Biased coin flip with sampled observation Let's say a coin flip has probability $H$ to come out as heads, and $1-H$ to come out as tails. Let's say we flip many times but for each flip we randomly decide if we will look at the result, by a probability of $L$, or $1-B$ ("B" means blind, i.e. not look).
Question: what's the probability of "flipping the coin $X$ times and not observing any heads"?
The following is my half-way solution and I cannot simplify it into a manually solvable expression. Will be great if someone can either correct me or point me in the right direction.

My half-way solution:

*

*Given $X$ flips, the probability of none of them actually being heads is $$P_0 = (1-H)^X$$

*The probability of only one of them actually being heads AND was NOT observed is $$P_1 = (1-H)^{X-1} * (H*B) * X$$ The "* X" is because of combination (instead of permutation).

*The probability of only two of them actually being heads AND was NOT observed is $$P_2 = (1-H)^{X-2} * (H*B)^2 * \frac{X*(X-1)}{2}$$

*...

The total probability would be $$P = P_0 + P_1 + P_2 + ... + P_X$$ But then I cannot simplify this series.
 A: Usually when there's a complicated sum of or'ed events, it's better to think about it the other way around as a product of and'ed events (especially when they're independent, like in this case).
At the $k$th flip, you do not observe a head with probability $B + (1-B)(1-H)$.
\begin{align*} 
\mathbb P(\text{not observing heads}) 
&= \mathbb P\Big(\bigcap_{k=1}^X \text{not observing heads at the $k$th flip}\Big)\\[4pt]
&= \prod_{k=1}^X \big(B + (1-B)(1-H) \big)\\[3pt]
&= (BH-H+1)^X.
\end{align*}
A: Since you did not mention otherwise, I will assume the decision to look is made independently of both the outcomes of the coin flips as well as all previous decisions.
Let's now define some important events. I apologize in advance, but I will override some of your notation.

*

*$A_i$ : the event that we do not see a heads on the $i$'th flip

*$L_i$ : the event that we decide to observe the $i$'th flip

*$B_i$ : the event that we decide not to observe the $i$'th flip

*$p$ : the probability of flipping a tails

*$n$ : the number of flips

By the law of total probability, we can then compute the probability of not seeing a heads on the $i$'th flip by
$$\Pr(A_i) = \Pr(A_i \mid L_i) \Pr(L_i) + \Pr(A_i \mid B_i) \Pr(B_i)$$
Of course, the probability of not seeing a heads given we do not look is 1, ie. $\Pr(A_i \mid B_i) = 1$. Furthermore, we know $\Pr(A_i \mid L_i) = p$ and $\Pr(L_i) = 1 - \Pr(B_i)$. Therefore,
$$\Pr(A_i) = p (1 - \Pr(B_i)) + \Pr(B_i) = p + (1 - p) \Pr(B_i)$$
Lastly, the probability to not see a head over all the flips can be expressed as
$$\Pr\left( \bigcap_{i = 1}^n A_i \right) = \prod_{i = 1}^n \Pr(A_i) = \left[ p + (1 - p) \Pr(B_i) \right]^n$$
