[net.rec.bridge] Statistics problem solved

bcn@mit-eddie.UUCP (Clifford Neuman) (08/22/83)

I was always taught that past probabilities can not be used to determine
the probability of a future occurence.  Just because a coin turned up
heads last time does not increase its probability of turning up tails
this time.  I'm not sure how relavant that was, but here are some more 
comments.

Lets try to look at this from a different point of view.  At each stage
we can visualise the possibilities once again as a set of drawers.  We start
by choosing a drawer at random from three cabinets.  The cabinet may contain
two gold, two silver, or gold and silver pieces.  We have also chosen a
drawer within this cabinet.  If the drawer contains a gold piece we know
that the cabinet does not contain two silver pieces.  At this stage we
have two possible cabinets.  One that contains a gold piece and a silver piece,
and another that contains two gold pieces.  One of the drawers containing
a gold piece is open so the remaining drawers contain a gold piece, a gold 
piece and a silver piece.  Thus the probability for finding a gold piece
when we open another drawer is 2/3.  But we have another constraint.  Since 
we are trying to determine the probability that the OTHER DRAWER OF THE SAME
CABINET contains a gold piece we have to change or model so that the drawer
that we opened on the selected cabinet is open in BOTH POSSIBLE CABINETS.  
This leaves us with two possible drawers, one of which contains a gold coin
and the other which contains a silver coin, thus a 50% probabilty of finding
another gold coin. 

I haven't quite convinced myself, but this seems correct.

	~ Cliff
	  decvax!genrad!mit-eddie!bcn
	  BCN@MIT-MC.arpa