2
$\begingroup$

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. :-(

$\endgroup$
1
  • $\begingroup$ @DavidMitra And does anything change when the coins are identical or not? I had a couple of lectures on cominatorics preceding to probability lectures and we were distinguishing all the time whether our coins, dice, balls etc. are identical or not. And because the probability exercises heavily use the combinatorics techniqes, I thought that we still need to examine whether our objects are identical or not... $\endgroup$
    – Jeyekomon
    Commented Oct 24, 2013 at 15:03

2 Answers 2

Reset to default
3
$\begingroup$

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 $P(HHH) = P(coin \: 1\: H)P(coin \: 2\: H)P(coin \: 3\: H) = \frac{1}{2} \frac{1}{2} \frac{1}{2} = \frac{1}{8}$.

$\endgroup$
7
  • $\begingroup$ I've already got this advice but I can't still understand what is the difference between TTH and THT - the difference between having two tails and one head in the pot and having two tails and one head in the pot. There is said that the coins are identical... $\endgroup$
    – Jeyekomon
    Commented Oct 24, 2013 at 14:50
  • 1
    $\begingroup$ @Jeyekomon It's easier to imagine if the coins are not identical. This will not have any effect on the outcome, but it's easier to see the difference between $HHT$ and $HTH$. $\endgroup$
    – Arthur
    Commented Oct 24, 2013 at 14:58
  • $\begingroup$ 2 throwings. If TH (or HT wich is in your view the same) then I win, if TT then you win. Deal? $\endgroup$
    – drhab
    Commented Oct 24, 2013 at 14:59
  • 1
    $\begingroup$ Ok, so you look at the pot and only count the number of heads. In that case you indeed get only four possibilities: $HHH, HHT, HTT, TTT$. To conclude from this that the probability of $HHH$ is $\frac{1}{4}$ you also need to know that these four possibilities are all equally likely. That is however not the case here. $\endgroup$
    – Arthur
    Commented Oct 24, 2013 at 14:59
  • 1
    $\begingroup$ Think about it. Does the probability of getting $3$ heads change if we mark the coins? $\endgroup$
    – Arthur
    Commented Oct 24, 2013 at 15:18
1
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ +1 - first sentence gives a correct answer with no need to describe the THH HHT difference. Simple and right. $\endgroup$
    – JTP - Apologise to Monica
    Commented Dec 13, 2013 at 2:24

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .