# A red urn contains 6 red marbles and 4 blue marbles,?

A red urn contains 6 red marbles and 4 blue marbles, and a blue urn contains 5 red marbles and 5 blue marbles. A marble is selected from an urn, the color is noted, and the marble is returned to the urn from which it was drawn. The next marble is drawn from the urn whose color is the same as the marble just drawn. Thus, this is a Markov process with two states: draw from the red urn or draw from the blue urn.

1) Write a transition matrix, 2) find the stationary matrix, and 3) calculate what percentage of marbles will be selected from the red urn over the long run.

Relevance

1) The transition matrix is, in Mathematica notation,

T = {{6/10, 4/10}, {5/10, 5/10}}.

The left eigenvector v of T for eigenvalue 1,

normalized so that v₁ + v₂ = 1 (i.e., so as to be a probability vector), is

v = {5/9, 4/9}.

The vector v gives the stationary probabilities of the states:

p(draw from red urn) = v₁ = 5/9,

p(draw from blue urn) = v₂ = 4/9.

So the answer to part (3) is v₁ = 5/9.

As for part (2), the stationary matrix S is the transition matrix modified so as to have each row sum equal to the stationary probability of its corresponding state (instead of having each row sum equal to 1, as T does).

The matrix S can be calculated by left-multiplying T by the diagonal matrix D having the entries of v along the diagonal:

S = DT = {{5/9, 0}, {0, 4/9}} . {{6/10, 4/10}, {5/10, 5/10}}

= {{1/3, 2/9}, {2/9, 2/9}}

• the blue branch one step

.5

.5

two step

.6

.4

the red branch one step

.6

.4

two step

.56 (= .36 + .20)

.44 (= .24 + .20)

three step

.6 (=.216 + .144 + .120 +.120)

.4 (=.144 + .096 + .080 + .080)

there's a reluctance on the part of the numbers to stray away from a value reaaal close to "60%"