How many times do we need to roll a fair die to get a better than evens chance of at least one six?

Question

"How many times do we need to roll a fair die to get a better than evens chance of at least one six?"

This is the question i need to answer as it is on a practice exam paper. I simply cannot understand how you can increase the chance of getting a 6?

Is it asking me how many times i need to roll a fair die in order to have more than a 50 percent chance of getting at least one 6? If it is then how can you increase your chance? I thought no matter how many times you roll it it will be a $${1}/{6}$$ chance of getting a 6 thus never will have a better than evens chance of at least one 6?

Im completely confused and have been for days now. I dont know if it is a badly asked question or if i am just not understanding it very well?! Could someone please explain what i would have to do.

would i have to use something like $$P(A^c)=1-P(A)$$?

Many thanks in advance for any help. It would be much appreciated as i cannot find even a similar style question to do anywhere online.

Answer 1

The question is asking about how likely it is that you'll get a six in any of the rolls you make.

The chance of not getting a $6$ is $\frac 5 6$. The chance of not getting any sixes in two rolls is $$\frac{5}{6}\cdot \frac{5}{6} = \left(\frac 5 6\right)^2$$

Similarly, the chance of not getting any sixes in $n$ rolls is $$\left(\frac 5 6\right)^n$$

So the chance of actually getting at least one six in $n$ rolls is $1 - \left(\frac 5 6\right)^n$; so you want to find the $n$ for which this number exceeds $\frac{1}{2}$.

A little trial-and-error shows that if $n = 4$, then the probability of getting at least one six is

$$1 - \left(\frac 5 6\right)^4 \approx 0.52$$

which works.

Answer 2

You can rethink the problem in the following way :

What is the probability of getting $\textbf{at least}$ a 6 if you roll a die n times ? The answer is :$$\sum_{k=1}^{n} \binom{n}{k} \left(\frac{1}{6}\right)^k \left(\frac{5}{6}\right)^{n-k} = 1 - \left(\frac{5}{6}\right)^n $$

So now, how many times should I roll to have more chances to get a 6 ?

$$ n = 1 \implies 1 - \left(\frac{5}{6}\right)^3 \approx 0.166666667$$ $$ n = 2 \implies 1 - \left(\frac{5}{6}\right)^3 \approx0.305555556$$ $$ n = 3 \implies 1 - \left(\frac{5}{6}\right)^3 \approx 0.421296296$$ $$ \textbf{n = 4} \implies 1 - \left(\frac{5}{6}\right)^4 \approx 0.517746914$$

Clearly 4 is definitely the right answer!

Answer 3

Yes, I think it is asking how many times you would have to roll a die to have a better than $50\%$ chance at rolling at least one $6$.

Note that the chance of not rolling a $6$ on an individual roll is $5/6$, so the chance of not rolling a $6$, say, $n$ times in a row would be $(5/6)^n$.

Thus, the chance of rolling a $6$ after $n$ rolls is $1 - (5/6)^n$; when does this exceed $0.5$?