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Textbook problem

In one of my homework problems, I'm struggling with the phrase: "with probabilities inversely proportional to the distances to the corner from the current position".

The three distances are 3 cm, 4 cm, and 5 cm, and I have to assign probabilities to these three events totaling 1 in accordance with this inverse proportionality rule.

I constructed the following equation:

(1/3)x+(1/4)x+(1/5)x=1

Then, I solved for x and got x=(60/47), and multiplied this with the (1/3), (1/4) and (1/5). My rationale was to find a constant factor so that I could assign probabilities to these three events totaling 1, but I do not think this is correct.

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    You need more than an equation. You need a matrix, where the $i,j$ entry is the probability of moving from vertex $i$ to vertex $j$. If, for example, $i=1$ then you need to find the probabilities of moving to $2$, $3$ and $4$. Those won't be equal. – 2017-01-22
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    It would be advisable to put the equation itself here. Google book links are notoriously fickle. – 2017-01-22
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    @TheCount Thanks for the suggestion. I just screenshot it and added it in. – 2017-01-22
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    @EthanBolker, thanks for your response. Although, that's what I'm having trouble with. I understand the need to construct a transition matrix; it's just determining a way to get the actual probabilities that's bothering me. – 2017-01-22
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    your reasoning is correct – 2017-01-22

1 Answers 1

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You are almost done.

The first row of the transition matrix should tell you the probabilities for going from corner $1$ to $2$, $3$ and $4$. Those must add to $1$ and should be in proportion $1/4:1/5:1/3$, the reciprocals of the distances. Since $$ \frac{1}{4} + \frac{1}{5} + \frac{1}{3} = \frac{47}{60} $$ (your arithmetic is correct) you have to scale the fractions by $60/47$. Then they will sum to $1$ and so represent probabilities. So the first row of your transition matrix will be $$ 0, \frac{15}{47} , \frac{12}{47} , \frac{20}{47} $$

(I wish the author of the problem had labeled the vertices $A,B,C,D$ rather than $1,2,3,4$.)

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    That makes perfect sense. I suppose what discouraged me was how ugly the calculations turned out to be following the nice 3-4-5 triangle result. I appreciate your help immensely! – 2017-01-22