I found an error in my previous algorithm. I now get an expected gain of 4.716526...
But I still have a doubt about one thing: suppose there are P red and N black cards remaining in the deck, what is the probability of drawing a red one ? Is it really P/(N+P) ??
At 03:00 25/07/2006, JoÃ£o Pedro Afonso wrote: >Hi to all. > >Nigel wrote: > > You have 52 playing cards (26 red, 26 black). You > > draw cards one by one. A red card pays you a dollar. > > A black one fines you a dollar. You can stop any time > > you want. Cards are not returned to the deck after > > being drawn. What is the optimal stopping rule in > > terms of maximizing expected payoff? Also, what is > > the expected payoff following this optimal rule? > > As I said in the previous post, I think I > achieved the solution for this problem. This is > a very interesting puzzle and I don't want to > spoil the solution for anybody in case I'm > right, so, for now, I'll only present the expected value for my strategy: > > E[v]= 1269479634238379/495918532948104 = > > ~ 2.55986... > > Can someone confirm this value or present a best one? > > > Cheers, >Joao Pedro Afonso