If you flip a fair coin 11 times what is the probability of getting at least one tails

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!

Your "naive first thought" is the clever (standard) solution.

To make it rigorous, let $\mathscr E_0$ be the event "A and B are tied after each has tossed 10 times;" let $\mathscr E_A$ and $\mathscr E_B$ be the events "A has more heads than B after 10 tosses each" and "B has more heads than A after 10 tosses each," respectively. Let $\mathscr F$ designate the event "A has more heads than B after all tosses are made."

Notice:

1. $\mathscr E_0,$ $\mathscr E_A,$ and $\mathscr E_B$ are mutually exclusive: no two have any outcomes in common and collectively they include all the possibilities. Therefore $$\Pr(\mathscr E_0) + \Pr(\mathscr E_A) + \Pr(\mathscr E_B)=1.$$

2. $\Pr(\mathscr F\mid \mathscr E_A) = 1$ (A has won by the first 10 tosses); $\Pr(\mathscr F\mid \mathscr E_B) = 0$ (A is behind after 10 tosses and therefore cannot win with the last toss); and $\Pr(\mathscr F\mid \mathscr E_0) = 1/2$ (if both are tied after 10 tosses, A's 11th toss is the tiebreaker).

3. $\Pr(\mathscr E_A) = \Pr(\mathscr{E_B})$ (after 10 tosses the game is symmetric -- both players are equally situated -- and therefore they have equal chances of being ahead at that point).

By the law of total probability,

$$\begin{aligned} \Pr(\mathscr F) &= \Pr(\mathscr F\mid \mathscr E_0)\Pr(\mathscr E_0) + \Pr(\mathscr F\mid \mathscr E_A)\Pr(\mathscr E_A) + \Pr(\mathscr F\mid \mathscr E_B)\Pr(\mathscr E_B)\\ & = \Pr(\mathscr E_0)\left(\frac{1}{2}\right) + \Pr(\mathscr E_A)(1) + \Pr(\mathscr E_B)(0)\\ &= \frac{1}{2}\left(\Pr(\mathscr E_0) + \Pr(\mathscr E_A) + \Pr(\mathscr E_A)\right)\\ &= \frac{1}{2}\left(\Pr(\mathscr E_0) + \Pr(\mathscr E_A) + \Pr(\mathscr E_B)\right)\\ & = \frac{1}{2}\left(1\right) = \frac{1}{2}. \end{aligned} $$

As an alternative approach, you wish to evaluate the double sum

$$\sum_{a \gt b} \binom{11}{a}\binom{10}{b} = \sum_{a \gt b} \binom{11}{11-a}\binom{10}{10-b} = \sum_{a^\prime \le b^\prime} \binom{11}{a^\prime}\binom{10}{b^\prime}.$$

(In case the algebra isn't obvious, the first equality exploits the Binomial coefficient symmetry and the second is the change of variable $a^\prime = 11-a,$ $b^\prime = 10-b.$ We can be a vague about the endpoints of the summations because whenever $a$ or $a^\prime$ is not in the range from $0$ through $11$ or $b$ or $b^\prime$ is not in the range from $0$ through $10$ the Binomial coefficients are zero.)

Because the indexes in the two sums on the left and right sides (1) never overlap and (2) cover all the possibilities (since either $a\gt b$ or $a\le b$ but never both), together they give the total probability, which is $1.$ Consequently, since those sums are equal, each is $1/2,$ QED.

This figure shows the rotational symmetry of the distribution under the mapping $(a,b)\to(11-a,10-b).$ The blue circles are rotated around the yellow dot into red triangles of exactly the same probability. The desired sum is the total of the blue circles, which therefore must be $1/2.$

What is the probability of getting at least one tails?

When flipping a coin 10 times what is the probability of at least one head?

What is the probability of flipping a coin and getting at least one tails?

What are the odds of losing a coin flip 11 times in a row?