How many times must I toss a coin in order that the odds are more than 100 to 1 that I get at least one head?

Question

How many times must I toss a coin in order that the odds are more than 100 to 1 that I get at least one head? I believe that it is 10 times as if the coin is flipped ten times there is only ten outcomes that include only 1 head out of 1024 total outcomes. Is this correct?

Answer 1

The chance of getting at least one head is $1 - (\frac{1}{2})^n$

This equation has to equal $99 \%$, which gives us the following:

$$1 - \Big(\frac{1}{2}\Big)^n = 0.99$$ $$\Big(\frac{1}{2}\Big)^n = 0.01$$ $$n \log\frac{1}{2} = \log0.01$$ $$n = \frac{\log0.01}{\log\frac{1}{2}}$$ $$n = 6.6438...$$

Obviously we can't have $6.6438...$ number of tries, which means it must go up to the nearest integer, which means we must have $7$ tries to have more than $99 \%$ to get at least one head.

Answer 2

Not quite. You want to compare getting at least one head to getting no heads at all; the probability of getting no heads at all during $k$ tosses is the same as getting all tails, i.e. $(1/2)^k$. The probability of getting at least one head is simply the complement of this, i.e. $1-(1/2)^k$. Thus you want $1-(1/2)^k$ to be at least 100 times greater than $(1/2)^k$. Thus we want $$ \frac{1-(1/2)^k}{(1/2)^k}=2^k-1>100; $$ can you go from here?

Answer 3

If we toss $n$ times, the probability to get at least one head is $$1-(\frac{1}{2})^n$$ For $n=7$, we have a probability of $0.992\cdots$ , which is larger than $0.99$ hence the answer is $7$. To be exact, the probability of a failure is $\frac{1}{128}$, for $n=6$, the failure-probability would be $\frac{1}{64}$, which is still larger than $\frac{1}{100}$