What is the probability of a run of at least 3 sixes when a die is thrown 5 times?
I think I have the answer but from what I have been told its not the correct answer. Would someone like to help?
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1Could you give your answer and the reasoning behind it? – hmakholm left over Monica Aug 25 '13 at 10:06
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@HenningMakholm Thanks to your comment below about ignoring a sequence, I've managed to reconcile my answer with the actual answer. I'm going to update it once everyone has had their shot. – Innocent Aug 25 '13 at 11:01
3 Answers
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Presumably a run means consecutive.
The run might begin with the first toss. This has probability $(1/6)^3$.
The run might begin with the second toss. This has probability $(5/6)(1/6)^3$.
The run might begin with the third toss. This also has probability $(5/6)(1/6)^3$.
Add these three numbers.
André Nicolas
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Should we not separate the following cases: 1. we get first 3 as sixes and then some other number. 2. first four sixes and then some other number. 3. first five sixes? – Innocent Aug 25 '13 at 10:18
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There is no need to. If you do it correctly, you will still get the right answer. Of course you must not forget about runs that start later like $2666$. My answer gets them all. – André Nicolas Aug 25 '13 at 10:24
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If you can change your answer, I'm going to accept it as a correct one. – Innocent Aug 25 '13 at 11:35
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;@InnocentRetard You are right, starting at $3$ is also $(5/6)(1/6)^3$. – André Nicolas Aug 25 '13 at 15:19
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I simply count all the cases:
Possibilities for three consecutive $6$s but no four consecutive $6$s:
The sequence be of the one of these three forms:
$ 666XY , X666X , YX666$
where the $X$'s can be chosen from $1,\dots ,5$ and the $Y$'s are also allowed to be $6$. So we get
$5\cdot6 + 5\cdot5 + 5\cdot6$
possibilities for this case.
Possibilities for four consective digits:
$6666X$ and $X6666$
So those are $5 + 5 $ possibilites.
And $1$ case where all the dice are $6$s.
So the probability is
$$p= \frac{5\cdot6 + 5\cdot5 + 5\cdot6 + 5 +5 +1}{6^5} = \frac{16}{6^4} = \frac{1}{3^4} $$
Brusko651
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Forget about 6 and just look for the probability of at least three consecutives. In the sequel s stands for 'toss has same result as preceding toss', d stands for 'toss has different result as preceding toss' We start with the second toss. Possibilities are:
ss with probability (1/6)x(1/6)
sdss with probability (1/6)x(5/6)x(1/6)x(1/6)
dss with probability (5/6)x(1/6)x(1/6)
ddss with probability (5/6)x(5/6)x(1/6)x(1/6)
summation gives 2/27. If you do this for 6 (or any other) specifically you get (2/27)x(1/6)=1/81. At this place you will find a formula for this.
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1@Kumar If e.g. the first toss is a $4$ then $\textbf{ss}$ stands for event $444\cdot\cdot$, and $\textbf{sdss}$ stands for $44kkk$ where $k\neq 4$, $\textbf{dss}$ stands for $4kkk\cdot$ and $\textbf{ddss}$ stands for $4klll$ where $k\neq4$ and $l\neq k$. So all distinct possibilities of occuring $3$ consecutives within $5$ tosses are captured so that summation of these probabilities gives the probability of occuring $3$ consecutives. Do you agree with that? – drhab Nov 03 '19 at 13:36
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Yes. I am fine. but I guess this example of yours should be in the answer. It would be slightly more easy for the reader to understand and comprehend.:) – Kumar Nov 03 '19 at 18:49
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@Kumar On this I agree with you. This answer is not very "OP-friendly" and an example would have improved that. – drhab Nov 04 '19 at 10:08