Open Board

[ Read Responses | Return to Index | Return HOME ]
[ Previous | Previous in Thread | Next ]

Solve a different problem ...

 Posted by ET Fan on 15 February 2005, at 10:28 a.m., in response to Re: My final take on this controversy., posted by Zenfighter on 11 February 2005, at 12:46 p.m.



... get a different answer. McDowell was was trying to give ~3 extra aces to the dealer and player, not 10. From his errata: "In this case, that means the player and the dealer get three additional aces each. The probability of the Ace "hitting the money" P(h) and ... P(d) become 0.10 while P(m) is reduced from 0.87 to 0.80."


Now, either you are taking his formula on faith and applying it directly -- which would be a HUGE MISTAKE, because in that case you are still adding EVs of events which are not mutually exclusive -- or you are simply solving a different problem from the one McDowell tried to solve in the errata, and I successfully solved on


Also, there is something wrong with your "Exact cost for the 80% of the hands where neither the player nor the dealer gets a first card ace" (-1.524617%). The cost for an ace completely missing the round is ~EOR(A)/6 since TC = -1/6. This is on the order of -0.1%. You are saying the cost for the ace missing half the hand is many times higher than the cost for the ace missing the hand completely.





The Open Board is maintained with WebBBS 2.24.092606.