At least one multiple of 3 in five dice rolls

Text

A fair die is rolled $5$ times. Giovanni bets $200$ euros that a number divisible by $3$ comes up at least once.

(a) Compute the probability of winning.

(b) If the odds are $1.1$, are they fair, favorable, or unfavorable to the house?

Solution

(a) The numbers on the die divisible by $3$ are $3$ and $6$, so $P(\text{mult. of }3)=\dfrac{2}{6}=\dfrac13$ and $P(\text{not mult. of }3)=\dfrac23$. With $5$ independent rolls, $P(\text{never}) = \left(\frac23\right)^5 = \frac{32}{243}$ so $P(\text{at least one}) = 1 - \frac{32}{243} = \frac{211}{243} \approx 0.868$

(b) The fair odds (that make the game fair) are $q_e = \dfrac{1}{0.868} \approx 1.152$. Since the offered odds $1.1$ are lower than $1.152$, the player receives less than what would be fair: the odds are unfavorable to the player (favorable to the house).

$\boxed{P\approx0.868;\ \text{odds }1.1\text{ favor the house}}$