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Suppose, as in this question, we cast a fair die until a 6 appears. We want the probability that it has to be cast more than 5 times

  • The answer given there (assuming independence of rolls) if $(5/6)^5$

I.E we want $P(\text{ no 6 in first 5 rolls})$.

How can I do this as $\frac{\text{ Number of sequences of 5 rolls with no 6}}{\text{ Number of sequences of 5 rolls}}$

Or, if I can't do it in that way, why can't I?

My attempt was: The number of sequences of $5$ rolls (assuming unordered) is $\binom{6+5-1}{5}$ (there are 6 numbers and we place 5 markers on them, which gives 5 walls + 5 markers =10 objects)

The number of sequences of 5 rolled with no 6: Now there are only 5 numbers to choose from so $\binom{5+5-1}{5}$

But this gives $$\frac{\binom{5+5-1}{5}}{\binom{6+5-1}{5}}=.5 \not= (5/6)^5$$

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    $\begingroup$ Sequences are ordered. The number of sequences of $5$ rolls is just $6^5$ since there are six possible outcomes for each roll. The number of favorable sequences is $5^5$ since there are five favorable outcomes (any number other than $6$) for each roll. $\endgroup$ – N. F. Taussig May 26 '18 at 18:28
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You can't "assume unordered" here. To understand why, start by thinking about the simpler version with two dice. The odds of getting no six is $$\frac{5}{6} \times \frac{5}{6}=\left(\frac{5}{6}\right)^2.$$

We can instead take $1-P(S)$ where $P(S)$ is the probability of getting at least one six. Picking any one die to list first, we can get a six with any of $(1,6),(2,6),(3,6),(4,6),(5,6),(6,6),(6,5),(6,4),(6,3),(6,2),(6,1).$ This is $$1 - \frac{11}{6^2} = \frac{36-11}{6^2} = \frac{25}{6^2} = \left(\frac{5}{6}\right)^2.$$

Both of these agree. So where does your method go wrong? If you look at unordered pairs with at least one six we get $(1,6),(2,6),(3,6),(4,6),(5,6),(6,6).$ There are only six of these, not eleven. In reality each of $(1,6)$ through $(5,6)$ are twice as likely as $(6,6)$ since they can occur in two ways, and not one.

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