1. ## what is certain?

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!

2. ## Re: what is certain?

b) The player does not record a score in his first 5 throws means that the first four throws were not a six. But, the fifth throw could be a six, but it may not be. It fact, the fifth throw could be anything. Part (a) specifically says that he DOES record a score for his 4th throw (so his 3rd throw must be a 6). Part (b) says that he does NOT record a score for the first 5 throws, but does not say if he records a score for his 6th throw or not. He might, he might not. We do not know.

3. ## Re: what is certain?

Originally Posted by s_ingram
b) if the player does not score in his first 5 throws, then he scores on his 6th throw.
this is your error. There is nothing in (b) that implies they start scoring on the 6th throw. It just says they don't score in the first 5. They might not score on 6 or 7 or 1000 either.

4. ## Re: what is certain?

Thanks guys, I see what you mean!

5. ## Re: what is certain?

You can use the geometric distribution to answer both parts of your problem, though I recommend the binomial distribution for part B. The geometric distribution models the number of trials it takes for the first "success" assuming the probability of a success is the same for each trial. The probability that the $k$th trial is the first success is given by the geometric probability mass function $P(N=k)=p(1-p)^k$, where $p$ is the independent probability of a success and $k = 1, 2, \ldots$.

In the context of your problem, each roll is a trial and a roll of 6 is a success. $p$ is the probability of rolling a 6, which is $\frac{1}{6}$.

For part B, you want the probability of not rolling a 6 on your first four rolls. According to the binomial distribution, the probability of $k$ successes in $n$ trials is $\binom{n}{k} p^k (1-p)^{n-k}$. Your job here is to find the binomial probability of 0 successes in 4 trials.
Using the geometric distribution, the probability that a 6 was not rolled on the first four rolls is probability that you roll your first 6 on roll number 5 or higher. So your answer is $P(N=5)+P(N=6)+\ldots$, which is the same as $1-\left[P(N=1)+P(N=2)+P(N=3)+P(N=4)\right] = 1 - P(N=1) - P(N=2) - P(N=3) - P(N=4)$.