Odds of the office raffle being rigged Hello mathemagicians,

We have this drawing for free tickets at my workplace and it just so happens that the same person has won twice in a row and this person has a close personal relationship with the raffle-drawer. I strongly suspect that the "random" raffle is not so random, but just to satisfy my own personal curiosity about how low the chances of this person winning twice in a row is I'd like to do the math but I ran into a snag, with which I'd like your help.

I'm trying to calculate the odds of this person winning twice in a row.

The first drawing had 31 entrants. All entrants have 1 ticket. 2 winners are drawn from this pool.

The second drawing had 25 entrants. All entrants, again, have 1 ticket. 1 winner is drawn from this pool. 

What are the odds of this person winning both the first drawing and second drawing? I know it's not as straightforward as $\frac{1}{31}\ast\frac{1}{25}$ but I don't know how to transform the first probability to accurately capture the two drawings for two winners. 

Thanks for your advice &amp; expertise.
We might also consider what is the chance that someone (not a specific individual chosen in advance) wins twice in a row.  This is a bit tricky since there are different numbers of entrants.  But assuming that all $25$ people in the second contest were also in the first contest, the probability of someone winning both times is 

$$1-\left(\frac{773}{775}\right)^{25}\approx .0626$$

So there is about a $6\%$ chance that someone would win twice in a row.
What my answer is about


I give the correct calculation to answer the question.
I point out that maybe the reason the event seems unlikely is because of the way we look at it.
I do some more calculation to show that similar events may be in fact common.


The correct calculation is the following :

For the first drawing, the probability of winning is the probability of winning the first draw, plus the probability of losing the first draw and winning the second: 

$$p_1 = \frac{1}{31} + \frac{30}{31}\times\frac{1}{30} \approx 0.0645$$

For the second drawing, the probability is strightfoward to compute : 

$$p_2 = \frac{1}{25} = 0.04$$

The total probability is the product of individual probabilities : 

$$P \approx 0.06*0.04 \approx 0.026 = 2.6\%$$

This seems low. 

However

When doing probabilities, you must be careful not to introduce bias! In your example, you include only the two drawings he won, but have not considered all the drawings that take place! 

I will now show you that the event you consider suspicious might not be. 

From your question, I will consider the following statement to be what you find suspicious :

"Mr. x won twice on 3 consecutive draws"

Now, say you have a drawing each week, with on average 30 people in each. Mr x participates every time.

The probability of him winning 2 times in 3 draws are : 

$$p = 3\times (\frac{1}{30})^2 = \frac{1}{300}$$

The probability of this event happening in 52 weeks is the following : 

$$P = 1- (1- \frac{1}{300})^{50} \approx 0.17 = 17\%$$

The term in parenthesis is the probability of not winning twice in any 3-week period, and there is 50 3-week periods in a year.

Those odds are a little bit less suspicious.

Furthermore

Now, if you are interested in the probability that anyone at some point in the year will win 2 times in 3 draws, the calculation is : 

$$ P = 1- (1-0.17)^{30} \approx 0.99 = 99\%$$

(we consider 30 persons playing each week for a year)

It is very likely that someone, at some point, will win twice out of 3 consecutive draws.

Here, I made some assumptions, but the idea is right.