We have three drawers with coins in them, and we know which drawers have which coins:

probability/gold.coins.drawers New

New entry, edited from material collected for the archive and not yet in a released version.

Problem

We have three drawers with coins in them, and we know which drawers
have which coins:

        Drawer #1 - 2 gold coins
        Drawer #2 - 1 silver and 1 gold coin
        Drawer #3 - 2 silver coins

A drawer is chosen at random, and a coin is chosen from it at random.
The coin is gold.  The coin is returned to the drawer, the drawer is
shaken thoroughly, and another coin is drawn at random from the same
drawer.  Again the coin is gold.  The coin is returned to the drawer,
the drawer is once more shaken, and a final random draw is made from
it.  What is the probability that the last coin drawn will be gold if

        a - the gold coins are identical
        b - the gold coins are distinguishable and the first two were
            different (that is if one coin had Mickey Mouse and the
            other had Donald Duck)
        c - the gold coins are distinguishable, and the first two
            were the same coin.

-- Charles Evans <cee1@Ra.MsState.Edu>

Show solution

Label the gold coins in drawer 1 "A" and "B", the gold coin in drawer
2 as "C", the silver coin in drawer 2 [Editor: "drawer 3" in the
original] as "D", and the silver coins in drawer 3 as "E" and "F".

The following outcomes are all equally likely:

  #    First draw      Second draw
  1       A                A
  2       A                B
  3       B                A
  4       B                B
  5       C                C
  6       C                D
  7       D                C
  8       D                D
  9       E                E
 10       E                F
 11       F                E
 12       F                F

Now, for the specific cases:

A: Cases 6, 7, 8, 9, 10, 11, and 12 are eliminated, because in all of
   them at least one of the drawn coins is silver.  This leaves you with
   5 equally probable cases.  There is a 4/5 chance that the third draw
   will produce a gold coin with a probability of 1, and a 1/5 chance
   that the third draw will produce a gold coin with a probability of
   1/2.  4/5 * 1 + 1/5 * 1/2 = 9/10.  The answer is *90%*.

B: Cases 1, 4, 5, 6, 7, 8, 9, 10, 11, and 12 are eliminated, because
   all of them either produce a single silver coin or identical gold
   coins.  Therefore, the probability of drawing a gold coin next is
   *100%*.

C: Cases 2, 3, 6, 7, 8, 9, 10, 11, and 12 are eliminated for either
   producing silver coins or differing gold coins.  There is a 2/3
   chance that the third draw will certainly produce a gold coin,
   and a 1/3 chance that the third draw will produce a gold coin with
   probability 1/2.  2/3 * 1 + 1/3 * 1/2 = 5/6.  The answer is *83%*
   (more or less).

-- Jonathan S. Haas <positron@neosoft.com>

[ Editor: in reply to an earlier posting that gave a: 90%, b is
  trivially 100%, c: 2/3. ]

There are three drawers each equally probable to be opened,
for each drawer drawing a coin twice may give four equiprobable
results (AA,AB,BA,BB).  Thus we have 12 equiprobable outcomes
for the experiment.
In a) all four double draws from drawer 1 and one double draw
from drawer 2 may yield a gold coin twice, thus the chance that
we have used drawer one is 4/5 and the chance that we have used
drawer 2 is 1/5.  So the chance to draw a gold coin again from
the same drawer is 4/5 * 1 + 1/5 * 1/2 = 9/10

b) really is trivial

c) is wrong.  Here two double draws from drawer 1 and one double draw
from drawer 2 may yield identical gold coins, thus the chance that
we have used drawer one is 2/3 and the chance that we have used
drawer 2 is 1/3.  So the chance to draw a gold coin again from
the same drawer is 2/3 * 1 + 1/3 * 1/2 = 5/6

Variant:
There exists N types of distinct gold coins, each gold coin in the
experiment is independently equally probable to be either of these.
As in c) you draw twice from a random drawer and get
identical gold coins, now what is the probability from again
drawing a gold coin from this drawer.

-- Stein Kulseth <stein.kulseth@nta.no>

[ Editor: no answer to the variant was found in the collected
  postings. ]