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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?

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  • $\begingroup$ 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. $\endgroup$
    – Bloc97
    Jul 31, 2019 at 21:12

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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?

For these answers I will assume fair dice and independence outcomes on all dice. Under those conditions, the probability of a win (by some player) in each of the respective games is:

$$\begin{equation} \begin{aligned} \mathbb{P}(\text{Win in Game 1}) &= \frac{1}{6} = 0.1666667, \\[10pt] \mathbb{P}(\text{Win in Game 2}) &= 1 - \Big( 1-\frac{1}{6} \Big)^6 = \frac{31031}{46656} = 0.665102, \\[10pt] \mathbb{P}(\text{Win in Game 3}) &= 1 - \Big( 1-\frac{1}{36} \Big)^6 = \frac{338516711}{2176782336} = 0.1555124. \\[10pt] \end{aligned} \end{equation}$$

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.

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