This is my first question here, please bear with me.
I want to figure out when the probability of observing at least $X$ heads is equal (or closest to equal) in the two events described below.
Event 1: Assume you have $N$ coins to flip. They end up head with probability $p$. After flipping all coins, you take the heads aside and flip the tails once again. What is the probability of observing at least X<N heads? This should just be a binomial distribution $E_1\sim B(N,2p-p^2)$, and $\mathbb{P}_1(x\geq X)=1-\sum_{k=0}^{X-1} \binom{N}{k}(2p-p^2)^k(1-2p+p^2)^{N-k}$
Event 2: Assume you have flipped $N$ coins and $n<X$ coins ended up head. Now you flip remaining $N-n$ coins again, what is the probability of observing at least $X$ heads (including the n heads from the previous flip). This is another binomial distribution, $E_2\sim B(N-n, p)$. Now I am interested in $\mathbb{P}_2(x\geq X-n)$ and how it depends on n.
(The probability of observing at least $X$ heads in Event 2 depends on $n$. If $n=0$, then the probability of observing at least $X$ heads is smaller in Event 2 than in Event 1. Likewise, if $n=X$, then the probability of observing at least $X$ heads is greater in Event 2 than in Event 1. )
My problem is: for which value of n are the probability of observing X heads equal (or closest to equal) in the two events? That is, when is $\mathbb{P}_1(x\geq X) \approx \mathbb{P}_2(x\geq X-n)$?