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The Question:

Suppose that we keep throwing a fair dice n times. Also, suppose that all throws are independent and the dice is fair, i.e. 1, 2, 3, 4, 5, and 6 have equal probabilities to show up in each throw. Compute the probability that

  1. we have the number 6 showing up in at least k throws;

  2. we have the number 6 showing up in at least k consecutive throws.


My work:

I believed 1. was pretty easy. Simply realize the probability of not rolling a 6 in any one throw is 5/6. Then since each roll is independent so multiply this to the power of k, which gives the probability of not rolling a 6 in k throws. Then subtract that product from 1 to get the probability of rolling a 6 in at least k throws. So the answer is $1-(5/6)^k$.

I am having trouble with 2. though. The question is asking for the probability of 6 showing up in k back to back throws of the n times the die is thrown. Since each die toss is independent I feel like the process and answer for finding 2. should be the same as 1. but this is obviously not the case.

So did I do 1. wrong and how should I proceed with 2.?

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  • $\begingroup$ The answer to 1 has to depend on $n$. For example, if $n=2,k=3$ the probability is zero as we don't have enough throws. $\endgroup$ – Ross Millikan Aug 5 '16 at 20:03
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There are several problems with your solution to 1. You are essentially picking $k$ throws to ask for a $6$ and ignoring the rest of the $n-k$ throws, which might give you a $6$. Then your figure $(5/6)^k$ is the chance that all those throws are non-sixes, so your figure $1-(5/6)^k$ is the chance of at least one six among the specific $k$ throws you picked.

What you have to do is pick a number of sixes to get, compute the probability of that many sixes, and sum over the allowable numbers of sixes. To get exactly $j$ sixes, you choose the $j$ rolls out of the $n$ (how many ways?), require that those are all sixes, and require that all the rest are non-sixes. Then sum these from $j=k$ to $j=n$

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  • $\begingroup$ I understand 1. now. I was misinterpreting the question to the probability of 6 showing up at all in a collective number of tosses, which was incorrect. I'm still stuck on 2. though so any help on that would be apprected. $\endgroup$ – kdrozd Aug 5 '16 at 22:14
  • $\begingroup$ An approximate approach for 2 that works as long as the probability is small is that you have $n-k+1$ places the run of sixes can start. The chance of each one is $\frac 1{6^k}$, so you have $(n-k+1)frac 1{6^k}$. This double counts events with two runs of $k$ sixes or those with a run of $k+1$, so the probability will be smaller than this. You can use inclusion/exclusion to do better. $\endgroup$ – Ross Millikan Aug 5 '16 at 23:30

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