Chance of flipping 50 heads over a span of 100 flips given more than 100 flips So I found that the chance of flipping 50 heads out of a string of 100 flips is 
$$0.5^{50} (1-0.5)^{50} \binom{100}{50},$$
My question is, how do the chances of having at least 1 string of 100 flips, with heads resulting 50 times, change if I am allowed to flip the coin 101 times?  In other words, I could get 50 out of 100 in flips 1-100 OR 50 out of 100 in flips 2-101 or both.  
What about if I were allowed to flip the coin 200 times, but needed to get at least one string of 100 flips resulting in 50 heads?  
My thinking is that there are 101 different 100 flip sequences in a 200 flip sequence, and each of those 101 sequences should have $$0.5^{50} (1-0.5)^{50} \binom{100}{50},$$ probability of yielding heads exactly 50 times, which would multiply the probability by 101 times, but since the 101 different 100 flip sequences are overlapping, rather than being independent of each other, does it change the odds?
 A: This answer gives a lower bound on the probability, not a complete answer.
Let $n_k$ be the number of heads in flips $k,k+1,\dots,k+99$. Note that $n_{k+1}$ is one of $n_{k}-1,n_{k},$ or $n_{k}+1$. Also, note that $n_{1}$ and $n_{101}$ are independent.
Now, if $n_1< 50$ and $n_{101}> 50$, there is an $n_k=50$, by the above. Similarly for $n_1>50$ and $n_{101}< 50$. So to not have some $n_k=50$, you'd need $n_1>50$ and $n_{101}>50$ or $n_{1}<50$ and $n_{101}<50$. Since $$P(n_1<50)=P(n_1>50)=P(n_{101}<50)=P(n_{101}>50)=\frac{1}{2}\left(1-0.5^{100}\binom{100}{50}\right)$$ this means the probability is at least:
$$\begin{align}P(n_k=50; k=1,\dots,101)&\geq 1-P(n_1<50)P(n_{101}<50)-P(n_1>50)P(n_{101}>50)\\
&=1-\frac{1}{2}\left(1-0.5^{100}\binom{100}{50}\right)^2\end{align}$$
This means the probability is at least $\frac{1}{2}$.
It's probably considerably higher.
This trick for the lower bound works only for a multiple of $100$ flips. If there are $100j$ flips, then you'd get that the probability is at least:
$$1-\frac{1}{2^{j-1}}\left(1-0.5^{100}\binom{100}{50}\right)^{j}$$
