How many times do you have to roll a pair of dice such that you've got a 50% chance of at least one roll being double sixes?

My own intuition (and mentioned in the video cited below as a common wrong answer) is 18: $\frac{1}{36}$ chance of rolling double sixes, therefore after 18 rolls you've got $\frac{18}{36}=0.5$.

I understand mathematically how the correct answer is $\frac{\log{\left(0.5\right)}}{\log{\left(\frac{35}{36}\right)}}\approx24.6$, but I'm missing how this number comes about intuitively. How could I explain this result to say, my family around the dinner table, without exponents or logarithms?

Relatedly, I made a spreadsheet calculating the number of rounds required to solve this problem for any $n$ (in this question $n=36$), and found that if you divide the correct answer (24.6) by the "intuitive" answer (18), the result appears to converge at 1.39. Does this value mean anything?


(The source of this question is a Vsauce2 video https://www.youtube.com/watch?v=Uyw7d579nxY&feature=youtu.be&t=141)

  • $\begingroup$ Not sure what you mean by saying $24.6$ is the correct answer. Obviously, you can't roll the dice $24.6$ times. The correct answer must be the least integer exceeding this, hence $25$. I've posted more detail below. $\endgroup$
    – lulu
    Oct 23, 2018 at 14:24

1 Answer 1


The probability that a given roll yields double sixes is $\frac 1{36}$. It follows that the probability that a given roll yields something else is $\frac {35}{36}$. Thus, the probability that $n$ given rolls fails to yield any double sixes is $\left(\frac {35}{36}\right)^n$. So you want the least $n$ such that $$\left(\frac {35}{36}\right)^n≤.5$$

You can now proceed by trial and error, using a calculator of course. Not hard to find $n=25$ as the solution. To do it analytically you can solve $\left(\frac {35}{36}\right)^x=.5$ using logs. We get $$x\times (\log {35}-\log {36})=\log {.5}\implies x\approx 24.61$$

which again leads us to $n=25$.

To your second question, since $$\lim_{n\to \infty} \left( \frac {n-1}n\right)^n=e^{-1}$$ you are just computing $-2\log {.5}=\boxed {1.386294361}$.


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