Probability of rolling unique sides on a bias die

Question

If you have a six-sided die labeled 1-5 with two sides labeled 5, what's the probability of rolling the die five times and getting a unique side that you haven't previously rolled?

I've been mulling this one over for some time, and can't seem to get around the idea that the probability seems to change depending upon when you roll a 5. There must be a logical work around.

Answer 1

There are 5! = 120 distinct events to consider, one event corresponding to a different permutation of $\{1,\ldots,5\}$. Each such event has probability $$ \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{3}. $$ Why? Because $P(1) = P(2) = P(3)=P(4) = \frac{1}{6}$, $P(5)=\frac{1}{3}$, and on each of the 120 distinct events mentioned above, each of $\{1,2,3,4,5\}$ occurs exactly once independently of the other rolls. Thus, the probability of the event you describe is

$$ 120 \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{6} \times \frac{1}{3} = \frac{120}{6^4\cdot 3} = \frac{5}{162}. $$

Answer 2

I'll provide an alternative approach to my answer above. Instead of thinking in terms of the probabilities $P(1),P(2),\ldots,P(5)$, one can approach this as a straight up counting problem and use the assumption that all six sides are equiprobable. If the sides were not equiprobable as in this problem, then you really need to directly compute probabilities, rather than just count events, like in my other answer.

So here goes. Label the sides $1,2,3,4,5_A, 5_B$. In five rolls, there are $6^5$ different possible sequences of sides. In order to have five distinct sides in five rolls, you'll need to see each of $1$,$2$, $3$, and $4$ exactly once and either $5_A$ or $5_B$ exactly once. That is, you'll need to see each of $1$,$2$, $3$, $4$, and $5_A$ exactly once you'll need to see each of $1$,$2$, $3$, $4$, and $5_B$ exactly once. There are $5!$ difference sequences of each of these two kinds. Thus, the probability of observing one of these sequences is $$ \frac{2\times 5!}{6^5} = \frac{5}{162}. $$

Answer 3

Here's one more way to get the probability. Use Will's labels: $1$, $2$, $3$, $4$, $5_A$, and $5_B$. The probability of getting five different results in six throws is $1 \times \frac{5}{6} \times \frac{4}{6} \times \frac{3}{6} \times \frac{2}{6} = \frac{120}{6^4}.$ If this occurs, exactly one label never turns up, and each label is equally likely to have not turned up.

For the sequence to meet the required conditions, the label that never turned up must be one of these two of the six sides: $5_A$, and $5_B$. Thus the desired probability is $\frac{2}{6} \times \frac{120}{6^4}$.