Probability: When 3 dice are rolled but one of them counts as double

Question

Let's say I have three dice with numbers $1$ to $6$ representing six players. If we hit the number 5 with one die, this means that player 5 will get one dollar. However, the catch is that for one of the dice, the player will get two dollars instead of one. For example, if we roll a 1 and a 4 with the two simple dice and a 1 with the double die, the result is that player 1 gets three dollars, while player 4 gets one dollar. The others get nothing.

My question is now, what is the probability that a player will earn:

• No dollar
• 1 dollar
• 2 dollars
• 3 dollars
• 4 dollars

after one roll of the three dice?

Answer 1

As you suggested, there are five possible outcomes. Listing the probabilities per die (simple, simple, double), we arrive at the following:

1. No dollar, with probability $\frac{5}{6}\frac{5}{6}\frac{5}{6} = \frac{125}{216}$

2. One dollar, with probability $\frac{1}{6}\frac{5}{6}\frac{5}{6} + \frac{5}{6}\frac{1}{6}\frac{5}{6} = \frac{50}{216}$

3. Two dollars, with probability $\frac{1}{6}\frac{1}{6}\frac{5}{6} + \frac{5}{6}\frac{5}{6}\frac{1}{6} = \frac{30}{216}$

4. Three dollars, with probability $\frac{1}{6}\frac{5}{6}\frac{1}{6} + \frac{5}{6}\frac{1}{6}\frac{1}{6} = \frac{10}{216}$

5. Four dollars, with probability $\frac{1}{6}\frac{1}{6}\frac{1}{6} = \frac{1}{216}$

Indeed, $\frac{125 + 50 + 30 + 10 + 1}{216} = 1$.

Answer 2

$\displaystyle P(\text{earns no dollar})=\left(\frac{5}{6}\right)^3=\frac{125}{216}$ (lose, lose, lose)

$\displaystyle P(\text{earns }1\text{ dollar})=\binom{2}{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)\left(\frac{5}{6}\right)=\frac{25}{108}$ (win, lose, lose) or (lose, win, lose)

$\displaystyle P(\text{earns }2\text{ dollars})=\left(\frac{1}{6}\right)^2\left(\frac{5}{6}\right)+\left(\frac{5}{6}\right)^2\left(\frac{1}{6}\right)=\frac{5}{36}$ (win, win, lose) or (lose, lose, win)

$\displaystyle P(\text{earns }3\text{ dollars})=\binom{2}{1}\left(\frac{1}{6}\right)\left(\frac{5}{6}\right)\left(\frac{1}{6}\right)=\frac{5}{108}$ (win, lose, win) or (lose, win, win)

$\displaystyle P(\text{earns }4\text{ dollars})=\left(\frac{1}{6}\right)^3=\frac{1}{216}$ (win, win, win)