# How do you calculate the probability that three people will all win a prize in two consecutive weekly raffles?

My company is having a series of holiday raffles. Employees earn entries by completing specified tasks, so any employee may have one or more entries within a given week. We do not know how many entries there are in total, but we can see how many entries any given employee has submitted in each time period.

After two weeks, three employees have won twice. Some people (meaning "those of us who haven't won") are curious about the probability that this could happen.

If we assume that there are 2000 total entries and 1000 employees have entered in each of the raffles and there are 10 prizes in each raffle, how do we calculate the probabilty that

1 person wins twice

2 people win twice

3 people win twice

Thanks in advance...

### 2 Answers

- 1 decade agoFavorite Answer
If there are 2000 total, and 1000 employees then each employee should've entered about 2, but each two has a chance of one of the ten so you get 20/2000. Simplify to 1/100. That's for one person, once. To win again they must get 1/100 again, so they have 1/10,000 For two people you have 1/100 X 1/100 for one person & 1/100 X 1/100 for the second person so it would be 1/100,000,000 for two people. Three people would do the same thing; 1/100 X 1/100 X 1/100 X 1/100 X 1/100 X 1/100 = 1/100,000,000,000,000. So if three (same) people win two in a row, it would be extremely rare.

- a_math_guyLv 51 decade ago
You really need the counts to get an accurate probability. For instance if 1 employee has 1001 entries and the other 999 employes have one each, then it is almost a sure thing that that person is going to win repeatedly.

But let's assume that (roughly speaking) everyone has two entries. Think of the first drawing as totally random, it doesn't matter whose name comes up first time around, it could've been Jack instead of Judy, whatever. Then you are asking: for the second time around, what's the chance that 3 out of the ten winners were winners the first time around? Number of ways to pick 10 winners randomly: 1000 C 10 where C stands for the binomial coefficient. Number of ways to pick three repeat winners and seven non-repeat winners: 10 C 3 * 990 C 7. The probability of this happening (exactly three repeats) is the ratio of these two, which simplifies to 1810 4364 75325 9356 /21950 79670 51641 844027 or approximately .0000824 7702804 or about 1 in 12 thousand.

It is sounding like it is rigged, hunh? One thousand employees, that's a big company! Anyway, I think the number of entries would make a BIG difference. Of course, those "kudos" could be handed in an unfair way, making the probability reasonable but nonetheless lopsided.

For k=2 people, I get about 1 in 263 chance and k=1 person about 1 in 11 chance. FYI about a 90.4% chance of no-repeats. IF they all have two tickets each.