
Re: (difficult)Theoretical gambling puzzle (solution?)
Posted:
Jul 25, 2006 8:11 AM


Hi again and again :)
Well, maybe I should stick to geometry... My current value is 2.6244755... (seems to converge towards yours :)
I assumed that the expected gain E(P,N) depends only on the quantities of red (P for positive) /black (N for negative) cards remaining in the deck.
E(0,N) is N (no more positive, stop now) E(P,0) is 0 (no more negative, draw all remaining cards to return to 0)
In the general case, we compare the current gain S = (NP) if we stop now to the expected gain if we continue C = E(P1,N)*P/(N+P) + E(P,N1)*N/(N+P). Here I have a little doubt about the probability of picking a P or a N, then E(P,N) = max(S,C)
 Eric
At 12:39 25/07/2006, Joao Pedro Afonso wrote:
> Hi Eric, :) > > I think you escaped from a good one. Now I have to read the > mapple script, but look what I was preparing to send in reply to > your first message, in the moment it arrived to the MathOrg (where > it is probably waiting now for the moderators approval): > >:) > >Eric Bainville wrote: >>Hi, >>Very fun problem! >>My strategy gives an expected gain of >>805867616040669 / 281474976710656 = 2.863016903... > > Grrrr! Your expected gain is bigger than mine!!! >( > >>Small numbers of cards can be checked by hand (P is a +1 card, N is >>a 1 card). >... >>2+2 cards: >>P: continue => >> PP: stop 2 (1/4) >> PN: continue => >> PNP: stop 1 (1/8) >> PNN: continue => PNNP 0 (1/8) >>N: continue => >> NP: continue => >> NPP: stop 1 (1/8) >> NPN: continue => NPNP 0 (1/8) >> NN: continue => NNPP 0 (1/4) >>=> 3/4 > > Hufff! Look careful the way you are doing your probabilities. :) > > > Cheers, >Joao Pedro Afonso >

