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

This is a random walk, with a loss at $H-T=-1$, and a win at $H-T=10$.  I think the probability is $1/11$, or simply $d_1/(d_1+d_2)$  
For the biased coin, let $W(n)$ be the chance of winning when you are $n$ heads ahead:
$$W(0)=\frac23W(1)\\
W(1)=\frac13W(0)+\frac23W(2)=\frac29W(1)+\frac23W(2)=\frac67W(2)\\
W(2)=\frac13W(1)+\frac23W(3)=\frac27W(2)+\frac23W(3)=\frac{14}{15}W(3)\\
W(3)=\frac13W(2)+\frac23W(4)=\frac{14}{45}W(3)+\frac23W(4)=\frac{30}{31}W(4)$$
It looks like $$W(n-2)=W(n-1)\frac{2^n-2}{2^n-1}$$
so you multiply all those fractions, stopping with $W(10)=1$
