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A dice is being rolled continuously. Suppose you roll a 3 ( or any number ). What is the probability that you will get the next "3" exactly after n rolls?

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    $\begingroup$ I suspect you mean "A die is being rolled continually". $\endgroup$ – Robert Israel Jul 28 '12 at 1:48
  • $\begingroup$ You are right.. I do mean continually. $\endgroup$ – Chip Jul 28 '12 at 7:48
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I will interpret "after $n$ rolls" as meaning that we want the probability that our next $n-1$ rolls do not yield a $3$, but the $n$-th roll does give a $3$..

The probability of a non-$3$ on any toss is $\frac{5}{6}$. So the probability of $n-1$ consecutive non-$3$'s followed by a $3$ is $$\left(\frac{5}{6}\right)^{n-1}\frac{1}{6}.$$

The reason that we simply multiply is that the results on successive tosses of the die are independent. The die does not remember what results it has come up with in the past.

We can also get the result by using a counting argument. Imagine tossing a die $n$ times, and recording the results. So a record of $4$ tosses might read $4,5,4,1$.

There are $6^n$ possible records of the results of $n$ tosses, all equally likely. There are $5^{n-1}$ records in which the first $n-1$ results are a non-$3$ and the last result is a $3$. So our probability is $$\frac{5^{n-1}}{6^n}.$$

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Well, if you do roll a $3$, then there would be a $1\over 6$ chance of rolling another $3$. So, for the next two rolls after rolling just once, if you roll two more times, then there will be a $1\over 36$ chance of rolling two $3$s. Here's a thing to help: Rolls to Get the Number | Chance the Number Will Appear Again Those Rolls$$2|{1\over6}$$$$3|{1\over 36}$$$$4|{1\over 216}$$$$5|{1\over1,296}$$

So, after one roll, the chance starts at the second roll and then multiplies by $1\over 6$ from the beginning as you go. This works for any number on a six-sided die.

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