# Probability of three dice in n rolls

Consider the bet that all three dice will turn up sixes at least once in n rolls of three dice. Calculate $f(n)$, the probability of at least one triple-six when three dice are rolled n times. Determine the smallest value of n necessary for a favorable bet that a triple-six will occur when three dice are rolled n times. (DeMoivre would say it should be about $216 \log 2 = 149.7$ and so would answer 150—see Exercise 1.2.17. Do you agree with him?)

As I understand:

$f(n)=1 - \text{(probability of no triple sixes in} \space n \space \text{rolls})$, i.e.

$f(n)=1- \left (\frac{5\cdot 5\cdot 5}{6\cdot 6\cdot 6}\right)^n$

Is that correct? What about the second question? Doesn't it depends on the bet, or what do they mean?

• I'm guessing a favourable bet would mean atleast a $0.5$ chance? – AnotherJohnDoe Oct 10 '17 at 14:26
• Please type out the question rather than posting a picture of it. – Théophile Oct 10 '17 at 14:37
• Théophile, okay, one minute – Max Oct 10 '17 at 14:42

The probability $\frac{5^3}{6^3}$ is the probability of the event "you get no sixes at all (in a particular throw)".
The probability for not getting "a triple six" (in one throw) is $1-\frac{1}{6^3}$. So
$$f(n) = 1 - (1-\frac{1}{6^3})^n$$
The first value of $n$ for which this is larger than $0.5$ is $n=150$, so I would agree with de Moivre.