Is it highly unusual to win a lottery with 10 tickets in 6 weeks out of 7? A woman in my local Rotary Club has won our weekly drawing 6 out of 7 weeks and I wanted to know the actual odds of this happening.  
On average 10 tickets are bought which I know makes her daily odds $1/10$. How to find the odds to win 6 out of 7?
 A: I suggest that the more meaningful probability is her chance of winning at least 6 out of the 7 drawings, not exactly 6.  Under the null hypothesis that the chance of winning any particular drawing is $\theta = 0.10$, the $p$-value of the result you observed is $$\Pr[X \ge 6 \mid H_0] = \binom{7}{6}\theta^6 (1-\theta) + \binom{7}{7}\theta^7 = \frac{1}{156250} \approx 6.4 \times 10^{-6}.$$  Still quite small, of course.
However, the above calculation presumes that exactly $10$ tickets were eligible for each drawing, but if this is only an average, then there is insufficient information from which to calculate a $p$-value.  If in some weeks the number of tickets was very small (e.g., only $2$ people entered one week, but $18$ people entered in another), then the $p$-value could be as large as $\approx 0.000198611$.  In the most extreme case, suppose for $6$ of the $7$ drawings, only $2$ people participated, and in the remaining drawing, $58$ participated, making the average still $10$.  The probability of winning at least $6$ of the drawings under the assumption that the drawings are fair, is $$\left(\tfrac{1}{2}\right)^6 \left(\tfrac{57}{58}\right) + 6\left(\tfrac{1}{2}\right)^5 \left(1 - \tfrac{1}{2}\right) \left(\tfrac{1}{58}\right) + \left(\tfrac{1}{2}\right)^6 \left(\tfrac{1}{58}\right) = \frac{5741}{338256} \approx 0.0169724,$$ or only about $1.7\%$.  This represents an absolute upper bound on the probability of observing such an outcome.  (I have made the assumption that there is no point to conducting a drawing if there is only one participant.)  That is still very unlikely, which would strongly suggest some kind of collusion taking place.
A: Most of the calculations assume that she was a person of interest BEFORE any of the draws, and there was a reason to be interested in those 7 draws. The question does not say that she was of special interest before, so those calculations are wrong. 
To give an example: if there are 10 tickets sold every week, and a person wins 2 weeks in a row, the probability is 1/100 for that person - unusual. However the probably of it happening to anyone in any given week is only 1/10. If this is a weekly draw happening for 2 years, it would be likely for this to have happened 10 times in the last 2 years, and 3 in a row should happen once. 
So to find the probability of your example happening randomly, you need to multiply the probability of a particular example happening by the number of ways it might happen. Every player is an extra chance, and every historical draw is an extra chance. You need to multiply the calculated probabilities by the number of people playing regularly (so times 10), but more significantly by the number of draws that there have been historically. Have there been 100 draws, 1000 draws? 
With 10 players and 100 draws, you need to multiply the probability calculated for her at this time by 1000. This makes the likelihood of this happening still unlikely, but not unlikely enough to be grounds for accusations of collusion.
If you then multiply this by the number of similar draws that happen in workplaces and clubs in just your city, it is bound to happen to someone sooner or later. 
Another method is to say that when she wins 3 in a row (should happen once every 100 draws), she becomes a person of interest, and you can calculate any further probabilities after that point. If her performance from this point remains highly unlikely (p<1%), maybe something is happening. 
As another answer noted, if there are multiple tickets bought by a single person, the odds can also quickly change, so make sure you take into account all the factors. 
A: If exactly $10$ tickets are sold each week and she buys one, the probability of winning $6$ of the next $7$ drawings is $\displaystyle {7 \choose 6}{\left( 1 \over 10\right)}^6{\left(9\over 10\right)}^1 = 6.3 \times 10^{-6}$ or roughly $1$ in $158,730$.
A: The odds are: 
$
\begin{pmatrix} 
7 \\
6 \\
\end{pmatrix} (\frac{1}{10})^6(\frac{9}{10}) = 6.3 \cdot 10^{-6} = 0.00063 \%$
Indeed low
