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 & expertise.
 A: 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.
A: Assuming the events are independent and everyone has the same chance of winning, then the odds of the same person winning twice is:
$$\frac{2}{31} \times \frac{1}{25} = \frac{2}{775}$$
In other words, they have a $2$ in $775$ or roughly $0.26\%$ chance of winning both. Unlikely, but not impossible.

Update: In light of Paw's answer above, I must make it clear here that this is the odds of a specific person winning twice in a row. The odds that someone wins twice in a row, can be calculated in two ways. Paw gives one way above, assuming all $25$ people in raffle two are in raffle one. However, there is another way to calculate this probability, that is arguably more general. First, we note that no two people can win twice in a row- i.e. the events are mutually exclusive. A nice property of mutually exclusive events $A$ and $B$ is that $P(A \vee B) = P(A) + P(B)$ i.e. the probability of one or the other is just their sum. From this, we get that the probability of someone winning twice in a row is:
$$\dfrac{2x}{775}$$
where $x$ is the number of people in both raffle number one and raffle number two. Note the maximum $x=25$, and for this we get:
$$\dfrac{2 \times 25}{775} = \dfrac{50}{775} \approx 0.0645$$
Which seems to agree reasonably well with Paw's answer (perhaps the large exponential caused some rounding errors).

Update 2: As A.S. points out in the comments above, this answer fails to incorporate any model of "rigging" or "close personal relationship". In this update, I will try to address these issues. However, as this is mathematics, I must still make some assumptions, and this answer is by no means perfect.
We start by noting that what we are after is the probability that the raffles were rigged, given the outcome of the close friend winning twice in a row. To begin, we introduce a baseline probability of the raffle being rigged, call it $p$. This baseline rig probability is how likely we think it is that the raffle will be rigged before we make any observations of the outcome. Of course, I must warn the reader that from a psychological point of view, it will be very hard to calculate this after said outcome was observed. This is because humans are notoriously bad for exaggerating patters. However, this is a different subject, so let's just assume as mathematicians that we have some baseline estimate for raffle-rigging $p$.
Now, let's assume that if the raffles weren't rigged, then everybody has an even chance of winning. Furthermore, let's assume that if the raffles were rigged, then the "close friend" of the organiser will definitely win them both. These assumptions are quite callous. E.g. they ignore the likely possibility that the organiser has many friendships, all of varying closeness. Worse still, the organiser mightn't even want to rig the draw for a friend: what if she was rigging it for some other reason? Or what if one raffle was rigged, the other not? The possibilities are endless, so we must draw the line somewhere and make assumptions. Hence the above.
OK, let $q$ be the probability that the organiser's friend won both raffles. Then we have, by our assumptions stated above:
$$q = p + (1 - p)\frac{2}{775}$$
Now let $r$ be the probability of the raffle being rigged given that the friend won both raffles. Now, we note that by our assumptions, the probability of the friend winning both raffles given that the raffles were rigged is $1$ i.e. absolute certainty. Then, by Bayesian inference, we have that:
$$r = \dfrac{p}{p + (1 - p)\frac{2}{775}} = \dfrac{775p}{773p + 2}$$
Using the formula, we can calculate for example, the conditional probability of a rigged raffle, given that the baseline probability was $0.01$ (or $1\%$):
$$\dfrac{7750.01}{7730.01 + 2} = 0.796$$
This is a substantial increase over the baseline of $0.01$. If the baseline was $0.001$, we still get a substantial conditional rig probability of around $0.28$- an even bigger increase. The percentage gain from initial rig probability to conditional rig probability increases monotonically as $p \rightarrow 0$. The increase is given by:
$$\dfrac{775}{773p + 2}$$
Which is just the above formula for $r$ divided by $p$. Clearly this is a maximum at $p=0$ and the increase is $\frac{775}{2}$- but of course with $p=0$, then the conditional probability is also just $0$, so we never experience this increase. So if the above assumptions approximately hold, which is a big "if", then we should expect a significant increase from the baseline probability of a rig.
A: The odds of winning the first contest is:
$\frac{1}{31}+\frac{30}{31}\cdot\frac{1}{30} = \frac{2}{31}$ - either they win the first or second draw - and there is no replacement between draws
The odds of winning the second contest is $\frac{1}{25}$, as you said.
The odds of winning both is the product of these, or
$\frac{2}{775} \approx 0.26\%$
So it's unlikely, but it's not necessarily an indication of cheating.
If it happens any more times, I'd start to think about it though...
A: The probability of winning the first raffle is: $1/31 + 30/31*1/30 = 2/31$.
The probability of winning the second raffle is: $1/25$.
The probability of her winning both: $2/31 * 1/25 = 2/775$ but that's the same probability of any outcome.
The probability of one of the two winners of the first raffle winning the second raffle is 2/25.  Which one shouldn't bet on but is hardly suspicious.
Okay, that the person is a close friend of the boss... Well, as 1/13 probability is hardly shocking, this is well within the realm of possibility.
A: 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.
