# Basic Probability Question of Two Teams choosing numbers

So this is probably a pretty easy question, but I wasn't sure how to go about it. Suppose there is a game with two teams. Each team has $k$ players. The game works by having each team member choose a unique number from 1 to $n$, and then the teams reveal what numbers they chose. What is the probability that the two teams choose the same numbers? (Player order doesn't matter)

My gut says its just ${n}\choose{k}$ but I feel that there is a trick or something there.

Also, what if the team members don't have to choose unique numbers?

What if the numbers are not uniformly choosen, but instead are chosen according to a probability function $P(x)$ (note P(x) doesn't add to 1 since 10 different numbers are chosen each game)?

• Can you clarify "probability that the two teams choose the same numbers"? Do you mean everyone has a corresponding pair, or at least 1 pair exists across teams? Commented May 21, 2017 at 14:45

The whole rules/team aspect needlessly complicates the topics. You just want to calculate the probability whether two sets of random numbers with uniform distribution are the same.

The result will depend on whether order matters and whether you allow duplicates. In the case where order doesn't matter and without duplicates, it becomes the classic "lottery" problem and you answer is indeed correct.

It's actually fun to go through the four different options (ordered/non-ordered, duplicates yes/no) and see the relationships. Let's use the German Lotto where you need to pick 6 numbers out of 49.

No order, no duplicate: We work through numbers one at a time: for the first number there is 1/49 chance that it hits one of the target numbers. We got 6 chances so the total likelihood of a match is 6/49. For the second number the individual chance for hit is 1/48 since the first number has been eliminated from the pool. On the other hand you have only 5 shots, so it's 5/48. Hence we get

$$p = \frac{6}{49} \cdot \frac{5}{48} \cdot \frac{4}{47} \cdot \frac{3}{46} \cdot \frac{2}{45} \cdot \frac{1}{45}$$

which is indeed the inverse of 49 choose 6.

No order, duplicates: This works the same except that after any match no number get removed from the pool so the likelihood is always a multiple of 1/49. We get

$$p = \frac{6}{49} \cdot \frac{5}{49} \cdot \frac{4}{49} \cdot \frac{3}{49} \cdot \frac{2}{49} \cdot \frac{1}{49}$$

Order, no duplicates: The difference between order and not-ordered is whether you have multiple shots at each match or not. If it's ordered you can only match to the next one in the sequence, so we get $$p = \frac{1}{49} \cdot \frac{1}{48} \cdot \frac{1}{47} \cdot \frac{1}{46} \cdot \frac{1}{45} \cdot \frac{1}{44}$$

Order, duplicates: That should be obvious by now, each single comparison step has a constant hit probability of 1/49 since there are no multiple comparisons and no elimination from the pool

$$p = \frac{1}{49} \cdot \frac{1}{49} \cdot \frac{1}{49} \cdot \frac{1}{49} \cdot \frac{1}{49} \cdot \frac{1}{49}$$