Probability of Winning Coin Tosses - Variable Number of Games

Question

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

Answer 1

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 $p^2$

Answer 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.

Answer 3

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 $$\sum\limits_{n=h}^{h+t-1} {h+t-1 \choose n}p^nq^{h+t-1-n} = \sum\limits_{n=0}^{t-1} {h+t-1 \choose h+n}p^{h+n}q^{t-1-n} = \sum\limits_{n=0}^{t-1} {h+t-1 \choose n}p^{h+t-1-n}q^{n}$$

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)

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

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