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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.

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

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 $1 – 2^{-n}$

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:

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

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 $1$.

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