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Person A and Person B each independently toss the same unbalanced coin and count the number of tosses it takes each of them to obtain the first head. Assume that the probability of obtaining a head with this coin is $\pi, 0 < \pi < 1, \pi \ne 1/2$. What is the probability that Person A will require more tosses of this coin than Person B to obtain the first head.

My thoughts: There are two ways to approach this question.

Way 1: Find the probability that the two people will require the same number of tosses of this unbalanced coin to obtain the first head. Then, take the complement of this space and divide by 2, since it is equally likely that # of tosses for A > # tosses for B as it is vice versa. This comes from the fact that they have the same marginal probabilities:

Let $X$ and $Y$ denote the number of throws for Person A and B, respectively. $p(x)=\pi(1-\pi)^{x-1}$ and $p(y)=\pi(1-\pi)^{y-1}$. Then, $P(X=Y)=\pi^2 + (\pi(1-\pi))^2 + \cdots$. After pulling out the $\pi$ and identifying this as a geometric series, I obtained $P(X=Y)=\frac{\pi}{2-\pi}$. Is this correct? If so, then we can just say $P(X>Y)=\frac{1}{2}(1-\frac{\pi}{2-\pi})$, right?

Way 2: This will be more rigorous, and I would like to truly learn this way in case the marginal probabilities for person A and person B were not identical. We can do a double summation. I have written down the following: $\Sigma^{\infty}_{j=1, j < i}\Sigma^{\infty}_{i=2}[(\pi(1-\pi)^{i-1}\cdot(\pi(1-\pi)^{j-1}]$, where $j$ indexes the # of tosses for person B and $i$ indexes the # of tosses for person A. Is this correct? If so, how can I derive a cleaner answer from this?

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    $\begingroup$ Note that there is a relatively easy way to solve this problem: write $P$ for the probability of person B "winning". Then $P$ is equal to the probability of person B winning on the first throw, plus the probability of them winning on some later throw. Winning first throw: $\pi(1-\pi)$ (since B needs a heads, and A needs a tails). Winning on later throws: $(1-\pi)^2P$ (since both need to throw tails, and then B needs to win afterwards, which is the same game). Thus $P = \pi(1-\pi) + (1-\pi)^2P$, and some algebra gives you the same answer. $\endgroup$ – Mees de Vries Sep 4 at 11:31
  • $\begingroup$ Thank you for your contribution. I appreciate the different perspective! $\endgroup$ – Edison Sep 4 at 20:34
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Your first way and result are correct. For the second way, we can exchange the summation order, and $i$ should start from $2$: $$\begin{align}p&=\pi^2\sum_{i=2}^\infty (1-\pi)^{i-1}\sum_{j=1}^{i-1}(1-\pi)^{j-1}= \pi^2\sum_{i=2}^\infty (1-\pi)^{i-1} \left(\frac{1-(1-\pi)^{i-1}}{\pi}\right)\\&=\pi\sum_{i=2}^\infty(1-\pi)^{i-1}-\pi \sum_{i=2}^\infty (1-\pi)^{2(i-1)}\\&=(1-\pi)-\frac{(1-\pi)^2}{2-\pi}=\frac{1-\pi}{2-\pi}\end{align}$$ which is the same with your first result.

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  • $\begingroup$ This is perfect. Thank you so much! $\endgroup$ – Edison Sep 4 at 5:18
  • $\begingroup$ Upon reviewing your answer, I have several follow-up questions. Why did you exchange the summation order and can you please explain the second equality? $\endgroup$ – Edison Sep 7 at 20:36
  • $\begingroup$ Because $\Sigma^{\infty}_{j=1, j < i}\Sigma^{\infty}_{i=2}[(\pi(1-\pi)^{i-1}\cdot(\pi(1-\pi)^{j-1}]$ is not correct. The inside summation index is $i$, and you try to use it in the outside summation. For the other question, are you asking about $$\sum_{j=1}^{i-1}(1-\pi)^{j-1}=\left(\frac{1-(1-\pi)^{i-1}}{\pi}\right)$$? $\endgroup$ – gunes Sep 7 at 20:39
  • $\begingroup$ I see- that makes sense. Thank you! $\endgroup$ – Edison Sep 7 at 20:41
  • $\begingroup$ Yes, but I found the series property to explain that equality. Thank you, though. $\endgroup$ – Edison Sep 7 at 20:42

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