probability question (dice) Suppose you toss three fair six-sided dice simultaneously until you observe a six. So if none of the three
  dice are a six, then you toss them simultaneously again. Find the probability that the first occurrence
  of a six requires n tosses. 


Since P(a six first toss) =$1/6+1/6+1/6=1/2$ shouldn't it be $n/2$ after $n$ tosses?
The probability of getting a six on the first toss is not $1/2$. If the reasoning that led you that conclusion were valid, you could argue that with six dice you’d be certain to get a six on the first toss, which clearly isn’t the case.

The easiest way to calculate the correct probability of getting a six on the first toss is to calculate the probability of not getting a six and subtract that from $1$. In order not to get a six, you must roll a not-$6$ on each of the three dice. The probability of rolling a not-$6$ on one die is $5/6$, and the dice rolls are independent, so the probability of rolling a not-$6$ on all three dice is $(5/6)(5/6)(5/6)=125/216$, and the probability of getting at least one six is therefore $1-125/216=91/216$, or a little over $0.42$.

Now in order to get a six for the first time on the $n$-th toss, you must get no sixes on each of the first $n-1$ tosses, and then you must get at least one six on the $n$-th toss.  The probability of getting no six on one toss is, as we just saw, $125/216$, so the probability of getting no six on each of $n-1$ tosses is $$\left(\frac{125}{216}\right)^{n-1}\;.$$ This needs to be followed by a successful toss, an event whose probability is $91/216$, so the probability of getting your first six on the $n$-th toss is $$\left(\frac{125}{216}\right)^{n-1}\left(\frac{91}{216}\right)=\frac{91\cdot125^{n-1}}{216^n}\;.$$
The first toss:

$$P(\text{no 6}) = \left(\frac{5}{6}\right)^3 = \frac{125}{216}$$

We generalize. The chance you throw a six after $n$ turns is:

$$P (\text{six in }n\text{th throw}) = \left(\frac{125}{216}\right)^{n-1} \frac{91}{216} = \frac{91 \cdot 125^{n-1}}{216^n}$$