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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?

Joe Roberts
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    If each drawing had exactly 10 tickets her chance of winning 6 out of 7 times would be 0.00063%. If just the average number of tickets is 10, we don't have enough information to determine the probability. – Reinstate Monica Feb 03 '15 at 21:00
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    Assuming you are talking about a real event: Can she buy more than one ticket? Every time she bought two tickets, she would have almost doubled her chance. – gnasher729 Feb 03 '15 at 21:23
  • @gnasher729 She would have exactly doubled her chance on each drawing, assuming the total number of tickets bought is fixed (that is, in addition to her buying one more ticket, someone else must have bought one fewer). And the cumulative effect would be a 57-fold increase in her chance of winning 6 out of 7 times, a huge increase but still highly unlikely. – Reinstate Monica Feb 03 '15 at 21:54
  • If this is real, be careful before making accusations. You need to take into account how many chances there have been for it to happen, which would likely be at leat 1000 chances. Combined with other effects like multiple ticket buying, it is very possible that this is not actually unusual. – Richard Feb 03 '15 at 23:09
  • You've told us that 10 tickets are bought each week, but you can't be sure that she only buys one. Perhaps she buys 9 each week or 10. So you don't really know her odds each day are 1 in 10, unless there's something else you didn't mention. What did she win? Perhaps it is fair considering she might be paying for the majority of the tickets. – Octopus Feb 04 '15 at 01:43
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    In addition to the factors that other people have pointed out, the answer depends on what you mean by "this happening." Do you mean that particular woman winning 6 out of the last 7 drawings? Or do you mean some person winning 6 out of 7 consecutive drawings sometime in the history of the rotary club? Or something in between? (Edit: I think this is what Richard is getting at.) – Trevor Wilson Feb 04 '15 at 01:59

4 Answers4

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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.

heropup
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    I wouldn't say that 1% is highly unlikely. Assuming there are hundreds of potential 1% events, it is extremely likely that several of them would indeed happen. – Yuval Filmus Feb 03 '15 at 22:40
  • @YuvalFilmus Yes, but that is under some extremely generous assumptions, far more than what the "average of 10 participants per drawing" would imply. It basically amounts to two participants per drawing. Under more reasonable assumptions, the $p$-value is extremely small. It isn't zero, but certainly it is small enough to suggest that reasons other than pure luck are involved. – heropup Feb 03 '15 at 22:44
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    Before suggesting wrongdoing, you need to multiply your probability by the number of chances there have been for this to happen to someone at sometime. Could easily be 1000 such chances with just this Rotary Club's raffle. You need to be very careful before suggesting collusion. See my answer for more details. – Richard Feb 03 '15 at 23:04
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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.

Richard
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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 $158,730$.

Simon S
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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

FlyingDutchman
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