# Fibonacci die probability

I was doing an exam review and I have absolutely no idea how to do this question. There is no solution to this so I'm hoping someone here knows how to do this.

The fibonacci numbers are defined as follows: f0=0, f1=1, and fn=fn-1 + fn-2 for n>=2. Let n be a large integer. A fibonacci die is a die that has fn faces. Such a die is fair: If we roll it, each face is on top with the same probability 1/fn. There are three different types of fibonacci dice.

• D1: fn-2 of its faces show the number 1 and the other fn-1 faces show the number 4.

• D2: Each face shows the number 3

• D3: fn-2 of its faces show the number 5 and the other fn-1 faces show the number 2.

Assume we roll each of D1, D2, and D3 once, independently of each other. Let R1, R2, and R3 be the numbers on the top face of D1, D2, and D3, respectively. Determine Pr(R1 > R2) and Pr(R2 > R3) and show that Pr(R3 > R1) = ((fn-1)(fn+1)) / (fn)2

• Would you please check that you copied the subscripts correctly on the last line of your question? – bof Mar 10 '14 at 3:32
• Please watch capitalization. $n \neq N$ in your fix. – Ross Millikan Mar 10 '14 at 3:44

$Pr(R_1 > R_2)$ is just the probability that $R_1 = 4$, which is $\frac{f_{n-1}}{f_n}$. Similarly, $Pr(R_2 > R_3) = \frac{f_{n-1}}{f_n}$. For $Pr(R_3 > R_1)$ we have

$$\frac{f_{n-2}f_n + f_{n-1}f_{n-2}}{(f_n)^2} = \frac{f_{n-2}(f_n + f_{n-1})}{(f_n)^2} = \frac{f_{n-2}f_{n+1}}{(f_n)^2}.$$

Possibly a typo in your question, or I've made a small mistake.

Hint: D2 beats D1 $\frac {f_{n-2}}{f_n}$ of the time because it beats the $1$'s and loses to the $4$'s. What about D3 vs D2? There are four possibilities when you compare D1 and D3. The fraction should remind you of a Fibonacci identity. The last $f_1$ looks like it should be $f_n$

• You're right. It was a typo. – Roman Mar 10 '14 at 3:38