# What is the probability that you throw four times a six within ten throws?

I have a probability problem which i cannot solve yet. I want to the manuals says that the correct answers is:

$${9 \choose3}\times \left(\frac 56\right)^6 \times\left(\frac16\right)^4$$

I understand the $$\left(\frac 56\right)^6 \times\left(\frac16\right)^4$$ part. Assume that the probability of success is $$\frac 16$$ and the probability of failure is $$\frac 56$$.

What I don't understand is the use of the term $$9\choose3$$. The first term is to take all the possible variations in sequence in account.

$$\frac{n!}{k!*(n-k)!}$$

But you throw $$10$$ times, so shouldn't $$n$$ be $$10$$? The times of success is $$4$$, so why isn't $$k = 4$$?

In my opinion the solutions should be:

$${10\choose 4}\times \left(\frac 56\right)^6 \times\left(\frac16\right)^4$$

But this is not correct.

Could somebody give me feedback on what is wrong with my thinking?

• The last throw is assumed to be a 4. Hence, you pick the remaining 3 throws out of the preceding 9 throws. Mar 6, 2020 at 12:32
• It is the answer on the question: "what is the probability that exactly 10 throws are needed if it is your aim to throw 4 times a six?" I wouldn't say that your thinking is wrong but rather that the question is not well posed. Mar 6, 2020 at 12:41
• It is clear now! Thank you for your effort :)
– Tim
Mar 6, 2020 at 12:43
• Can you give type up the exact wording of the problem from wherever you got it? Based on the wording that you currently have, your thought process is right. Mar 6, 2020 at 14:09
• Hi, the exact problem statement is: Imagine throwing a die until you have thrown 6 for the fourth time. calculate the chance that you will succeed in 10 times. I think the problem was my in the understanding of the problem. Thanks for your effort! Ter
– Tim
Mar 7, 2020 at 15:17