We throw a die $10$ times. What is the probability of having the number 6 appear for the first time at the $6$th throw?

I have tried $$\frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{5}{6} \cdot \frac{1}{6} \cdot (6!-1)$$

but it's wrong.

$(6!-1)$ is the expression to say that there is only 1 way to order the 6
Should I consider the last 4 throws ? any help ?

What is the probability that the first $3$ throws are distinct and the $4$th will be equal to one of the first $3$ ones?

Here I did : $$\frac{6}{6} \cdot \frac{5}{6} \cdot \frac {4}{6} \cdot \frac{1}{2}$$
I don't know if it's correct.

  • 2
    $\begingroup$ "Should I consider the last 4 throws" No, because these results do not have any impact on the first 6 throws. I don´t understand why you have added the factor $(6!-1)$? What is the reasoning? $\endgroup$ – callculus Oct 2 '17 at 14:07
  • $\begingroup$ You can always check your result if it fullfills the properties. For instance, $(5/6)^{5}\cdot 1/6\cdot (6!-1)=48.16$ cannot be right since $0\leq p\leq 1$. See the axioms of probability here: en.wikipedia.org/wiki/Probability_axioms $\endgroup$ – callculus Oct 2 '17 at 14:33
  • $\begingroup$ But if one the first throw i have a 6 o the 7th throw, that means that I have fail 1 more time. So the number of total outputs is greater no? $\endgroup$ – Romain B. Oct 2 '17 at 15:10
  • $\begingroup$ Why do you want to regard the fail cases? In general you have used probabilities and not the favorable and possible outcomes. If you would use the outcomes, the number of favorable outcome would be $5^5\cdot 1$. And the number of possible outcome would be $6^6$. $\endgroup$ – callculus Oct 2 '17 at 16:07
  • $\begingroup$ why not 6^10 in total possible outcomes ? $\endgroup$ – Romain B. Oct 2 '17 at 17:42
  1. For your first question: $$\left(\frac56\right)^5\cdot\frac16$$ You want to get anything but six on the first five rolls and six on the sixth roll. After that, you do not care whether you get a six or not, so the probability of succeeding is one after the sixth throw. You don't need to use the factorial at all in this question.

    If you really want to consider the four dice rolls after you got your first six on the sixth roll, you can look at this similar problem with coin flips (There are a lot fewer outcomes, which is easier to work with.)

    Say you have an unfair coin so that you have a $\frac13$ chance of getting tails and a $\frac23$ chance of getting heads. What is the probability that you get your first tail on the fourth try out of six coin flips?

    There are $2^6$ possible outcomes, but you only have the favorable outcomes of $HHHTHH$, $HHHTHT$, $HHHTTH$, and $HHHTTT$. Now, the probability of any of these happening is the sum of their probabilities since these are disjoint events. Our probability is therefore $$\left(\left(\frac23\right)^3\cdot\frac13\cdot\left(\frac23\right)^2\right)+ \left(\left(\frac23\right)^3\cdot\frac13\cdot\frac23\cdot\frac13\right)+\left(\left(\frac23\right)^3\cdot\frac13\cdot\frac13\cdot\frac23\right)+\left( \left(\frac23\right)^3\cdot\frac13\cdot\left(\frac13\right)^2\right)$$ We can factor out the $\left(\frac23\right)^3\cdot\frac13$ from all of these, which is why I wrote the probabilities in that way. This leaves us with $$\left(\frac23\right)^3\cdot\frac13\cdot\left(\left(\frac23\right)^2+2\cdot\frac23\cdot\frac13+\left(\frac13\right)^2\right)$$ You should notice that the stuff in parentheses is equal to $\left(\frac23+\frac13\right)^2=1^2=1$, leving us with $\left(\frac23\right)^3\cdot\frac13$, which is what would happen if you only looked at the first four coin flips. This should make sense for several reasons, chief among them being the coin flips after the fourth coin flip have no effect on the coin flips before them. Likewise, the probability of getting your first six on the sixth dice roll after getting your first six on the sixth roll is one.

  2. You answered your second question correctly.

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