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Original Question:

Paul, Dave and Sarah are rolling a fair six sided die. Paul will go first, always followed by Dave, who is always followed by Sarah, who is always followed by Paul, and so on... What is the probability that Sarah will be the first one to roll a six?

Here's my work, I found the following formula in s, the number of sequences Sarah did not roll a six, terms

(5/6 * 5/6 * 5/6)^s * (5/6 * 5/6 *1/6)

(5^3/6^3)^s * (5^2/6^3)

(5^3s/6^3s) * (5^2/6^3)

$\dfrac{5^{(3s+2)}}{6^{(3s+3)}}$ = $\dfrac{5^{(3s+2)}}{6^{(3)(s+1)}}$

$\dfrac{5^{(3s+2)}}{216^{(s+1)}}$

And then we would use the summation formula for $\displaystyle \sum_{s=0}^{\infty} \frac{5^{3s+2}}{216^{s+1}}$.

Would this be an acceptable answer or do I need to find an exact value

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  • $\begingroup$ try to find the probability of P and D not rolling a 6. Then use a simple geometric distribution $\endgroup$ – David L Feb 21 '14 at 2:12
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It is the relative probability that you will have to calculate.

Answer

Probability that Sarah will be the first one to roll the winning number = $$\frac{(5/6)*(5/6)*(1/6)}{(1/6)+(5/6)*(1/6)+((5/6)*(5/6)*(1/6)}$$ $$=\frac{25}{91}$$

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Your summation is correct. You can evaluate it by writing is as $$ \frac{25}{216} \sum_{s=0}^{\infty} \frac{5^{3s}}{216^s} = \frac{25}{216} \sum_{s=0}^{\infty} \left(\frac{125}{216}\right)^s = \left(\frac{25}{216}\right) \frac{1}{1-\frac{125}{216}}=\frac{25}{91}. $$

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$\displaystyle \sum_{s=0}^{\infty} \frac{5^{3s+2}}{216^{s+1}}$ can be simplified into $\displaystyle \sum_{s=0}^{\infty} (\frac{125}{216})^{s} \frac{25}{216}$, which can be futher simplified too.

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