From: grabiner@math.harvard.edu (David Grabiner)
Newsgroups: rec.games.bridge
Subject: Re: Which inference is better, WAS - "finesse or play for the drop"
Date: 20 Nov 92 03:23:39 GMT
 55K@irvine.com><1ee8ukINN8ij@agate.berkeley.edu><1992Nov18.224411.4092@u.washi
 ngton.edu><GRABINER.92Nov19121848@boucher.harvard.edu> <davidm.722213002@questor>
Organization: /etc/organization
In-reply-to: davidm@questor.Rational.COM's message of 19 Nov 92 22:43:22 GMT

In article <davidm.722213002@questor>, David Moore writes:

> grabiner@math.harvard.edu (David Grabiner) writes:

Please check your attributions: I didn't make either of the statements
below.  Eddie Grove made the first one, and Bryant Fujimoto made the
second.

>>>>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. 

It won't be 1 unless you shuffle the deck several times.  Even a good
riffle shuffle (randomly merging two piles) will break only half the
pairs.  Seven shuffles are needed to come close to randomizing a deck,
if the previous order is fixed.

> Let's suppose you get 10 such pairs. 

That's probably a bit too many for the average shuffler; three normal
shuffles might be expected to leave about 10 pairs, but many people
shuffle more than three times.

> Let us also suppose that the
> probability of the K being over the Q in a pack is 10% (before the
> shuffle). 

I don't think this is likely.  If the deck came fresh from the factory,
was sorted by hand, or was just used for a winning game of solitaire,
the probability that the king is over the queen is either 0% or 100%.
But in a deck that was last used for a game of rubber bridge, the main
reason that would make the king would lie over the queen would be if it
covered a queen on the last deal.  I don't think that a king covers a
queen on 1/3 of the 25% of hands that have the king lying over the
queen.

> 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 incorrect.  If the king lies over the queen, and
they aren't split, the king definitely lies over the queen.  Otherwise,
it is 50-50, but you only get 50% of the *remaining* cases, which happen
98% of the time by your figuring.  You get .02 + .5*.98=.51, and even
closer to .50 if you make more reasonable assumptions.

And don't forget that 2-2 breaks are also more likely in a poorly
shuffled deck; if the spade trick containing the ace stayed together as
a clump of four cards, it's better than 50-50 that the other two spades
are split.  You should still play for the drop.

--
David Grabiner, grabiner@zariski.harvard.edu
"We are sorry, but the number you have dialed is imaginary."
"Please rotate your phone 90 degrees and try again."
Disclaimer: I speak for no one and no one speaks for me.