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Suppose that $A$ tosses a coin which lands heads with probability $p_A$ and $B$ tosses a coin which lands heads with probability $p_B$. They toss their coins simultaneously over and over again, in a competition to see who gets the first head. The one to get the first head is the winner, except that a draw results if they get their first heads together.

Calculate P(A wins).

Let $X_A$ be the # of tosses that it takes $A$ to get the first head, $X_A \ge 1$.

Let $X_B$ be the # of tosses that it takes $B$ to get the first head, $X_B \ge 1$.

$P(A $ wins$)$

$= P(X_A \lt X_B)$

$=\sum_{i=1}^{\infty}P(X_A<X_B|X_A=i)P(X_A=i)$

$=\sum_{i=1}^{\infty}(1-p_B)^i(1-p_A)^{i-1}p_A$

$=\frac{p_A}{1-p_A}\sum_{i=1}^{\infty}(1-p_B)^i(1-p_A)^i$

$=\frac{p_A}{1-p_A}\sum_{i=1}^{\infty}((1-p_B)(1-p_A))^i$

$=\frac{p_A}{1-p_A}\left(\sum_{i=0}^{\infty}((1-p_B)(1-p_A))^i - ((1-p_B)(1-p_A))^0\right)$

$=\frac{p_A}{1-p_A}\left(\frac{1}{1-(1-p_A)(1-p_B)}-1\right)$

Textbook Answer:

$\frac{(1-p_A)p_B}{1−(1−p_A )(1−p_B )}$

They're not the same, I tested with $p_A = 0.2$, $p_B=0.5$

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A_for_ Abacus
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    The book's answer does not pass a basic sanity check. For $p_A = 0.1$ and $p_B = 0.8$, we expect $B$ to win the majority of the time. But using the book's answer, we have $$P(A\text{ wins}) = \frac{0.9\cdot 0.8}{1-0.9\cdot 0.2} \approx 0.878...$$ which is bizarre. Your answer can be rewritten $$P(A\text{ wins}) = \frac{(1-p_B)p_A}{1-(1-p_A)(1-p_B)}$$ which gives a more reasonable answer of $0.02439...$ – bames Mar 01 '18 at 09:49

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Literally had to search every corner of the internet for this solution. I'll just leave it here.

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A_for_ Abacus
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  • And this is the answer that you got originally! :) – bames Mar 01 '18 at 10:06
  • Huh? My answer doesn't simplify into the textbook answer though. – A_for_ Abacus Mar 01 '18 at 10:08
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    HOLY CRAP, I WROTE THE TEXTBOOK ANSWER TO PART 2 IN THE QUESTION! AGEAGWAHAWH – A_for_ Abacus Mar 01 '18 at 10:10
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    In case you didn't see it, your answer simplifies to the answer above like this: \begin{align} \frac{p_A}{q_A}\left(\frac{1}{1-q_Aq_B} - 1\right) &= \frac{p_A}{q_A}\left(\frac{1 - (1-q_Aq_B)}{1-q_Aq_B}\right)\ &= \frac{p_A}{q_A}\left(\frac{q_Aq_B}{1-q_Aq_B}\right) \ &= \frac{p_Aq_B}{1-q_Aq_B}\end{align} – bames Mar 01 '18 at 10:14