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This question already has an answer here:

Below is a problem that I did. I have reason to believe that the correct answer is $\frac{1}{2}$. Please check my work and find any and all errors.
Thanks,
Bob

Problem:
Suppose person 1 flips a fair coin $5$ times and that person 2 flips a fair cont $4$ times. What is the probability that person 1 got more heads than person 2.
Answer:
Let $p_1(x)$ be the probability that person 1 got $x$ heads. Let $p_2(x)$ be the probability that person 2 got $x$ heads. Let $p$ be the probability we seek. \begin{align*} P_1(0) &= \frac{1}{32} \\ P_1(1) &= \frac{5}{32} \\ P_1(2) &= \frac{5}{16} \\ P_1(3) &= \frac{5}{16} \\ P_1(4) &= \frac{5}{32} \\ P_1(5) &= \frac{1}{32} \\ % P_2(0) &= \frac{1}{16} \\ P_2(1) &= \frac{1}{4} \\ P_2(2) &= \frac{3}{8} \\ P_2(3) &= \frac{1}{4} \\ P_2(4) &= \frac{1}{16} \\ \end{align*} \begin{align*} p &= p_1(1) \left( p_2(0) \right) + p_1(2)\left( p_2(0) + p_2(1) \right) + \\ & p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + \\ & p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ % p_1(1) \left( p_2(0) \right) &= \frac{5}{32} \left( \frac{1}{16} \right) = \frac{5}{512} \\ p &= \frac{5}{512} + p_1(2)\left( p_2(0) + p_2(1) \right) + \\ &p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + \\ &p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p_1(2)\left( p_2(0) + p_2(1) \right) &= \frac{5}{32} \left( \frac{1}{16} + \frac{1}{4} \right) \\ p_1(2)\left( p_2(0) + p_2(1) \right) &= \frac{5}{32} \left( \frac{5}{16} \right) = \frac{25}{512} \\ p &= \frac{5}{512} + \frac{25}{512} + p_1(3) \\ & \left( p_2(0) + p_2(1) + p_2(2) \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{30}{512} + \\ & p_1(3) \left( \frac{1}{16} + \frac{1}{4} + \frac{3}{8} \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{30}{512} + p_1(3) \left( \frac{11}{16} \right) + \frac{5}{32} \left( \frac{1}{16} + \frac{1}{4} + p_2(2) + p_2(3) \right) + p_1(5) \\ p &= \frac{30}{512} + \frac{5}{16} \left( \frac{11}{16} \right) + \frac{5}{32} \left( \frac{1}{16} + \frac{1}{4} + \frac{3}{8} + \frac{1}{4} \right) + \frac{1}{32} \\ \end{align*} \begin{align*} p &= \frac{30}{512} + \frac{55}{256} + \frac{5}{32} \left( \frac{15}{16} \right) + \frac{1}{32} \\ p &= \frac{140}{512} + \frac{5}{32} \left( \frac{15}{16} \right) + \frac{1}{32} \\ p &= \frac{140}{512} + \frac{75}{512} + \frac{16}{512} \\ p &= \frac{231}{512} \\ \end{align*}

I am hoping that somebody can tell me where I went wrong above.

Based upon the comments I got from the group, here is the correct solution:

Answer: \newline Let $p_1(x)$ be the probability that person 1 got $x$ heads. Let $p_2(x)$ be the probability that person 2 got $x$ heads. Let $p$ be the probability we seek. \begin{align*} P_1(0) &= \frac{1}{32} \\ P_1(1) &= \frac{5}{32} \\ P_1(2) &= \frac{5}{16} \\ P_1(3) &= \frac{5}{16} \\ P_1(4) &= \frac{5}{32} \\ P_1(5) &= \frac{1}{32} \\ % P_2(0) &= \frac{1}{16} \\ P_2(1) &= \frac{1}{4} \\ P_2(2) &= \frac{3}{8} \\ P_2(3) &= \frac{1}{4} \\ P_2(4) &= \frac{1}{16} \\ \end{align*} \begin{align*} p &= p_1(1) \left( p_2(0) \right) + p_1(2)\left( p_2(0) + p_2(1) \right) + \\ & p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ % p_1(1) \left( p_2(0) \right) &= \frac{5}{32} \left( \frac{1}{16} \right) = \frac{5}{512} \\ p &= \frac{5}{512} + p_1(2)\left( p_2(0) + p_2(1) \right) + \\ &p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p_1(2)\left( p_2(0) + p_2(1) \right) &= \frac{5}{16} \left( \frac{1}{16} + \frac{1}{4} \right) \\ p_1(2)\left( p_2(0) + p_2(1) \right) &= \frac{5}{16} \left( \frac{5}{16} \right) = \frac{25}{256} \\ \end{align*} \begin{align*} p &= \frac{5}{512} + \frac{25}{256} + p_1(3) \left( p_2(0) + p_2(1) + p_2(2) \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{55}{512} + \frac{5}{16} \left( \frac{1}{16} + \frac{1}{4} + \frac{3}{8} \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{55}{512} + \frac{5}{16} \left( \frac{11}{16} \right) + p_1(4) ( p_2(0) + p_2(1) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{55}{512} + \frac{55}{256} + \frac{5}{32} \left( \frac{1}{16} + \frac{1}{4} \right) + p_2(2) + p_2(3) ) + p_1(5) \\ p &= \frac{165}{512} + \frac{5}{32} \left( \frac{5}{16} + p_2(2) + p_2(3) \right) + p_1(5) \\ p &= \frac{165}{512} + \frac{5}{32} \left( \frac{5}{16} + \frac{3}{8} + \frac{1}{4} \right) + \frac{1}{32} \\ p &= \frac{165}{512} + \frac{5}{32} \left( \frac{11}{16} + \frac{1}{4} \right) + \frac{1}{32} \\ p &= \frac{165}{512} + \frac{75}{512} + \frac{1}{32} = \frac{256}{512} \\ p &= \frac{1}{2} \\ \end{align*}

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marked as duplicate by lulu, Paul Frost, the_candyman, Lord_Farin, kjetil b halvorsen Dec 16 '18 at 22:01

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  • 3
    $\begingroup$ This is true more generally. See this question $\endgroup$ – lulu Dec 16 '18 at 13:07
  • $\begingroup$ @lulu I was thinking that the general answer was $\frac{1}{2}$. I am now hoping somebody will tell me where I went wrong above. I am thinking it is an error in arithmetic. $\endgroup$ – Bob Dec 16 '18 at 13:20
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    $\begingroup$ Well, that certainly seems likely...it looks like a fairly error prone computation. I don't see anything conceptually wrong with it. $\endgroup$ – lulu Dec 16 '18 at 13:24
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    $\begingroup$ I just did the same computation on a spreadsheet and got $.5$ so there's just an arithmetic slip up somewhere. $\endgroup$ – lulu Dec 16 '18 at 13:29
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    $\begingroup$ You have written $$P_1(2)(P_2(0)+P_2(1))=\frac{5}{\color{red}{32}}\left(\frac{1}{16}+\frac{1}{4}\right)$$ fixing the $32$ to $16$ yields correct computation. $\endgroup$ – Galc127 Dec 16 '18 at 15:18

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