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Re: (difficult)Theoretical gambling puzzle (solution?)
Posted:
Jul 25, 2006 8:11 AM
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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 = (N-P) if we stop now
to the expected gain if we continue C = E(P-1,N)*P/(N+P) + E(P,N-1)*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
>
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