# Adding even positioned terms in infinite series

A die is thrown until a $6$ appears. If the casting stops on an odd number of times Mike wins. Otherwise, Tom wins. What is the probability that Mike wins? What is the probability that Tom wins? Also show that this game favors mike for all $p$.

Attempt: $$\Pr(\text{Mike wins}) = \sum_{k=1}^\infty \left(\frac 5 6\right)^{k-1} \frac 5 6\text{ for }k = 1,3,5,\ldots,\infty$$

$$\Pr(\text{Tom wins}) = \sum_{k=1}^\infty \left(\frac 5 6\right)^{k-1}\frac 5 6\text{ for }k = 2,4,6,\ldots, \infty$$

I am stuck here.

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It's a geometric series with common ratio $(5/6)^2$ – Gerry Myerson Jan 29 '12 at 1:41
You don't need any infinite series to solve this problem. See my answer below. – Michael Hardy Jan 29 '12 at 2:36

With a fair six-sided die:

For $k\ge0$, the probability that the first 6 occurs on the $(2k+1)$'st toss is $({5\over6})^{2k}\cdot{1\over6}$. So the probability that Mike wins is $$\sum_{k=0}^\infty ({\textstyle{5\over6}})^{2k}\cdot{\textstyle{1\over6}}= \sum_{k=0}^\infty ({\textstyle{25\over36}})^{ k}\cdot{\textstyle{1\over6}}= {1/6\over1-{25\over36}}={6\over11}.$$

The probability that Tom wins is $1-{6\over 11}={5\over11}$.

You can generalize this if the die has $p$ probability of casting 6. Just replace $5/6$ with $1-p$ and $1/6$ by $p$ in the above. The probability that Mike wins is: $$\sum_{k=0}^\infty ({\textstyle{(1-p)}})^{2k}\cdot{\textstyle{p}}= \sum_{k=0}^\infty ({\textstyle{(1-p)^2}})^{ k}\cdot{\textstyle{p}}= {p\over1-{(1-p)^2}}={p\over 2p-p^2}={1\over 2-p}.$$

Note ${1\over 2-p}>{1\over 2}$.

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You can solve the problem without infinite series, as I explained in my answer. – Michael Hardy Jan 29 '12 at 2:38

Here's a simple way to do it that involves no infinite series:

\begin{align} & \Pr(\text{Mike wins}) = \Pr(\text{6 on 1st trial}) + \Pr(\text{first 6 on a later odd-numbered trial}) \\ \\ & = \frac 16 + \Pr(\text{non-6 on 1st trial})\cdot\Pr(\text{6 on a later odd-numbered trial}\mid\text{non-6 on 1st trial}) \\ \\ & = \frac 1 6 + \frac 5 6 \cdot \Pr(\text{6 on an even-numbered trial})\tag{(explanation below)} \\ \\ & = \frac 1 6 + \frac 5 6 \cdot \Pr(\text{Tom wins}) \\ \\ & = \frac 1 6 + \frac 5 6 \cdot (1 - \Pr(\text{Mike wins})). \end{align} The reason this works is that onece you've got a non-6 on the first trial, you're starting over again, so getting a 6 on an odd-numbered trial after the first trial is the same as getting a 6 on a trial that bears an even number if you start counting after that initial non-6.

So $$p= \frac 1 6 + \frac 5 6 (1-p)$$

Therefore $p= \dfrac{6}{11}$.

(This is the probability that Mike wins, so the probability that Tom wins is $\dfrac{5}{11}$.)

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