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$A$ tosses a coin $n$ times and $B$ tosses it $n$ times. Find the probability that the number of $A$'s tosses that land heads are greater than the number of $B$'s tosses that land heads.

I have learned that if the number of $A$ and $B$ tossing heads is equal, it is the random walk model and the number can be calculated by Bernoulli equation.

But, how can you calculate the probability that $A$'s tosses that land on heads are greater than the number of $B$'s tosses that land on heads?

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    $\begingroup$ $1 = P(A > B) + P(A < B) + P(A=B)$. If the coin is fair, then by symmetry $P(A>B)=P(A<B)$. It remains to compute $P(A=B)$. $\endgroup$
    – angryavian
    Commented Jun 26, 2021 at 2:00
  • $\begingroup$ Probability of $A$ getting exactly $k$ heads $~: ~k \in \{0,1,2, \cdots, n\}$ is $P_k = \binom{n}{k}2^{-k}2^{-(n-k)} = \binom{n}{k}2^{-n}.$ Thus, chance of $A$ and $B$ both getting exactly $k$ heads is $(P_k)^2.$ Note that $\sum_{k = 0}^n \left[\binom{n}{k}^2\right] = \binom{2n}{n}.$ $\endgroup$
    – user2661923
    Commented Jun 26, 2021 at 3:00

1 Answer 1

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Let $P_k$ be the probability that $A$ gets $k$ heads in $n$ tosses. For fair coin $P_k=\binom{n}{k}\frac{1}{2^n}$. The probabilities for $B$ are the same, so the probability that both get the same number is $P=\frac{1}{4^n}\sum\limits_{k=0}^n \binom{n}{k}^2$. This gives a final result $\frac{1-P}{2}$ as the probability that $A$ gets more heads than $B$.

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  • $\begingroup$ $$P=2(2n-1)!/n!(n-1)!4^n$$ in closed form. $\endgroup$
    – Suzu Hirose
    Commented Jun 26, 2021 at 4:11

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