Tossing coins - basic probability example Oh no, another coin tossing problem? Yes. I've read more than a dozen of coin tossing questions here but I didn't find anything helpful.
Let's have an experiment: I have $3$ identical coins and a pot to throw them in.
So I throw all the coins into the pot. What is the probability of having $3$ heads?
Well, you can see several situations when you look in the pot. All the possible outcomes of this experiment are (H = head, T = tail):
$$
\Omega = \{ \text{TTT}, \text{TTH},\text{THH},\text{HHH}\}
$$
And then the probability $P$ is:
$$
P(\text{HHH}) = \frac{1}{|\Omega|} = \frac{1}{4}
$$
I was told that it's wrong but I just can't figure out why. :-(
 A: Actually $\Omega = \{TTT,TTH,THT,THH,HTT,HTH,HHT,HHH\}$, and the probability is $\frac{1}{8}$. An easier way to see this without writing down all the possibilities is by noting that the results of each coin are independent. Then it follows that $P(HHH) = P(coin \: 1\:  H)P(coin \: 2\:  H)P(coin \: 3\:  H) = \frac{1}{2} \frac{1}{2} \frac{1}{2} = \frac{1}{8}$.
A: The probability of getting exactly three heads in a row is $1\over 8$. For the first throw there is a $1\over 2$ chance, the second throw there is a $1\over 2$ chance, and the third throw there is a $1\over 2$ chance. The reason is because each coin throw is independent of the last one. The reason that $TTH$ is different from $THT$ is because the order of the event taking place matters. In other words it is a permutation. For example, some people think that when you are consistently throwing dice and $7$ does not come up for many throws, then the probability of $7$ coming up is greater because it hasn't happened in a bit. However, this thinking is incorrect because each throw is independent from the last.
