# Two different dice games. What are the odds the game will be won?

Based on the images above, is it correct that the first game should be won 1 out of six times, the second game every time, and the 3rd game 1 out of six times?

• A quick glance at the problem hints at using the Binomial distribution to compute the cumulative probability. Hint: 100% Probability does not exist in a binomial distribution. Commented Jul 31, 2019 at 21:12

\begin{aligned} \mathbb{P}(\text{Win in Game 1}) &= \frac{1}{6} = 0.1666667, \\[10pt] \mathbb{P}(\text{Win in Game 2}) &= 1 - \Big( 1-\frac{1}{6} \Big)^6 = \frac{31031}{46656} = 0.665102, \\[10pt] \mathbb{P}(\text{Win in Game 3}) &= 1 - \Big( 1-\frac{1}{36} \Big)^6 = \frac{338516711}{2176782336} = 0.1555124. \\[10pt] \end{aligned}
So as you can see, the first game is win-probability of exactly $$1/6$$, the third game has approximately (but not exactly) this win probability, and the second game has a win-probability of about $$2/3$$. This is consistent with the description in the image.