Number of throws likely to have occurred the most times to get a 6 This question has me stumped and I've been trying to figure out how to answer it for a while now - was hoping someone here could share some insight!
'The probability of getting a 6 on any throw of a fair die is 1/6. A player throws the die as many times as necessary to get the first six and notes the number of throws needed. The player repeats this experiment many times, on each occasion noting the number of throws needed. 
What number of throws is likely to have occurred the most times?'
 A: The Negative Binomial Distribution describes the number of trials ran with probability $p$ until number $r$ of successes are found. You want the most likely outcome, which would be the mode of the distribution (Note that this is not the same as saying "the average number of attempts required" or anything like that). In your experiment, you have $p = 5/6$ (probability of not getting a 6) and $r = 1$ (number of 6's required to finish).
As you can see in the Wikipedia link, the most likely outcome is 0 experiments until termination -- i.e. stopping on the very first throw. This makes sense if you think about it: the probability of stopping on the $n^{th}$ throw is the combined probability of (1) getting a 6 on that throw ( = $1/6$), and (2) getting not a 6 on every throw before ( = $(5/6) ^n$). This is obviously strictly less likely than simply getting a 6 on the first throw, which is just simply $1/6$.
If you want the average number of trials required, this is the expected value. Wikipedia tells us that the Negative Binomial expected value = $\frac{pr}{1-p} = \frac{5/6}{1 - 5/6} = \frac{5/6}{1/6}  = 5$, indicating that we get a 6 on the $6^{th}$ throw.
