# dice game

#### wtrow

##### New member
You have a single die. If you roll a 2 or 4 you win the game, if you roll a 6 you lose. Otherwise, keep rolling until you get a 3 or 6. If you get a 3 you win and a 6 you lose. What is the probability you win?

I know on the first roll the probability is 2/6 you win, 1/6 you lose, and 3/6 you continue. Onwards its 1/6 win, 1/6 loss, and 4/6 roll again. We've never done a continuous problem like this one so I'm completely lost.

#### Denis

##### Senior Member
1/3 + 1/2(1/2) = 1/3 + 1/4 = 7/12

#### soroban

##### Elite Member
Hello, wtrow!

You have a single die.
If you roll a 2 or 4 you win the game; if you roll a 6 you lose.
Otherwise, keep rolling until you get a 3 or 6.
If you get a 3, you win. if you roll a 6, you lose.
What is the probability you win?

I know on the first roll the probability is 2/6 you win, 1/6 you lose, and 3/6 you continue.
Onwards its 1/6 win, 1/6 loss, and 4/6 roll again. . Correct!

Let's baby-step through this . . .

Win on 1st roll
. . $$\displaystyle P(\text{win 1st roll}) \:=\:\tfrac{2}{6} \:=\:\tfrac{1}{3}$$

Win on 2nd roll
$$\displaystyle P(\text{cont. 1st roll}) \,=\,\tfrac{3}{6} \,=\,\tfrac{1}{2}$$
$$\displaystyle P(\text{win 2nd roll}) \,=\,\tfrac{1}{6}$$
. . $$\displaystyle P(\text{Win 2nd roll}) \:=\:\tfrac{1}{2}\cdot\tfrac{1}{6}$$

Win on 3rd roll
$$\displaystyle P(\text{cont. 1st roll}) \,=\,\tfrac{1}{2}$$
$$\displaystyle P(\text{cont. 2nd roll}) \,=\,\tfrac{2}{3}$$
$$\displaystyle P(\text{win 3rd roll}) \:=\:\tfrac{1}{6}$$
. . $$\displaystyle P(\text{Win 3rd roll}) \:=\:\tfrac{1}{2}\cdot\tfrac{2}{3}\cdot\tfrac{1}{6}$$

Win on 4th roll
$$\displaystyle P(\text{cont. 1st roll}) \,=\,\tfrac{1}{2}$$
$$\displaystyle P(\text{cont. 2nd roll}) \,=\,\tfrac{2}{3}$$
$$\displaystyle P(\text{cont. 3rd roll}) \,=\,\tfrac{2}{3}$$
$$\displaystyle P(\text{win 4th roll}) \,=\,\tfrac{1}{6}$$
. . $$\displaystyle P(\text{Win 4th roll}) \:=\:\tfrac{1}{2}\cdot\left(\tfrac{2}{3}\right)^2\cdot\tfrac{1}{6}$$

Do you see the pattern?

$$\displaystyle \text{We have: }\;P(\text{Win}) \;=\;\frac{1}{3} + \frac{1}{2}\!\cdot\!\frac{1}{6} + \frac{1}{2}\!\cdot\!\frac{2}{3}\!\cdot\!\frac{1}{6} + \frac{1}{2}\!\left(\frac{2}{3}\right)^2\!\frac{1}{6} +\frac{1}{2}\!\left(\frac{2}{3}\right)^3\!\frac{1}{6} + \cdots$$

. . . . . . . . . . . . . $$\displaystyle =\;\frac{1}{3} + \frac{1}{2}\!\cdot\!\frac{1}{6}\underbrace{\bigg[1 + \frac{2}{3} + \left(\frac{2}{3}\right)^2 + \left(\frac{2}{3}\right)^3 + \cdots \bigg]}_{\text{geometric series}}$$

$$\displaystyle \text{The sum of the geometric series is: }\;\frac{1}{1-\frac{2}{3}} \:=\:\frac{1}{\frac{1}{3}} \:=\:3$$

$$\displaystyle \text{Therefore: }\;P(\text{Win}) \;=\;\frac{1}{3} + \frac{1}{12}(3) \;=\;\frac{1}{3} + \frac{1}{4} \;=\;\frac{7}{12}$$

. . . . exactly as Denis promised!