Find the probability of getting two sixes in $5$ throws of a die. 
In an experiment, a fair die is rolled until two sixes are obtained in succession. What is the probability that the experiment will end in the fifth trial? 

My work:
The probability of not getting a $6$ in the first roll is $\frac{5}{6}$
Similarly for the second and third throw. Again the probability of getting a $6$ is fourth roll is $\frac{1}{6}$. So the probability of ending the game in the fifth roll is $\frac{5^3}{6^3}\times\frac{1}{6^2}=\frac{125}{6^5}$.
But the answer is not correct. Where is my mistake? Help please.
 A: Let $X$ denote any value between $1-5$, then the optional sequences are:


*

*$XXX66$

*$X6X66$

*$6XX66$


Calculate the probability of each sequence:


*

*$P(XXX66)=\frac56\cdot\frac56\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^3}{6^5}$

*$P(X6X66)=\frac56\cdot\frac16\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^2}{6^5}$

*$P(6XX66)=\frac16\cdot\frac56\cdot\frac56\cdot\frac16\cdot\frac16=\frac{5^2}{6^5}$


Add up the above probabilities:
$$\frac{5^3+5^2+5^2}{6^5}\approx2.25\%$$
A: The other solutions work perfectly for a modest number of rolls.  If you were interested in a larger number, you might find a recursion helpful.
Let $P(n)$ be the probability that your game ends in exactly $n$ rolls.  Thus your problem is asking for $P(5)$.  We note that $P(1)=0,\;P(2)=\frac 1{6^2}$. For $n>2$ we remark that the first roll is either a $6$ or it isn't.  If it is, then the second roll can't be a $6$.  That leads to the recursion $$P(n)=\frac 16\times \frac 56\times P(n-2)+\frac 56\times P(n-1)$$
Very easy to implement this. As a consistency check,  we quickly get $P(5)\sim 0.022505144$ which is in line with the direct solutions.
A: Here is simplified solution.
In order experiment to end at 5th trial, the last two rolls must be 6. So we have,
NNN66 $Pr=5^3/6^5$
6NN66  $Pr=5^2/6^5$
N6N66  $Pr=5^2/6^5$
$(N <6)$
Add all probabilities. 
PS. The last two roll must be (6,6), so we get $\frac {1}{6^2}$
The third roll can't be 6, so we get $\frac {5}{6}$
And first two rolls can be anything except  (6,6), so we get $\frac {35}{6^2}$
Answer is $\frac {175}{6^5}$
A: So both the fourth and the fifth rolls need to be sixes:
$$P(4^{th},5^{th}\mbox{ rolls are sixes}) = \frac{1}{6^2}$$
There are following possible combinations of rolls for the game to not end until the fifth roll:
$$(XXXOO), (XOXOO),(OXXOO)$$
where $X,O$ represent non-six and six, respectively.
Thus, the probability is:
$$P(\mbox{end in fifth trial}) = \frac{1}{6^2}\left( \frac{5^3}{6^3} +2 \cdot \frac{5^2}{6^2} \cdot \frac{1}{6} \right) = \frac{175}{6^5}$$
A: Condition A: The first two rolls musn't be both 6:
    $$A=1-(\frac{1}{6}\times\frac{1}{6})=\frac{35}{36}$$
Condition B: The third roll mustn't be 6:
    $$B=1-\frac{1}{6}=\frac{5}{6}$$
Condition C: The last two rolls must be 6:
    $$C=\frac{1}{6}\times\frac{1}{6}=\frac{1}{36}$$
Probability of A, B and C being true at the same time:
    $$A\bigcup B\bigcup C=A\times B\times C=\frac{175}{6^5}=\frac{175}{7776}$$
A: the odds of getting two sixes, in any order, within 6 dice (or is it die??) can be easily determined by using the binomial distribution
the odds would be 1/6 * 1/6 * 5/6 ^ 4
that opening statement very simply finds the chances of getting two sixes, but there is more than one way to get two sixes, so you need to multiply that value by
2!/6!(6-2)!
goes to 15..
so 15 * 1/6 * 6 * 5/6 * 4 = about 0.2009
so basically 1/5 
