# Probability of of an event happening at least once in a sequence of independent events?

I want to find the probability of flipping heads at least once if you flip a coin two times. The possible outcomes (we don't care about the order) are (each equally likely) TT, TH, HT, HH. Three out of four have an H in them, so the probability is $\large \frac 34$. Is this correct? Is there a better and efficient way (especially when dealing with a higher number of flips? Please use only very basic terminology and concepts from probability because I've never taken a class. Thanks.

• Two coins two times, or one coin four times? Jun 24, 2013 at 18:08
• You state flipping a coin "two times", in which case, there are four possible outcomes: TT, TH, HT, HH, with probability $3/4$ that a head will appear at least once. Jun 24, 2013 at 18:10
• Yes I made a mistake haha, you are correct I meant the TT, TH, HT, HH
– Ovi
Jun 24, 2013 at 18:11
• Answer can't be correct, the answer has to be a dyadic rational. Jun 24, 2013 at 18:11
• @AndréNicolas Yea you're right I just made a mistake when writing the possible outcomes, should've been TT, TH, HT, HH
– Ovi
Jun 24, 2013 at 18:12

We have the following equally likely outcomes:

T T
T H <--
H T <--
H H <--

In three of the four outcomes, a Heads appears: Probability of at least one head is indeed $\dfrac 34$. Put differently, note that the probability that no heads appears is 1 out of four or $\frac 14$. So the probability of at least one head is $1$ minus the probability of getting NO heads, which is $1$ minus the probability of getting all tails: is $1 - \frac 14 = 1 - \frac{1}{2^2} = 1 - \frac 12 \cdot \frac 12$

In the above demonstration, it is quite easy to list out the "probability space": which is essentially, all possible outcomes.

The "at least one head" qualifier is handy, because it allows you to simplify the determination of probability for any arbitrarily large numbers of flips.

If you flip a coin $n$ times, and want to know the probability of getting at least one head, note the outcome of getting all tails is the "complement" of the set of outcomes in which you get at least one head. The probability of getting all/only tails, when flipping a coin $n$ times is equal to $$\underbrace{\dfrac 12 \cdot \frac 12 \cdot \cdots \cdot \frac 12}_{\large n\; \text{times}} = \left(\frac 1{2}\right)^n$$

So...\begin{align} \text{probability of getting at least one head}\; & = 1 -\text{probability of not getting any heads}\; \\ \\ & = 1 - \;\text{probability of getting all tails}\;\\ \\& = 1 - \left(\dfrac{1}{2}\right)^n\end{align}

• I don't understand why the probability of getting all tails is 1/2? If we have TT, TH, HT, and HH, "TT" is the only one that is all tails, so wouldn't it be 1/4? Dec 26, 2019 at 17:33

in problems like this one it's usually much easier to calculate the probability of the opposite event, so in this case - what is the probability of flipping a coin $n$ times without any heads? answer - it's $2^{-n}$ because each time you flip you need to get tails therefore the probability you seek is $1 - 2^{-n}$

The quickest way to find the probability of of an event happening at least once in a sequence of independent trials is to find the probability that it never happens in that sequence of trials, and then subtract from $1$.

You probably know that if $p$ is the probability of an event (say, flipping heads at least once) happening, then $1-p$ is the probability of that event not happening. Now let's apply this to your problem.

If you do not flip heads at least once in $n$ trials, then all $n$ trials must be tails. The probability of getting tails in one trial is $\frac12$, and since the trials are independent (what you get on flip 1 does not affect what you get on flip 2 does not affect what you get on flip k, etc.) the probability of getting tails in $n$ trials is $(\frac12)^n$. This is your $1-p$, so your $p=1-(\frac12)^n$.

I think it's helpful to generalize this question to a multi-sided coin (a coin with $m$ sides, which can be thought of as a die, if you will). I will call "side A" the desired outcome (heads for the coin). For two tosses, we can add the probabilities of two independent events.

Probability of getting side A on the first toss:

$\frac 1m$

Probability of not getting "side A" on the first toss, but getting it on the second toss=

$p(not side A) \cdot p(side A) = (1-\frac 1m) \cdot \frac 1m$

Total probability of getting side A in two tosses is the sum of the two events:

$\frac 1m + (1-\frac 1m) \cdot \frac 1m$

If we continue with more tosses ($n$ total), the probability grows. The probability of not getting "side A" on the first, second, ... or $n-1$th toss, but getting it on the $n$th toss =

$(1-\frac 1m)^{n-1} \cdot \frac 1m$

Total probability of getting "side A" in $n$ tosses is the sum of the independent events:

$\frac 1m \cdot [1 + (1-\frac 1m) + (1-\frac 1m)^2 + ...(1-\frac 1m)^{n-1}]$

• As you've shown the probability of getting at least one side A in $n$ tosses is $p=\frac{1}{m}\cdot\sum_{k=0}^{n-1}(1-\frac{1}{m})^k$ which features a geometric series of ratio $(1-\frac{1}{m})$. So $p= \frac{1}{m}\cdot\frac{1-(1-\frac{1}{m})^n}{1-(1-\frac{1}{m})} =1-(1-\frac{1}{m})^n$. This gives a hint for a simpler method, as suggested by @MichaelHardy: The probability to have no side A in one toss is $(1 - \frac{1}{m})$, for $n$ tosses it's $(1 - \frac{1}{m})^n$. The probability to have at least one side A in $n$ tosses is the complement $1-(1 - \frac{1}{m})^n$. Both methods work. Oct 20, 2018 at 15:22