home *** CD-ROM | disk | FTP | other *** search
- Newsgroups: rec.games.bridge
- Path: sparky!uunet!rational.com!questor!davidm
- From: davidm@questor.Rational.COM (David Moore)
- Subject: Re: Which inference is better, WAS - "finesse or play for the drop"
- Message-ID: <davidm.722213002@questor>
- Sender: news@rational.com
- Organization: Rational
- References: <1992Nov16.131237.19210@ms.uky.edu> <lsimonse.722011812@vipunen.hut.fi><BxwHCq.55K@irvine.com> <1ee8ukINN8ij@agate.berkeley.edu><1992Nov18.224411.4092@u.washington.edu> <GRABINER.92Nov19121848@boucher.harvard.edu>
- Date: Thu, 19 Nov 1992 22:43:22 GMT
- Lines: 27
-
- grabiner@math.harvard.edu (David Grabiner) writes:
-
-
- >>>This is wrong, of course. Unless you use a perfect random source for
- >>>your shuffles, the conditional distribution of the opponents' cards
- >>>will tend to depend upon your cards.
-
- >> Would someone please explain this last point more fully? I am afraid
- >> it isn't obvious to me.
-
- Try this experiment. Sort a deck of cards into order (2C 3C .... KS AS) and
- then shuffle them. Now count the number of times a card is followed by the
- next higher card; that is, they have "stuck together" during the shuffle.
-
- If the shuffle was totally random, the average number of such occurences in
- a deck would be 1. In practice, the number will be quite high, even if you
- make a good attempt at shuffling.
-
- Let's suppose you get 10 such pairs. Let us also suppose that the probability
- of the K being over the Q in a pack is 10% (before the shuffle). Then the
- probability of a finess working is raised to roughly 20%*10% + 50%=52%, and
- the probabiliuty of it failing is lowered to 48%. (This calculation is actually not
- quite right; the fact that the K and Q are not with you and your partner raises the
- probability that they stuck together, so that the answer should be higher, but as we
- are guessing at the other numbers, this is of little importance)
-
- So, now the finesse and the drop are about equal probabilities.
-