The Fox and the NumbatDate: 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 (exactly) 1643/2592, about 0.633873, which agrees well with your 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/ |
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