# Coin flipping and probablities

I was given the next question: Harry and Liam flip coins. Harry flips the coin 8 times whereas Liam flips the coin 7 times. What is the probability that Harry got more "tails" than Liam? (the probability for the coin to land on "tail" is 1/2)

The second part of the question asks if there is a number $p$ for which the probability that I was asked to find earlier is smaller than 0.2

My question is, how do I approach the first part of the question? My idea was to use the Law of total probability whereas the different scenes are Liam getting different numbers of "tails". Yet, I couldn't manage to find the use of the fact that the probability for the coin to land on "tail" is 1/2.

Firstly about your last sentence. Realize that the information probability for the coin to land on "tail" is 1/2 gives you the full information about the probablity of the flipping coin process. This is because there is only one other option that can happen, namely the coin lands on "head", we know that the probablity of the latter to happen is $1/2$ as well (due to the fact that the sum of all probablities equals to 1) Moreover, assuming that the each flip of the coin is independent of any other flip (which is a classical assmuption in these flipping coin riddles), we can deduce from the initial statement the probablities of arbitrary flipping scenario of the coin, say $n$ flips in a row will result in something (e.g. probablity to get $n-1$ "heads" and the last flip to be "tail" equals $(1/2)^n$).
So the above consideration is a hint how to approach the first part of the question. To be more specific, assume that we flipped the coin $n$ times in a row. The probability of any specific outcome is $(1/2)^n$. This is due to the fact that the probability of "tails" is the same as of "heads" and that the flips are independent of each other (i.e. we multiply the possibility that the first flip will be something with the probability that the second flip will be something and so on till we multiply all the probabilities of all the $n$ cases, each of which has probability to happen equal to $1/2$).
Say Harry got 2 "tails" which has the probability to happen equal to $\binom{8}{2}(1/2)^8$ (the binomial coefficient pops up because arbitrary sequence with precisely two "tails" satisfies the 2 "tails" condition). Then Liam to have less "tails" means Liam got either zero or one "tail". The probability of the first case is $\binom{7}{0}(1/2)^7$ and of the second is $\binom{7}{1}(1/2)^7$. It is natural to assume that the flips of the two guys does not influence each other so there will be multiplication of probabilities again. The chance that Harry gor more "tails" than Liam when we know that Harry got precisely 2 "tails" is $$\binom{8}{2}(1/2)^8 (\binom{7}{0}(1/2)^7 + \binom{7}{1}(1/2)^7) .$$ The plus sign arises because this example has two separate cases: (Harry = 2 "tails" and Liam = 0 "tails") and (Harry = 2 "tails" and Liam = 1 "tail") which we add together to get the probability of the event.