# An interesting probability calculation problem

#### Blain Waan

##### New Member
Hello all!

I'll be really glad if you respond to this question. I am working with an interesting probability calculation problem. In my problem there are ratings for 32 chess players. They are divided into two groups to form two chess teams in such a way that each team contains all types of players. However, each team should contain only 4 players and the whole design for selecting the players for the two teams has been made according to the diagram attached with this question (which is very important to see how the design goes).

32 players are randomly divided into two groups first, each containing 16 players. Each of these 16 players are further randomized into 4 groups each containing 4 members. Then within each group, the players are arranged in ascending order of their ratings. Then the players with lowest rating in the 1st group, 2nd lowest in the 2nd group, 3rd lowest in the 3rd group and highest in the 4th group are selected to from a team.

Given this design, what is the probability that player i, (i=1,...,32) will be assigned to Chess Team-1? (Please see the attached image to have a better understanding!)

#### Mean Joe

##### TS Contributor
For the highest-ranked player (I guess i=1):
There are 8 groups, and 1 group where she would have to fall into to be selected to Team #1. So I guess 1/8

For the 2nd-ranked player (i=2):
There are 8 groups. She could fall into the same group as the i=1 player and be selected, if it's the second-from-left group in the diagram. Or she could fall into the group on the far-left and the i=1 player would have to not be in it.
So I guess (1/8)*(1/8) + (1/8)*(7/8)

And then a pattern should come up.

#### BGM

##### TS Contributor
I would try to claim that each player has the identical probability of $$\frac {1} {8}$$ to become one of the four members of the first chess team.

I have just verified for the two players with the highest two ranks, but actually this should be true by symmetry.

Ok maybe I try to follow up Joe's idea:

For the $$i$$th-ranked player, there are $$i - 1$$ and $$32 - i$$ players having higher and lower ranks than him respectively.

So the probability of the $$i$$th-ranked player selected as team 1 would be

$$\frac {1} {8} \times \left[\binom {31} {3}\right]^{-1}\times \sum_{j=0}^3\binom {i - 1} {j} \binom {32 - i} {3-j}$$

(of course have slight changes of the formula for the extreme ranks)

But the latter part is just the sum of hypergeometric pmfs so it sum up to 1.