If I toss $n$ coins and you toss $n + m$, what is the probability you'll have more heads?

Question

My intuition is that it's the probability that you get at least 1 head in your extra $m$ tosses. Basically on the first $n$ tosses we have equal chance of winning so let's not even count those and only bet on the $m$ extra ones you have.

I haven't been able to formalize this. My closest attempt is that this probability is

$$P(M + N_2 > N_1) = P(M >N_2-N_1)=E[I(M>N_2-N1)]$$

where $M$ is the random variable for the number of heads in your extra coins, $N_2$ is for the heads in your first $n$ tosses and $N_1$ for mine. I don't know what to do with that indicator variable but I believe the solution lies that way.

Answer 1

Let $M_2 \sim Bin(n+m, p)$

Let $N_1 \sim Bin(n,p)$

\begin{align}P(M_2 > N_1)&=\sum_{i=0}^nP(M_2>N_1|N_1=i)P(N_1=i)\\ &= \sum_{i=0}^n\sum_{j=i+1}^{m+n}\binom{m+n}{j}p^j(1-p)^{m+n-j}\binom{n}{i}p^i(1-p)^{n-i}\\ &=\sum_{i=0}^n\sum_{j=i+1}^{m+n}\binom{m+n}{j}\binom{n}{i}p^{i+j}(1-p)^{m+2n-i-j}\end{align}

Answer 2

Let event $A$ be the number heads in $n$ flips and let event $B$ be the number of heads in $n+m$ flips. Then we wish to calculate $\Pr(B>A)$. So we condition on the value of $A$ to get $$ \begin{align*} \Pr(B>A)&=\sum_{i=0}^n \Pr(A=i)\cdot Pr(B>i\;|\;A=i)\\ &=\sum_{i=0}^n \sum_{j=i+1}^{n+m} \Pr(A=i)\cdot \Pr(B=j\;|\;A=i)\\ &=\sum_{i=0}^n \sum_{j=i+1}^{n+m} \Pr(A=i)\cdot \frac{\Pr(B=j\;\cap\;A=i)}{\Pr(A=i)}\\ &=\sum_{i=0}^n \sum_{j=i+1}^{n+m} \Pr(A=i)\cdot\Pr(B=j). \end{align*} $$