|
|
![[parent]](http://images.planetmath.org:8080/images/uparrow.png) Viewing Message
|
|
|
| ``Re: New Year Ponder This!!!''
by balakrishnan_v on 2009-01-06 08:21:25 |
|
| My solution:
Let t be the last value attained. If you stop, the you will get t-kc where k is the number of tickets bought.
If you don't stop, you get beststrat(t')-(k+1)c where t' is a random variable between 0 and 1. beststrat(t') is the maximum gain that you can get given the previous value was t'.(modulo the losses of all the kc tickets previously)
We will abbreviate beststrat() as f()
We will chose the second one if E(f(t'))-(k+1)c>=t-kc or t0=E[f(t')]-c>=t
so we have
f(t)=E[f(t')]-c=t0 if t<=t0
f(t)=t if t>=t0
This implies that E[f(t')]-c=integral_{0}^{t0} t0 dt+ integral_{t0}^{1} t dt -c = (1+t0^2)/2-c = t0
solving gives t0=1-sqrt(2c) which is f(0) which is what we need(as we start with no money)
For c>1/2 , we have t0 to be negative. and the strategy is to stop at the beginning itself giving us 0 profit.
So the answer is
f(t)=1-sqrt(2c) for c<=1/2
f(t)=0 for c>1/2
|
| | [ reply | up | top ] | |
|
|
|
|
|
|
|
