) + C) = n ln(No/n)

Maximum is chieved when dh/dn = 0,
h(t) = n(t)
n ln(No/n) = n
No/e = n_max


Answer:
Maximum number of halves h_max = No/e,
is achieved when numeber of whole cookies is
n_max = h_max = No/e too.

Answer 2

Well if all the whole cookies are selected and none of the halves are selected until all the wholes are selected, then you can have a maximum. You'll have N halves in the jar. The other N halves would be eaten. After this the number of halves would begin to decrease.