I have been thinking of this problem for the post 3-4 hours, I have come up with this problem it is not a home work exercise

Let's say I have 3 coins and I toss them, Here order is not important

so possible sample space should be

0 H, 1 H, 2 HH, 3 HHH (H being heads) TTT, HTT, HHT, HHH

since P(T) and P(H) =1/2;

Here we have fair coins only, Since each and every outcome is equally likely, answer should be

1/4 (is this correct)

and if that is correct, all of the probabilities don't add up to one, will I have to do the manipulation to make it add up to one, or I am doing anything wrong.

EDIT In my opinion, with order being not important, there should be only 4 possible outcomes. All of the answers have ignored that condition.

  • 1
    $\begingroup$ are you looking for the probability to get exactly 2 heads or at last 2 heads in 3 tosses? $\endgroup$
    – kafman
    Apr 17, 2013 at 22:45
  • 1
    $\begingroup$ The events of $0$, $1$, $2$, and $3$ heads are not equally likely. $\endgroup$ Apr 17, 2013 at 22:46
  • $\begingroup$ @StringerBell exactly 2 heads, updated the question $\endgroup$
    – Max
    Apr 17, 2013 at 22:48
  • $\begingroup$ @SammyBlack Can you please explain why? Both head and tall have got 1/2, so all of the setting should have 1/8. $\endgroup$
    – Max
    Apr 17, 2013 at 22:50
  • $\begingroup$ Do you feel like the following sufficiently captures your concept of order not important: "Let's say we flip three separate coins, and then arrange them so that the heads are first and then the tails are last. What is the probability that we end up with the sequence $\rm HHT$?" $\endgroup$ Apr 17, 2013 at 23:07

7 Answers 7


The sample space has size $2^3 = 8$ and consists of triples $$ \begin{array}{*{3}{c}} H&H&H \\ H&H&T \\ H&T&H \\ H&T&T \\ T&H&H \\ T&H&T \\ T&T&H \\ T&T&T \end{array} $$

The events $$ \begin{align} \{ 0 \text{ heads} \} &= \{TTT\}, \\ \{ 1 \text{ head} \} &= \{HTT, THT, TTH\}, \end{align} $$ and I'll let you figure out the other two.

The probabilities are, for example, $$ P(\{ 1 \text{ head} \}) = \frac{3}{8}. $$

This is called a binomial distribution, and the sizes of the events "got $k$ heads out of $n$ coin flips" are called binomial coefficients.

  • $\begingroup$ I know that, but (for some stupid reason) I have a put up a restriction here, order not being important. So HTH, HHT are same. $\endgroup$
    – Max
    Apr 17, 2013 at 22:55
  • 1
    $\begingroup$ You are saying "order is not important" but this regards counting the number of heads. Order is still important in terms of the different possibilities. $\endgroup$ Apr 17, 2013 at 22:57
  • $\begingroup$ No I am saying 2 heads and 1 tail is 2 Heads and 1 tail, why am I being forced to calculate their combinations. May be I am being absurd, But I think it is a valid point. $\endgroup$
    – Max
    Apr 17, 2013 at 22:59
  • $\begingroup$ Try actually flipping these coins. Do 100 sets or so. You'll get about 3 times more "2 heads" than "3 heads". $\endgroup$ Apr 17, 2013 at 23:10
  • $\begingroup$ Thanks for the answer, but just to simplify it for anyone who would like to work it out in decimals. First total possibilities 8 = 2 x 2 x 2 Second Probability of Head 50% (0.5) so 3 coin flips 1.5 = 0.5 + 0.5 + 0.5 That gives you the probability of 1 head so double it for 2 heads is 3 = 1.5 x 2 (Heads) So 0.375 = 3/8 or 37.5% I hope that helps anyone who doesn't want to write out all the possibilities by hand. $\endgroup$
    – Jon
    Dec 7, 2018 at 16:50

Consider all the possible ways to get two heads, $\rm HHT, HTH \; and \; THH$. There are $2 \cdot 2 \cdot 2 = 8$ possible combinations in total. Therefore, the answer is $3/8$.

Your answer is wrong because the number of ways of changing around $\rm HHT$ (3) is not the same as the number of ways of changing around $\rm HHH$ (1). Can you see why this would invalidate your argument?

General solution: Binomial distribution. The probability of getting $k$ successes (here $2$) in $n$ trials (here $3$) is given by:

$$ \Pr(x=k) = \binom{n}{k} p^k (1-p)^{n-k}$$

Where $p$ is the probability of success (here, $p=1/2$), and $\binom{n}{k} = n!/(k!(n-k)!)$. This gives us:

$$ \binom{3}{2} \left(\frac12\right)^2 \left(1 - \frac12\right)^{3-2} $$ $$ 3 \cdot \frac14 \cdot \frac12 $$ $$ \frac38 $$


The outcomes you are looking for are either THH, HTH or HHT. Taking a look at for example THH: the possibility to toss T or H is $0.5$. Thus the possibility to throw T and then H and then H is $\frac{1}{2} \cdot \frac{1}{2} \cdot \frac{1}{2} = \frac{1}{8}$. But since we have three ways to "achieve" the desired result, the possibility of throwing exactly tow heads in three tosses is $\frac{1}{8} + \frac{1}{8} + \frac{1}{8} = \frac{3}{8}$.

If you continue you like this, you'll find that the possibilites of all possible outcomes (THH, TTH, TTT, HTT ...) add up to 1 indeed.


I believe there is a 50% chance of getting a toss of more then 1 head in 3 tosses. If you look at a chart you will see that the possible outcomes are HHH, HHT, HTH, HTT, THH, THT, TTH and TTT. As you can see if you add up all the possible outcomes with more then 1 head you will get 4/8. Therefore I believe it is an even chance of getting a toss of three dice with more than 1 head.

  • $\begingroup$ This already had an accepted anser. You are not contributing anything new $\endgroup$
    – Shailesh
    Nov 3, 2015 at 9:57

You are looking for two heads and nothing more. List the possibilities:


There are 3 cases out of the 8 with two heads:


The answer is $\frac38$.


Saying "order is not important" does not mean that each scenario (0, 1, 2, 3 heads) is equally likely. The probability of getting one head, for example, means that you got one head first, one head second, or one head third. There are 8 possible outcomes, so the probability of one head must be 3/8.


Let's say I have 3 coins and I toss them, Here order is not important so possible sample space should be 0 H, 1 H, 2 HH, 3 HHH (H being heads) TTT, HTT, HHT, HHH

So far, okay.

Since each and every outcome is equally likely

This statement is not true. If you flip three coins, then HHT is going to occur more often than HHH. The four outcomes are not equally likely.

The formula to calculate the probabilities is given in the answer by George V. Williams:

Pr(x = k) = (n C k) * [ p^k ] * [ (1-p)^(n-k) ]


(n C k) = n!/(k! * (n-k)!)

In the case of three fair coins, n = 3 and p = 0.5:

TTT (k=0 and HHH (k=3) both have probability 1/8 each.

HTT (k=1) and HHT (k=2) each have probability 3/8 each.


If order is not important, then there are four outcomes, but with different probabilities.

If order was important, then there would be eight outcomes, with equal probability.

  • 1
    $\begingroup$ Your answer would be much easier to read if it were properly formatted using MathJax. It might also be helpful if you explained why HHT and HHH are not equally probable (as far as I am concerned, they are equiprobable, but I would count THH, HTH, and HHT as distinct outcomes). $\endgroup$
    – Xander Henderson
    Jan 17, 2018 at 0:56

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