Question

Tammy is playing a game where she is tryipg to roll a three with a standard die. If she gets a three in any of her first 4 rolls, she wins; otherwise she loses. What is the probability that Tammy wins the game? Round your answer to the nearest tenth of a percent.

Answer 1

We can partition the events as follows:

$P(\text{win}) = P(\text{win at round 1})+P(\text{win at round 2})+P(\text{win at round 3})+P(\text{win at round 4})$

Let's compute each term.

Winning at round 1

To win with the first die, Tammy must roll a 3. This happens with probability 1/6. So, we have

$P(\text{win at round 1})=\dfrac{1}{6}$

Winning at round 2

To win with the second die, Tammy must not roll a 3 with the first roll (probability 5/6), and then roll a 3 with the second roll (probability 1/6). So, we have

$P(\text{win at round 2})=\dfrac{5}{6}\cdot\dfrac{1}{6}$

Winning at round 3

To win with the third die, Tammy must not roll a 3 with the first two rolls (probability 5/6), and then roll a 3 with the third roll (probability 1/6). So, we have

$P(\text{win at round 3})=\dfrac{5}{6}\cdot\dfrac{5}{6}\cdot\dfrac{1}{6}$

Winning at round 4

To win with the fourth die, Tammy must not roll a 3 with the first three rolls (probability 5/6), and then roll a 3 with the fourth roll (probability 1/6). So, we have

$P(\text{win at round 4})=\dfrac{5}{6}\cdot\dfrac{5}{6}\cdot\dfrac{5}{6}\cdot\dfrac{1}{6}$

So, we have

$\displaystyle P(\text{win}) = \dfrac{1}{6}+\dfrac{1}{6}\cdot\dfrac{5}{6}+\dfrac{1}{6}\cdot\left(\dfrac{5}{6}\right)^2+\dfrac{1}{6}\cdot\left(\dfrac{5}{6}\right)^3 = \dfrac{1}{6}\sum_{i=0}^3 \left(\dfrac{5}{6}\right)^i\approx 0.5$

Note

You can do this exercise more quickly by observing

$P(\text{win}) = 1-P(\text{lose})$

And you lose if you never roll a 3 in 4 rolls (each failure has a probability of 5/6). In fact, you have

$1-\left(\dfrac{5}{6}\right)^4 \approx 0.5$