From: davidm@questor.Rational.COM (David Moore)
Subject: Re: Which inference is better, WAS - "finesse or play for the drop"
Organization: Rational
Date: Fri, 20 Nov 1992 18:35:43 GMT


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

>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.  

Sorry, I have had the same thing happen to me, so I know how annoying it
can be.

>> 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.

Right, by "shuffle" I meant the total process, not just one riffle. Computer
dealt hands, for example, should be completely random.

(Question; has anyone come up with a set a sensitive statistics for estimating
the randomness of a shuffle? )
>> 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.

And many don't. Your estimate that 3 riffles would leave 10 pairs seems reasonable
to me.

>> 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.

This is the estimate I am least sure is reasonable.However,  I would think that the
1/3 is a reasonable estimate here. In duplicate, however, whenever both
cards are in the same hand, the K will often be played before the Q, and as
the token shuffle at the end of the play will do little to disturb the order,
I still think 10% to be a reasonable guesstimate. It would be nice to have
real data. 

>> 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.

I guess its time to do the calculation properly; neither of the above is correct.

Lets assume the probability of the two cards sticking together is 0.02.
In this case, if you have the King, the opponents always have the Q.

If they do not stick, the probability of the opponents having the Queen is 50%.

Now, you are sitting at the table and you have the K and the opponents have the Q.
What is the probability the cards stuck? We apply Bayes' rule:

                                        P(stuck)P(split|stuck)
           P(stuck|split)    =   -------------------------------------------------
                                 P(stuck)P(split|stuck)+P(unstuck)P(split|unstuck)

                                        0.02*1.0
                             =   ------------------
                                 0.02*1.0+0.98*0.50

                             =    0.02/0.51

                             =    0.04  (rounding)

So the probability of the K lying over the Q is 0.04+0.5*0.96 = 0.52. So my answer 
was correct even though I kludged the math.

However, my intention in the original post was not to suggest people should always
take the finesse, but to lend support to the idea that any additional information
should be used instead of just going with the odds. 

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