this question has me puzzled:

in order to start scoring you must roll a 6 on your dice. After this, subsequent throws count towards your score.

So, 2,4,3,6,4,6,2,5 gives recorded scores of 4,6,2,5.

Calculate the probability that

a) the first score recorded is that of the player's fourth throw

b) the player does not record a score in his first 5 throws.

a) P(throw 6) = 1/6. P(no six) = 5/6

this means no 6, no 6, 6....

P(4th throw) = $\frac{5}{6} \times \frac{5}{6} \times \frac{1}{6} = \frac{25}{216}$

b) if the player does not score in his first 5 throws, then he scores on his 6th throw. This means his 5th throw was a 6.

P(no score on first 5 throws) means: no 6, no 6, no 6, no 6, 6 .... = $(\frac{5}{6}) ^4 \times \frac{1}{6} = \frac{5^4}{6^5} = \frac{625}{7776}$

Now this is the wrong answer. The correct answer is $ \frac{5^4}{6^4} = \frac{625}{1296} $ because the question assumes that the 6 on the 5th roll is a certainty if we are to score on the 6th throw (or not score on the first 5). But, given the result from part a) where we include the probability of rolling the 6, it seems strange that in part b) we do not include the probability of rolling the 6.

Perhaps you will tell me that this is more a question for the English grammar forum and you would have a point! None the less, I am believing that there is a mathematical explanation!