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Hillary and Trump play a game of coin toss. The coin is fair such that $mathrm{Pr}(x=H) = mathrm{Pr}(x=T) = 0.5$. If it gets a Head (H), Hillary wins, otherwise, Trump wins.
They agree in advance that the first player who has won $3$ rounds will collect the entire prize. Coin flipping, however, is interrupted for some reason after $3$ rounds and they got $1$ Head and $2$ Tails. Suppose that they continue to toss the coin afterwards, what is the probability that Henry Hillary will win the entire prize?

I saw this question also on Chegg and I got the correct answer, but I drew a tree of the rest of the remaining outcomes. We get: 3T, 3HT, and 3HH. Hillary winning would be HH, so $mathrm{Pr}(HH) = 3/9 = 1/3$. But I want to know the proper way to do this problem. For example, what formula is used?

The second the coin changed to be biased, I have no idea what to do. In addition, how would you derive the solution is the coin is biased? Ex: Pr(x=H)=0.75? I believe this is an Expected Value problem...looked through some other questions and I couldn't quite figure out the solution.

Any solutions or links to duplicate questions (with solutions) would be great. Thank you in advance!

P.S. I saw this question but it didn't quite help me...
Fair and Unfair coin Probability

Hillary wins if the next two flips are $HH$, which is probability $frac 14$. Otherwise Trump wins. If the coin is not fair and falls heads with probability $p$ her chance of winning is $p^2$

Drawing a tree (or even a partial tree) is the way I'd recommend approaching this sort of problem, at least if you're dealing with a reasonably-sized case set:

If you're dealing with an unfair coin, I find it helps to draw the probabilities on the edges. Where we start, you can see that we flip the coin and get heads with probability $p$. To get heads again (and for Hillary/Henry) to win, we have to get heads again, so we multiply by $p$ (independent events). This gives us a probability of $p^2$ to get two heads and for Hillary to win.

This is essentially the problem of points, first solved satisfactorily by Fermat and Pascal in 1654 and laying the foundations for probability theory.

You asked for a formula in general: if Hillary needs $h$ more points to win and Trump needs $t$ more points, and the probability of each heads is $p$ and each tails $q=1-p$, then the probability of Hillary winning the match is any of $$sumlimits_{n=h}^{h+t-1} {h+t-1 choose n}p^nq^{h+t-1-n} = sumlimits_{n=0}^{t-1} {h+t-1 choose h+n}p^{h+n}q^{t-1-n} = sumlimits_{n=0}^{t-1} {h+t-1 choose n}p^{h+t-1-n}q^{n}$$

If the match was originally the first to reach $m$ points and the score so far has Hillary on $a$ points and Trump on $b$ points then you have $h=m-a$ and $t=m-b$ making the probability of Hillary winning the match (using the third expression above)

$$sumlimits_{n=0}^{m-b-1} {2m-a-b-1 choose n}p^{2m-a-b-1-n}q^{n}$$

Applying these to your original example with $m=3, a=1, b=2, h=2, t=1, p=q=frac12$ gives a result of $frac14=0.25$