# Two people $A$ and $B$ flip a coin $10$ times…

I have not studied probability theory for a long while so I forgot the intuition under the computaion problems. Now I'm facing this exercise:

Two people $A$ and $B$ flip a coin $10$ times. Let's call $H$ the event of getting a head and $T$ the event of getting a tail. Suppose the coin is fair, so $P(H)=P(T)=\frac12$.

The $A$ flips are: $HTHHTHTTTH$ and the $B$ flips are $HHHHHHHTTT$.

• Which of the two flips were more likely to happen? (which one had a greater probability).
• What is the probability of getting at least one Head in the $10$ flips?

I would appreciate some hints, or maybe titles of some good books concerning this kind of exercises. Thanks a lot.

• Start with small sequences. Say $A$ flipped $H$ and $B$ flipped $T$. Which is more likely? What about if $A$ flipped $HT$ and $B$ flipped $TT$? – Arthur Dec 29 '15 at 18:44

Since each coin flips are independent, $A$ and $B$ both have the same chance of happening: $\frac{1}{2^{10}}$(Each unique sequence has the same chance of happening, and there are a total of $2^{10}$ sequences.).

If you want at least $1$ Head, suppose the contrary. Getting $0$ Head: You need to get $10$ Tails: The chance of that is $\frac{1}{2^{10}}$. The chance of getting at least is therefore $1-\frac{1}{2^{10}}$

If each person is flipping 10 times that is 20 flips in total, otherwise each person is flipping 5 times that is an ambiguity in the beginning worth nothing.

The easiest way to compute the probability of at least one Head is to consider subtracting 1 from the probability of all tails, which is the only other outcome allowed usually though technically some coins could land on their side this isn't usually considered.

As for the quoted portion, consider that each is just one of the 1024 possibilities and thus each is equally likely assuming that one doesn't care about order which is another ambiguity to the question as some may just count the number of Heads or number of Tails rather than want a specific order here.

Hints:

In this context any fixed sequence of $n$ heads and tails has probability $\left(\frac12\right)^n$ to occur.

To find the probability of getting at least one head it is handsome to find the probability of the so-called complement: no heads at all.