What is the probability of getting $6$ $K$ times in a row when rolling a dice N times?
I thought it's $(1/6)^k*(5/6)^{n-k}$ and that times $N-K+1$ since there are $N-K+1$ ways to place an array of consecutive elements to $N$ places.
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It it at least $k$-times or exactly $k$ times? What you've calculated is the probability that $6$ occurs exactly $k$ times, but nothing about being in a row. – Michael Burr Jun 25 '18 at 09:59
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At least k times. I thought that if I multiply this with N-K+1 that's the probability they would be in order? – Lazar Šćekić Jun 25 '18 at 10:01
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But the $(5/6)$ says that all the other positions are not $6$'s. This is not what you want either. – Michael Burr Jun 25 '18 at 10:01
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Oh right, that's what I forgot about. – Lazar Šćekić Jun 25 '18 at 10:03
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That is the probability of getting 6 K times in a row and no other 6s. But what about 6-6-5-6, for example - should that count as two 6s in a row ? Or 6-6-6-5 - does that count as two 6s in a row, even though there are actually three 6s in a row ? – gandalf61 Jun 25 '18 at 10:05
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@MichaelBurr, so I should just instead of that (5/6)^(n-k) put all the cases up to Nth? – Lazar Šćekić Jun 25 '18 at 10:06
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@gandalf61 no, just one array of 6s, and it's said at least k times. – Lazar Šćekić Jun 25 '18 at 10:30
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So what I think is that the question says that we have to choose a set of 6 places from N given places, so it's just N-6 ways. – Sri Jun 25 '18 at 10:35
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@Sri not really, since you need to get "6" k-times. – Lazar Šćekić Jun 25 '18 at 10:37
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Oh I am sorry it's N-k+1 ways then – Sri Jun 25 '18 at 10:43
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Is that the correct answer?? – Sri Jun 25 '18 at 10:43
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This question might help to solve this. – drhab Jun 25 '18 at 11:43
3 Answers
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In the following for convenience I write $p$ for $\frac{1}{6}$ and $q$ for $\frac{5}{6}$
Let $A_{K,N}$ denote the event that among the first $N$ throws there is a consecutive row of $K$ sixes.
Let it be that the $F$-th throw is the first throw that does not give a six, so that $P\left(F=r\right)=p^{r-1}q$.
Then $P\left(A_{K,N}\mid F=r\right)=1$ if $r>K$ and $P\left(A_{K,N}\mid F=r\right)=P\left(A_{K,N-r}\right)$ otherwise.
Further $P\left(A_{K,N-r}\right)=0$ if $K>N-r$.
So for $K\leq N$ we find:
$$P\left(A_{K,N}\right)=\sum_{r=1}^{\infty}P\left(A_{K,N}\mid F=r\right)P\left(F=r\right)=\sum_{r=1}^{K}P\left(A_{K,N-r}\right)p^{r-1}q+p^{K}$$
Based on this equality with induction it can be proved that:
$$P\left(A_{K,N}\right)=p^{K}\left(1+\left(N-K\right)q\right)\text{ for } N\in\left\{ K,K+1,\dots,2K\right\} $$(which is not valid for $N>2K$).
See also this question that was posed by me once at a time.
drhab
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This is really just a long comment, but I thought that it would be a mess to write it as a comment:
Please clarify your question with, at least, the following data (some of the answers are in the comments):
Does the run of $6$'s need to be the only $6$'s in the string? If $k=2$, would $6656$ be allowable?
Can the run of $6$'s be longer than $k$? If $k=2$, would $6665$ be counted?
Can you have more than one run of $6$'s in the string? If $k=2$, would $66566$ be allowable?
Michael Burr
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N is smaller than 2*K, so there can't be two sequences, and only sequences with consecutive "6" are considered as a success. I think I'm on the edge of the solution, so I'll write it down here as soon as I think I've got it. – Lazar Šćekić Jun 25 '18 at 11:07
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In this case, please add the assumption that $N<2K$ to your problem statement, so that you have a complete problem statement. – Michael Burr Jun 25 '18 at 19:10
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So, let's say that X is a random variable for tracing the number of 6s. Let's say that C is the condition for the 6s to be consecutive. Since the task is to find the probability of gaining at least K subsequent 6s, we can look for probability of event A-at least K subsequent 6s fell this way: P(A)=P(X=K|C)+P(X=K+1|C)+...+P(X=K+N|C)+..., where P(X=K|C) is the probability of K 6s falling under the condition that they are consecutive. Since the event that K 6s have fallen is independent from the event that all the 6s are in order, we can say that: P(X=K|C)=P(X=K)*P(C). Is this the right way to go?
Edit: Doesn't seem right because the events ARE independent, because K 6s need to fall in order for them to be consecutive.
Lazar Šćekić
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