Probability of getting 10 effective heads What is the probability of success satisfying both conditions ?

Condition 1 : The toss ends (success) when $N_H - N_T = 10$ where $N_T$ is number of tails and $N_H$ number of heads.
Condition 2 : During every toss, number of heads must be greater than or equal to tails.

Example:

*

*$HHTTHHHHHHHHHH$ is a valid toss.

*$HTTHHHHHHHHHHH$ is not a valid toss.

Solving this using only first condition is quite easy, but satisfying both conditions is tough.
What if a biased coin is used? P(head) = 2/3 and P(tail) = 1/3

One can model this as a Markov process. In particular, note that you only care about the difference $H-T$, which, at each round, either increases by $1$ or decreases by $1$ with equal probability. You have twelve states, since $H-T$ is always within $[-1,10]$, with $-1$ and $10$ both being absorbing states (as they end the sequence of flips). It is thus possible to calculate the probability of reaching these states using the usual methods.
If one wants to be a little more clever, they can note that the expected value of $H-T$ is $0$ at any step. Along with the fact that $H-T$ is bounded and any sequence of flips will almost surely stop at some point, this suffices to show that if $p$ is the probability of getting $10$ flips, we have
$$0=10\cdot p+-1\cdot (1-p)$$
where, we essentially take the expected value of the stopping state. Solving gives $p=\frac{1}{11}$. I'll leave plugging in the necessary technical details to do this manipulation as an exercise to the reader.
