Probability of rolling doubles more than once in ten rolls with a single a pair of dice?

Question

This is a question I made up just trying to practice probability calculations. I searched, but I didn't find any duplicates--most questions want to know probability of getting doubles once, or of getting an exact combination (I.e. snake eyes, or double sixes).

I came up with $.8385$.

My reasoning was: $P(\text{doubles})=6/36=1/6$, so $P(\text{not doubles})=5/6$. We have ten rolls, so the probabilty of not getting doubles more than once is $(5/6)^{10}=.1615$ so $P(\text{doubles more than once})=1-.1615=.8385$.

Is this the right answer, and if not, could someone help me understand where I went wrong?

Thanks!

Answer 1

You are almost there, you also have to subtract the cases where we get exactly one double, as the question is asking for the probability of more than one double.

Now, $P(\text{one double}) = {10\choose 1} \cdot \left(\frac 16 \right)\cdot \left(\frac 56\right)^9 \approx 0.323$

and so the desired result is $0.8385-0.323=0.5155$.

Answer 2

Ok, so this is a good example of a binomial random variable. Let $X$ be the number of doubles obtained in $10$ rolls of a pair of dice. So, the sample space of $X$ is $\{0,1,2,...10\}$. On any particular roll, the probability of rolling a double is $p=\frac{1}{6}$. The binomial probability formula tells us that in $n$ trials, the probability of $k$ successes is $$\mathrm{P}( X=x) =\frac{n!}{( n-x) !x!} p^{x}( 1-p)^{n-x}$$ So in our case, $p=\frac{1}{6}$ and $n=10$. So the probability of at least $2$ doubles is $$\mathrm{P}( X\geq 2)=\sum ^{10}_{x=2}\frac{10!}{( 10-x) !x!}\left(\frac{1}{6}\right)^{x}\left( 1-\frac{1}{6}\right)^{10-x}$$ Or equivalently, $$1-\sum ^{1}_{x=0}\frac{10!}{( 10-x) !x!}\left(\frac{1}{6}\right)^{x}\left( 1-\frac{1}{6}\right)^{10-x}$$ So doing the calculations, $$1-\left(\left(\frac{5}{6}\right)^{10} +10\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)^{9}\right)\approx 0.515.$$