Transition Matrix

**cnmath16** · January 12th 2009, 03:58 PM

52. Three people -- John, Joan, and Kim -- throw a ball to each other. There is a probability of 1/3 that John will throw the ball to Joan. There is a probability of 1/2 that Joan will throw the ball to Kim. There is a probability of 1/4 that Kim will throw the ball to John.

b) Assuming that the ball starts with Joan, what is the probability that she will have it back after 2 throws?
_____ ___ ___
[__0 _____ 1/3 ____ 2/3 ]2 (this matrix squared)...
[ _ 1/2 ____ 0 ______ 1/2 ]
[__ 1/4 ____ 3/4 _____ 0 ]

..which gives you
[__1/3_____ 1/6 ____ 1/6 ]
[ _ 1/8 ____ 13/24___ 1/3 ]
[__ 3/8 ____ 1/12_____13/24 ]

Therefore, there is a 13/24 probability that Joan will have the ball.

Second part of question.. Assuming that the ball starts with Kim, what is the probability that Joan will have it after 3 throws??

THANKS

**Soroban** · January 12th 2009, 07:08 PM

Hello, cnmath16!

Three people: John, Joan, and Kim, throw a ball to each other.
There is a probability of 1/3 that John will throw the ball to Joan.
There is a probability of 1/2 that Joan will throw the ball to Kim.
There is a probability of 1/4 that Kim will throw the ball to John.

a) Assuming that the ball starts with Joan,
what is the probability that she will have it back after 2 throws?

$\begin{array}{cccc} & \text{John} & \text{Joan} & \text{Kim} \\ \text{John} & 0 & \frac{1}{3} & \frac{2}{3} \\ \\[-4mm] \text{Joan} & \frac{1}{2} & 0 & \frac{1}{2} \\ \\[-4mm] \text{Kim} & \frac{1}{4} & \frac{3}{4} & 0 \end{array}\quad=\quad A$

Then: . $A^2 \;=\;\begin{pmatrix}\frac{1}{3} & \frac{1}{2} & \frac{1}{6} \\ \\[-4mm] \frac{1}{8} & {\color{blue}\frac{13}{24}} & \frac{1}{3} \\ \\[-4mm] \frac{3}{8} & \frac{1}{12} & \frac{13}{24} \end{pmatrix}$

Therefore, there is a $\tfrac{13}{24}$ probability that Joan will have the ball after 2 throws.

Correct!

b) Assuming that the ball starts with Kim,
what is the probability that Joan will have it after 3 throws?

We want $A^3.$

$A^3 \:=\:A^2\cdot A \:=\:\begin{pmatrix}\frac{1}{3} & \frac{1}{2} & \frac{1}{6} \\ \\[-4mm] \frac{1}{8} & \frac{13}{24} & \frac{1}{3} \\ \\[-4mm] \frac{3}{8} & \frac{1}{12} & \frac{13}{24}\end{pmatrix}$ $\begin{pmatrix}0 & \frac{1}{3} & \frac{2}{3} \\ \\[-4mm] \frac{1}{2} & 0 & \frac{1}{2} \\ \\[-4mm] \frac{1}{4} & \frac{3}{4} & 0 \end{pmatrix}$ . $=\:\begin{pmatrix}\frac{7}{24} & \frac{17}{72} & \frac{17}{36} \\ \\[-4mm] \frac{17}{48} & \frac{7}{24} & \frac{17}{48} \\ \\[-4mm] \frac{17}{96} & {\color{blue}\frac{17}{32}} & \frac{7}{24} \end{pmatrix}$

If the ball starts with Kim, Joan has the ball in 3 throws with probability $\frac{17}{32}$

**cnmath16** · January 13th 2009, 05:06 AM

Thanks very much for your reply..
One quick question.

For A^2, you have 1/2 in the top middle of the matrix. Is this a typo? Should the first row be 1/3 1/6 1/6... Not 1/3 1/2 1/6?

And will this change the final answer?

**HallsofIvy** · January 13th 2009, 06:00 AM

1/2 is correct. It is (0)(1/3)+ (1/3)(0)+ (2/3)(3/4)= (1/3)(3/2)= 1/2.

You can trust Soroban!

**Soroban** · January 13th 2009, 09:07 AM

Hello, cnmath16!

A product of two transition matrices is another transition matrix.

This means: every row must add up to one (1).

That's how I checked my work.
(And it took me several tries to get it right!)

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