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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?

$$\lim_{n\to\infty}\frac{2\log\left(0.5\right)}{n\log\left(\frac{n-1}{n}\right)}\approx1.3874032818...$$


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

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  • $\begingroup$ 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 '18 at 14:24
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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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