# Probability of 3 Heads in 10 Coin Flips

What’s the probability of getting 3 heads and 7 tails if one flips a fair coin 10 times. I just can’t figure out how to model this correctly.

• In how many ways can you order the string $HHHTTTTTT$?.
– Pedro
May 30, 2012 at 22:42
• Pedro has only 6 T's, but he meant 7 T's, like this: HHHTTTTTTT Jun 12, 2017 at 0:31
• I really like the above two comments... They help the person understand how to solve this on their own (assuming they are a Math student who was asked to solve this - which is most likely who 95% of the people who come to this page are) - teaches them a way to think to solve the problem instead of here is the answer on a platter. (The other answers are useful for people like me, who has been away from this for 25 years and just needed the formula) Jun 29, 2020 at 20:35

Your question is related to the binomial distribution.

You do $n = 10$ trials. The probability of one successful trial is $p = \frac{1}{2}$. You want $k = 3$ successes and $n - k = 7$ failures. The probability is:

$$\binom{n}{k} p^k (1-p)^{n-k} = \binom{10}{3} \cdot \left(\dfrac{1}{2}\right)^{3} \cdot \left(\dfrac{1}{2}\right)^{7} = \dfrac{15}{128}$$

One way to understand this formula: You want $k$ successes (probability: $p^k$) and $n-k$ failures (probability: $(1-p)^{n-k}$). The successes can occur anywhere in the trials, and there are $\binom{n}{k}$ to arrange $k$ successes in $n$ trials.

• I am bit confused about this. Don't we use permutation we need to "arrange"? Why are we using combinations here? Sep 16, 2016 at 20:30
• @MuhammadAdeelZahid might be a bit late, but: $\left(\dfrac{1}{2}\right)^{3} \cdot \left(\dfrac{1}{2}\right)^{7}$ is the probability of one positive outcome whereas $\binom{10}{3}$ gives you the amount of possible outcomes. Since the positive outcomes are equally likely, it's simply adding them all up. Nov 14, 2017 at 15:02

We build a mathematical model of the experiment. Write H for head and T for tail. Record the results of the tosses as a string of length $10$, made up of the letters H and/or T. So for example the string HHHTTHHTHT means that we got a head, then a head, then a head, then a tail, and so on.

There are $2^{10}$ such strings of length $10$. This is because we have $2$ choices for the first letter, and for every such choice we have $2$ choices for the second letter, and for every choice of the first two letters, we have $2$ choices for the third letter, and so on.

Because we assume that the coin is fair, and that the result we get on say the first $6$ tosses does not affect the probability of getting a head on the $7$-th toss, each of these $2^{10}$ ($1024$) strings is equally likely. Since the probabilities must add up to $1$, each string has probability $\frac{1}{2^{10}}$. So for example the outcome HHHHHHHHHH is just as likely as the outcome HTTHHTHTHT. This may have an intuitively implausible feel, but it fits in very well with experiments.

Now let us assume that we will be happy only if we get exactly $3$ heads. To find the probability we will be happy, we count the number of strings that will make us happy. Suppose there are $k$ such strings. Then the probability we will be happy is $\frac{k}{2^{10}}$.

Now we need to find $k$. So we need to count the number of strings that have exactly $3$ H's. To do this, we find the number of ways to choose where the H's will occur. So we must choose $3$ places (from the $10$ available) for the H's to be.

We can choose $3$ objects from $10$ in $\binom{10}{3}$ ways. This number is called also by various other names, such as $C_3^{10}$, or ${}_{10}C_3$, or $C(10,3)$, and there are other names too. It is called a binomial coefficient, because it is the coefficient of $x^3$ when the expression $(1+x)^{10}$ is expanded.

There is a useful formula for the binomial coefficients. In general $$\binom{n}{r}=\frac{n!}{r!(n-r)!}.$$

In particular, $\binom{10}{3}=\frac{10!}{3!7!}$. This turns out to be $120$. So the probability of exactly $3$ heads in $10$ tosses is $\frac{120}{1024}$.

Remark: The idea can be substantially generalized. If we toss a coin $n$ times, and the probability of a head on any toss is $p$ (which need not be equal to $1/2$, the coin could be unfair), then the probability of exactly $k$ heads is $$\binom{n}{k}p^k(1-p)^{n-k}.$$ This probability model is called the Binomial distribution. It is of great practical importance, since it underlies all simple yes/no polling.

You are looking for

$$\frac{\text{Number of Relevant Outcomes}}{\text{Number ofTotal Outcomes}}.$$

The number of total outcomes is $2^{10}$. The number of relevant outcomes is the number of ways you can get exactly three heads in a string of 10 coin flips, or ${10}\choose{3}$. So the answer is

$$\frac{{10}\choose{3}}{2^{10}}.$$

• @Newbie1923 This approach is good here since all outcomes are equally likely. But if the coin is unfair not all outcomes will be equally likely, so you will have to do the binomial model instead. May 30, 2012 at 22:58

If you want them in that order then $\dfrac{1}{2^{10}} =\dfrac{1}{1024}$

If order does not matter then ${10 \choose 3}\times \dfrac{1}{2^{10}}=\dfrac{120}{1024} =\dfrac{15}{128}$

You need to model this with a binomial distribution, with $n=10$ and $p=0.5$.

• Perhaps you could add a bit to this answer? Jun 2, 2012 at 9:48

If this is in a continuum and you just need the probability of 3 of 10, then the practical answer is shown on the grey line here. I see it is about 15%. Other tips on Statistical Ideas web log, including treatment of any broader trials prior to the selection (e.g., 10 in this case).

Using Pascal's Triangle: Total of elements in row 10 = 1024 Choose element 4 of row 10 (this represents the number of ways three heads can occur): 120

Probability = 120/1024

• Hi The Bobster, please use appropriate formatting as described in the ChatJax tutorial to your right of the page. Nov 14, 2013 at 5:40
• This answer provides only information that all of the other answers also provided - and the other answers are both older by a year and a half and much more complete.
– Did
Nov 14, 2013 at 6:03

What I would do is to first find the total # of outcomes. $$\text{Outcomes}=2^{10}=1024$$ Then you have to arrange 3 heads and 7 tails within 10 open slots. We can name our Heads: $H(A)$, $H(B)$, and $H(C)$. Now, you can arrange the heads in such ways that $H(A)$ will essentially have 10 open slots, $H(B)$ will have 9 open slots and $H(C)$ will have 8 open slots. In other words there are: $$10 \times 9 \times 8 = 720$$ ways to arrange these heads. However, this accounts for the fact that all heads are different. Therefore the number of positions in which $H(x)$ can be arranged in is $$3 \times 2 \times 1 = 6,$$ and so there are: $$\frac{720}{6} = 120$$ scenarios of which 3 heads and 7 tails are given. Therefore $$P= \frac{120}{1024} = \frac{15}{128} \approx 11.7\%.\tag{getting 3 heads and 7 tails}$$

I think the easiest way to understand this problem is to do some rephrasing. Once we do that you can see that it is a simple combination problem.

Original question: What is the probability of getting only three heads with 10 coin flips?

There are 2 possibilities for each coin flip and 10 flips so the total number of outcomes is $2^{10}=1024.$

Now we want to know the total number of outcomes that result in only 3 heads with 10 coin flips. This is our desired outcome. This is where I find rephrasing very helpful.

Rephrase to: There are 10 canidates named Flip1, Flip2, Flip3,...Flip 10 running for the 3 council positions called 'Heads'. How many different groups of council members could there be if they all had eqaul chance of winning .

We can quickly see that this is a combination problem because all three positions are the same and we can't distinguish them apart. The number of combinations is:

$10!/((10-3)!*3!) = 120$ combinations

Therefore the total probability is $120/1024 = 15/128$

I hope this helps people see the similarity between this question and the more common combination problem format.