What is the probability of flipping a coin 11 times and getting heads 3 times?

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.

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Welcome to the coin flip probability calculator, where you'll have the opportunity to learn how to calculate the probability of obtaining a set number of heads (or tails) from a set number of tosses. This is one of the fundamental classical probability problems, which later developed into quite a big topic of interest in mathematics. For example, maybe you like Batman, and know of one of his many villains, Two-Face? You'd think that his name comes from the fact that half of his face is burnt, but no! (Okay, maybe a little bit.) He has a lucky coin that he always flips before doing anything. As this coin has two faces on it, his coin toss probability of getting a head is 1. Better not get on the wrong side (or face) of him!

🙋 If you're interested in the probability of runs in coin flips, visit our dedicated coin toss streak calculator.

Classical probability

The probability of some event happening is a mathematical (numerical) representation of how likely it is to happen, where a probability of 1 means that an event will always happen, while a probability of 0 means that it will never happen. Classical probability problems often need to you find how often one outcome occurs versus another, and how one event happening affects the probability of future events happening. When you look at all the things that may occur, the formula (just as our coin flip probability formula) states that

probability = (no. of successful results) / (no. of all possible results).

Take a die roll as an example. If you have a standard, 6-face die, then there are six possible outcomes, namely the numbers from 1 to 6. If it is a fair die, then the likelihood of each of these results is the same, i.e., 1 in 6 or 1 / 6. Therefore, the probability of obtaining 6 when you roll the die is 1 / 6. The probability is the same for 3. Or 2. You get the drill. If you don't believe me, take a dice and roll it a few times and note the results. Remember that the more times you repeat an experiment, the more trustworthy the results. So go on, roll it, say, a thousand times. We'll be waiting here until you get back to tell us we've been right all along. Go to the dice probability calculator if you want a shortcut.

But what if you repeat an experiment a hundred times and want to find the odds that you'll obtain a fixed result at least 20 times?

Let's look at another example. Say that you're a teenager straight out of middle school and decide that you want to meet the love of your life this year. More specifically, you want to ask ten girls out and go on a date with only four of them. One of those has got to be the one, right? The first thing you have to do in this situation is look in the mirror and rate how likely a girl is to agree to go out with you when you start talking to her. If you have problems with assessing your looks fairly, go downstairs and let your grandma tell you what a handsome, young gentleman you are. So a solid 9 / 10 then.

As you only want to go on four dates, that means you only want four of your romance attempts to succeed. This has an outcome of 9 / 10. This means that you want the other six girls to reject you, which, based on your good looks, has only a 1 / 10 change of happening (The sum of all events happening is always equal to 1, so we get this number by subtracting 9 / 10 from 1). If you multiply the probability of each event by itself the number of times you want it to occur, you get the chance that your scenario will come true. In this case, your odds are 210 * (9 / 10)4 * (1 / 10)6 = 0.000137781, where the 210 comes from the number of possible fours of girls among the ten that would agree. Not very likely to happen, is it? Maybe you should try being less beautiful!

How to calculate probability?

"Hey man, but girls and coins are two different things! I should know, I've seen at least one of each." Well, let me explain that these two problems are basically the same, that is, from the point of view of mathematics. Whether you want to toss a coin or ask a girl out, there are only two possibilities that can occur. In other words, if you assign the success of your experiment, be it getting tails or the girl agreeing to your proposal, to one side of the coin and the other option to the back of the coin, the coin toss probability will determine the answer. It all boils down to getting your hands on a coin that is weighted appropriately. Mathematically, we talk about the binomial probability distribution.

Let's look at a step-by-step example to see how to calculate the probability of an event using the coin toss probability calculator:

What is the probability of flipping a coin 10 times and getting heads 3 times?

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