Conditional Probability problem on Dice

Question

A fair dice is rolled four times. What is the probability of obtaining two or more 6's given that no two successive outcomes were the same?

This is quite a straightforward problem, that I thought of myself, and want corroboration on my answer to it.

Is my thinking right, if I say that considering the four outcomes numbered $1, 2, 3, 4$, there are exactly three ways to get two $6$'s: $(1, 3), (1, 4), (2, 4)$ and the probability of getting $6$ any of these outcomes, i.e. involving two $6$'s at these numbered outcomes and not a $6$ at the others, is $\frac{1}{6} \cdot \frac{1}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} = \frac{25}{1296}$. Hence, the total probability is $ \frac{3 \cdot 25}{1296} = \frac{25}{432}$

Please provide some inputs if I am correctly aligned.

Answer 1

It says given no two successive outcomes are same. So $6^4$ is not the correct sample space. It should be $6 \cdot 5^3$.

Also for the favorable outcomes, when you have positions of two sixes with a single gap, that is positions - 1 and 3 or 2 and 4, there are $2 \cdot 5\cdot 5$ favorable outcomes. But when the positions of sixes are 1 and 4, there are only $5 \cdot 4$ favorable outcomes.

So the answer should be $ \ \displaystyle \frac {2\cdot5\cdot5 + 5\cdot4}{750} = \frac{7}{75}$

Answer 2

If no two successive rolls are the same, then you can only get at most $2$ sixes in $4$ rolls:

• first and third rolls with probability $\frac16 \times \frac55 \times \frac15 \times \frac55$
• first and fourth rolls with probability $\frac16 \times \frac55 \times \frac45 \times \frac15$
• second and fourth rolls with probability $\frac56 \times \frac15 \times \frac55 \times \frac15$

Add these up to get $\frac7 {75}$