Question

In a dice game played with three fair six-sided dice, the objective is to collect 6s. after the first roll, any dice showing a 6 are reserved, and the other dice are rolled again. what is the probability that a second roll will result in two dice showing a 6?

Answer 1

Since we're talking about "a second roll", we must consider all the possible outcomes of the first.

Case 1: the first roll had no 6s.

This happens with probability

`\left((5)/(6)\right)^3`

(you want one of the other 5 numbers in all dice). In this scenario, the second roll involves all three dice, and we want exactly two 6s. Using Bernoulli's distribution, we know that this happens with probability

`\displaystyle\binom{3}{2}\cdot\left((1)/(6)\right)^2\cdot (5)/(6)`

Case 2: the first roll had one 6.

This happens with probability

`\displaystyle\binom{3}{1}\cdot\left((5)/(6)\right)^2\cdot (1)/(6)`

In this scenario, the second roll involves two dice, and we want them both to show 6. This happens with probability

`\left((1)/(6)\right)^2`

Case 3: the first roll had two 6s.

In this case, the second roll would involve the only remaining dice, so there's no way you can have two 6s in the second roll

Case 4: the first roll had three 6s.

In this case there is no second roll, because all three dice have already shown 6, and are thus reserved.

So, if we sum all possible scenario, multiplied by the probability that that scenario would occur, we have this global probability for the first scenario

`\displaystyle \left((5)/(6)\right)^3\cdot\binom{3}{2}\cdot\left((1)/(6)\right)^2\cdot (5)/(6) = \binom{3}{2}\cdot\left((1)/(6)\right)^2\cdot\left((5)/(6)\right)^4`

and this global probability for the second scenario

`\displaystyle\binom{3}{1}\cdot\left((5)/(6)\right)^2\cdot (1)/(6)\cdot \left((1)/(6)\right)^2 = \binom{3}{1}\cdot\left((5)/(6)\right)^2\cdot \left((1)/(6)\right)^3`

Their sum is

`\displaystyle\binom{3}{2}\cdot\left((1)/(6)\right)^2\cdot\left((5)/(6)\right)^4 + \binom{3}{1}\cdot\left((5)/(6)\right)^2\cdot \left((1)/(6)\right)^3 = \left((1)/(6)\right)^2\cdot \left((5)/(6)\right)^2 \left(\binom{3}{2}\left((5)/(6)\right)^2+\binom{3}{1}\cdot(1)/(6)\right)`

Both binomial coefficients evaluate to 3, so we can simplify the expression as follows:

`\displaystyle\left((5)/(36)\right)^2\left(3\left((5)/(6)\right)^2+(1)/(2)\right)`

This approximates to 0.05, or 5%