A coin is flipped eight times where each flip comes up either heads or tails. How many possible outcomes

a) are there in total?

b) contain exactly three heads?

c) contain at least three heads?

d) contain the same number of heads and tails?

  • 4
    $\begingroup$ Where are you stuck exactly? $\endgroup$ Nov 30, 2013 at 17:08
  • $\begingroup$ I want the answer $\endgroup$
    – ali
    Nov 30, 2013 at 17:13
  • 19
    $\begingroup$ I want everlasting peace on earth. $\endgroup$ Nov 30, 2013 at 17:30
  • 4
    $\begingroup$ Here's the answer, 42. $\endgroup$
    – kintoki
    Nov 30, 2013 at 17:39
  • 1
    $\begingroup$ Related. $\endgroup$
    – Julien
    Nov 30, 2013 at 21:13

3 Answers 3


First, you need to decide what counts as a possible outcome.

For example, are $HHTTTTTT$ and $TTTTTTHH$ consider different outcomes (the heads and/or tails appear in different orders, where order matters?); alternatively, since they both result in $2$ heads and $6$ tails, are they considered the same outcome (i.e., order doesn't matter).?

I will assume that the order of the results from each toss does matter, so that in the example above, we consider the two outcomes to be different.

Then, for each part of your problem, it will be helpful to answer $(a):$

(a): How many such outcomes are possible, given 8 coin tosses?

For each toss, there are two possible outcomes: heads, or tails.

Using the rule of the product, we have that after $8$ tosses, the possible outcomes/sequences is equal to $$\underbrace {2\times 2 \times \cdots \times 2}_{\large 8\; \text{tosses}} = (2)^8= \bf 256$$

(b) Now of these $256$ possible outcomes, we need to determine how many of them are sequences containing

  • exactly three heads (hence, exactly 5 tails), no fewer, no more. So there are eight coins, and we need to compute how many ways we can choose three to be heads: $$ \binom{8}{3} = \dfrac{8!}{3!\,5!} = \bf 56$$

  • at least three heads (hence, at most 5 tails): here we count sequences with exactly 3 heads, and also count sequences with more than three heads. What doesn't count are those sequences containing (zero heads, exactly one head, or exactly two heads). We could compute this as $$\binom 83 + \binom 84 + \cdots + \binom 88$$ but it would be far easier to compute its equivalent, by subtracting from 256 the number of sequences with no head, only one head, and only 2 heads: $$ 256 - \left(\binom 80 + \binom 81 + \binom 82\right)= 256 - (1 + 8 + 28)= \bf 219$$

  • the same number of heads and tails: An outcome "counts" if and only if it contains exactly 4 heads (and hence, exactly 4 tails). Here, we need only compute the number of ways to choose exactly four heads, since the the other four will necessarily then be tails. $$\bf \binom 84 = \dfrac{8!}{4!\,4!}= \bf 70$$

  • 1
    $\begingroup$ Things are not so obvious. Suppose you come to the table and observe the outcome after the coins are thrown. The only thing you can observe is the number of heads. And three heads is a distinct configuration different from two or four heads. So in this case, there is only one outcome with three heads. Either it happened or it didn't. Of course, this says nothing about probability, the binomial distribution, or any of the other answers from the "in the box" thinkers here. $\endgroup$ Sep 24, 2018 at 23:59
  • $\begingroup$ Hey, novice here. I'm confused by this answer after staring at it. It seems like you said you assume the order does matter, which is fine, but then you went on to use the formula for combinations without repetition, which is supposed to be for choosing things in an unordered way, right? For (b), when you write out 8!/(3!*5!), having just reviewed the derivation for the "n choose k" formula, it looks like the 3! in the denominator is dropping the double-countings of equal-valued outcomes, a step done because "n choose k" is supposed to be for unordered results. What am I misunderstanding? $\endgroup$ Nov 8, 2020 at 12:45

Relate this to binary numbers. A coin outcome is 0 or 1. So you have base 2 (binary) numbers 00000000 to 11111111. An 8-bit number can express $2^{8}=256$ possible states. In the same way, an 8 digit base-10 number can express 0 - 99999999, which is 100000000 = $10^{8}$ numbers.

There's only one way to get 0 heads, which is ${8}\choose{0}$. There are 8 ways to get 1 head: 10000000 01000000 00100000 00010000 00001000 00000100 00000010 00000001 . In other words, you have 8 coins and you have to choose one of them to show a head: that's ${8}\choose{1}$. If you want 2 heads, then it's ${8}\choose{2}$, etc. To get 3 or more heads you can add ${8}\choose{3}$ + ${8}\choose{4}$ + $\cdots$ + ${8}\choose{8}$.


Think about in combinations.

If you experiment with 1, 2, 3, and 4 coins, you find that there are always $2^n$ possibilities, where n is the number of coins. Now consider Pascal's triangle. The sum of the numbers is exactly what we want. So just use combinations, i.e. just 3 heads is 8C3, the number of possibilities is $2^8$, at least 3 heads is just 2^8-8C1-8C2. You should be able to do the rest.

If you want to know why the combinations actually work, think about it like this. You are counting the combinations with 3 heads. It's like arranging these where T is tail and H is head:


This is just 8 total objects, and 3 to put into the "spots". I mean, it's like putting 3 H's into 8 slots, the T's.


Not the answer you're looking for? Browse other questions tagged or ask your own question.