[net.math] fallacy in 75% argument

rainbow@ihuxe.UUCP (08/24/83)

It never ceases to amaze me the thinking processes people go through to
arrive at the wrong answers. A response to the second coin problem stated
the following:
                  case 1)  top=g bot=g
                           top=g bot=s
                           top=s bot=s
  
                  case 2)  top=g bot=g
                           top=s bot=g
                           top=s bot=s
                                          
    "If a gold coin is found in the bottom drawer in case 1, a gold coin
     will be in the top drawer 100% of the time. If a gold coin is found
     in the bottom drawer in case 2, a gold coin will be in the top
     drawer 50% of the time. Since both cases are equally likely:
                       
                          .50(1)+.50(.5)=.75 "
  
*****************************  NOT TRUE  *******************************
  
Since case 2 has more gold coins in the bottom drawers, this case occurs
more often. Exactly twice as often. Hence:
  
                          .33(1)+.67(.5)=2/3 !!
  
Next time dont let anyone talk you out of the correct answer which is and
always will be 2/3.