# Probability of Winning Coin Tosses - Variable Number of Games

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

• Who is "Henry"? – David G. Stork Dec 15 '18 at 3:06
• You can continue tossing coins even after an overall winner is determined. By the fifth coin-flip there is going to be exactly one person with at least 3 wins and that person is the overall winner. Use binomial distribution to calculate the probabilities of a player having won exactly 3, exactly 4, or exactly 5 of the flips over the course of the series and add those probabilities together. If told that certain outcomes have already happened, then only do a binomial distribution over the still remaining games to play until 5 happen. – JMoravitz Dec 15 '18 at 3:12
• Thank you for highlighting the actual question, and for pointing out the typo. Henry is supposed to be Hillary (the original question has Henry). – Connor Watson Dec 15 '18 at 3:12
• This is not in fact an expected value problem. – JMoravitz Dec 15 '18 at 3:12
• Hi all, I made a clarification in the question. But, I am not looking for the probability that any player wins. I'm looking for the probability that Hillary wins, which actually isn't determined by an exact number of coin flips. As shown, the first player to win 3 rounds wins, which means the game can end on the next flip (if trump gets a tails), being a total of either 4 or 5 rounds. The probability shifts when the coin is biased. – Connor Watson Dec 15 '18 at 3:23

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 $$p^2$$

• This is nice and concise, the tree below helps clarify why this is! – Connor Watson Dec 16 '18 at 8:19

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 helped a lot! – Connor Watson Dec 16 '18 at 8:19

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 $$\frac14=0.25$$

• Amazing, thank you! Will have to check this out. – Connor Watson Dec 16 '18 at 8:19