CSvax:Pucc-H:Pucc-I:ags@pur-ee.UUCP (09/01/83)
At the risk of beating a dead horse, I would like to offer what I hope is
a different view of the coins and cabinets problem. Consider the following
variations on the theme:
(1) The first version is the usual one. You opened a drawer at random and
found a gold coin. What is the probability that the other drawer in
the same cabinet also contains a gold coin? Call this unknown probability
P1.
(2) The second version has you drawing a silver coin first. What is the
probability that the other drawer in the same cabinet also contains a
silver coin? Let this unknown probability be P2.
(3) This time you select one of the six drawers at random and withdraw a coin
without looking at it. What is the probability that the other drawer in
the same cabinet contains a coin of the same kind you now hold? Call this
probability P3.
I hope everyone agrees at least that P1 = P2.
P3 can be expressed in terms of P1 and P2 as follows: The coin in your hand
is equally likely to be gold or silver. If it's gold, the probability of a
match is P1; otherwise it is P2. Therefore P3 = 0.5 * P1 + 0.5 * P2 = P1 = P2
(i.e. all three problems are equivalent).
But what is P3? (See if this sounds familiar)
There are three cabinets, each equally likely to be chosen. Two of them contain
like coins and one contains unlike coins. Therefore P3 = 2/3.
Therefore P1 = 2/3 also.
Dave Seaman
ags@PUCC