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### The Fox and the Numbat

```
Date: 07/19/99 at 23:16:45
From: Geoff Kissane
Subject: Probability / Game Theory (?)

The Fox and the Numbat

This is a mathematical activity that is a (mini) board game. The board
is effectively as shown below:

Fox     1    2    3    4    5    6
Home   12   11   10    9    8    7
Numbat

The fox starts 7 squares behind the numbat (as shown). A single die is
rolled (repeatedly). If the die shows a 1, 2, 3 or 4, then the numbat
moves the corresponding number of spaces towards its home. If,
however, the die shows a 5 or 6, then the fox moves that many spaces.

The numbat wins by reaching its home.
The fox wins by reaching (or passing) the numbat.

What is each animal's probability of winning?

N.B. for the uninitiated: a numbat is a small marsupial, and is the
animal emblem of Western Australia. The numbat is sometimes called a
banded anteater (due to the bands of colour on its body and its
penchant for eating ants.)

This activity appears in a mathematics book and was used by a staff
member as a probability activity. The class was asked to play a number
of games, pool their results and then estimate the probability of each
animal winning. The result was (out of 100 trials):

Numbat wins   68 times
Fox wins      32 times

This seemed to imply that the correct probabilities may have been 2/3
and 1/3. The text, however, indicated that the two animals had
approximately equal chances of winning.

The mathematics involved seemed a little daunting to us and so we
appeal to you for help.

What is the probability of each animal winning (assuming a fair die,
etc.)? Is there a relatively easy proof of this result? Your
assistance will be greatly appreciated.

Thanking you in anticipation,
Geoffrey Kissane
(Head of Mathematics, Mandurah Catholic College, Western Australia)
```

```
Date: 07/20/99 at 10:51:51
From: Doctor Anthony
Subject: Re: Probability / Game Theory (?)

Does the Numbat have to get the exact correct number (12) to end up at
home? If so this means it could hang around a long time before being
safe. This would greatly increase the chances for the fox.

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 07/20/99 at 20:06:01
From: Geoff Kissane
Subject: RE: Probability / Game Theory (?)

No. As I understand it - and certainly as it was played - the numbat
is allowed to get "too many" and it is still safe.
```

```
Date: 07/21/99 at 11:42:45
From: Doctor Nick
Subject: Re: Probability / Game Theory (?)

Hi Geoffrey -

This is a neat problem, but not an easy one to solve. The way I solved
it was by treating the game as a "Markov chain." Basically, we can
consider the game to be in one of a finite number of states
(determined by the positions of the animals), and on each turn, there
is a certain probability of moving from each state to one of the other
states. This allows a representation of the game as a matrix of
probabilities, and a computation with the matrix yields the result.
The hard part is the generation of the matrix, which for this game is
a 23 by 23 square matrix. Once the matrix is made, the result can be
found by raising the matrix to the 9th power (9 from the fact that the
game can last at most 9 moves). Fortunately, there is some nice
software available to do this (Mathematica).

The result I got was that the numbat's probability of winning is
experimental data. This is assuming that the numbat doesn't have to
land exactly on 12 (so, e.g., if the numbat's at 10 and gets a 4, the
numbat wins). If the numbat has to land exactly on its home, its
probability of winning decreases quite a bit, I think. I'll work
_that_ probability out and get back to you.

For information on Markov chains, take a look at just about any
college-level probability text. I can give specific recommendations if
you'd like.

Feel free to write back if you'd like more details of the method I
used, or any other questions.

Have fun,
- Doctor Nick, The Math Forum
http://mathforum.org/dr.math/
```

```
Date: 07/23/99 at 07:27:45
From: Doctor Anthony
Subject: Re: Probability / Game Theory (?)

I thought I would complete the calculation of the probabilities of
this game. Below are the details of what I found.

There are various states of the game, and you can use probability
matrices to show the probabilities as the game proceeds. The notation
(x,y) represents the state when the fox is at position x and the
numbat at position y. The fox wins if x >= y and the numbat wins when
y >= 12 before x >= y.

The possible 'states' are 'a' to 'p' as listed below:

a    b    c     d     e    f    g    h     i     j    k    l     m
(1,7)(1,8)(1,9)(1,10)(1,11)(6,7)(6,8)(6,9)(6,10)(6,11)(7,8)(7,9)(7,10)

n      o       p
(7,11)(x wins)(y wins)

You make up the matrix with these 'states' along the top and down the
lefthand side. Label the top line 'FROM' and the lefthand column 'TO'
and fill in the probabilities of going from say (1,7) to (6,7). Note
that only one of x or y can change in one throw of the die. Most
entries in the probability matrix will be zeros and each column must
sum to 1. Multiply the matrix by itself a number of times to see the
probabilities after 1, 2, 3, ... throws of the die. The starting
vector is a 1 at the (1,7) (state 'a') point and with zeroes
everywhere else in the starting vector.

FROM

a   b   c   d   e   f   g   h   i   j   k   l   m   n   o   p
----------------------------------------------------------------
a| 0   0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
b|1/6  0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
c|1/6 1/6  0   0   0   0   0   0   0   0   0   0   0   0   0   0
d|1/6 1/6 1/6  0   0   0   0   0   0   0   0   0   0   0   0   0
e|1/6 1/6 1/6 1/6  0   0   0   0   0   0   0   0   0   0   0   0
f|1/6  0   0   0   0   0   0   0   0   0   0   0   0   0   0   0
g| 0  1/6  0   0   0  1/6  0   0   0   0   0   0   0   0   0   0
h| 0   0  1/6  0   0  1/6 1/6  0   0   0   0   0   0   0   0   0
i| 0   0   0  1/6  0  1/6 1/6 1/6  0   0   0   0   0   0   0   0
j| 0   0   0   0  1/6 1/6 1/6 1/6 1/6  0   0   0   0   0   0   0
k| 0  1/6  0   0   0   0   0   0   0   0   0   0   0   0   0   0
l| 0   0  1/6  0   0   0   0   0   0   0  1/6  0   0   0   0   0
m| 0   0   0  1/6  0   0   0   0   0   0  1/6 1/6  0   0   0   0
n| 0   0   0   0  1/6  0   0   0   0   0  1/6 1/6 1/6  0   0   0
o|1/6  0   0   0   0  1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3  1   0
p| 0  1/6 1/3 1/2 2/3  0  1/6 1/3 1/2 2/3 1/6 1/3 1/2 2/3  0   1
----------------------------------------------------------------
Sum 1  1   1   1   1   1   1   1   1   1   1   1   1   1   1   1

If you have the energy you multiply this matrix by itself a few times
to see what is happening. The long-term probabilities will be given by
the entries in the o (fox wins) and p (numbat wins) rows.

I used a computer program to find the 12th power of the matrix. That
will definitely give the final probabilities, since the game must
finish within 8 or 9 throws. Having determined the final form of the
matrix I multiplied by the starting column vector, which has a 1 in
the top position and 0 everywhere else. This product gives a column
vector with 0 everywhere except in the last two positions, the last
but one representing the probability that the fox will win and the
bottom position the probability that the numbat will win.

I get   Prob(Fox wins) = 3011/7776 = 0.387217

Prob(Numbat wins) = 4765/7776 = 0.612783

- Doctor Anthony, The Math Forum
http://mathforum.org/dr.math/
```
Associated Topics:
College Probability

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