# Sampling without replacement (Board game problem)

#### Orjanbd

##### New Member
Hi!

I am trying to figure out this probability problem for a board game with a drafting mechanic.

There is 26 characters. 10 is of the kind called A and 16 is of the kind called B. 6 characters are to be randomly picked for a drafting pool.

1. What is the chance of exactly four of the six being of kind A? How do you calculate this?
2. What is the chance of more than four of the six being of kind A? How do you calculate this?

A helping hand here is greatly appreciated #### pbl24

##### New Member
This seems to reduce to a counting problem. Suggestions for 1) is to determine the total number of possibilities and divide that into the number of possibilities that have 4/6 being 'A'. For 2), similar to above, but would be either 5 or 6 of type 'A'.

#### Orjanbd

##### New Member
I guess you can call it a counting problem, but there are so many possibilities (720 I believe) . And I am wondering if there is a way to calculate this? Just counting is not very practical.

#### pbl24

##### New Member
There are actually far more possible outcomes than 720. If you imagine 6 slots, there are 26 options for the first slot, 25 for the second, 24 for the third, and so on. However, you need to account for situations in which the 6 slots are made up of the same items, just in different order. This nets out as being the number of combinations: C(26, 6) = 230,230. For part 1), you then need to figure out how many combinations there are that consist of 4 A's and 2 B's. This number will be the number of combinations, of the total 230230, that are made up 4 A's and 2 B's. I realize that the magnitude of the numbers seems big, so it might be useful to reduce it down to something that you can manage and sketch out a probability tree and generalize (e.g., a total of 2 A's and 3 B's and try to find the chance of 1 A and 2 B's).

Once you know the number of combinations of 4 A's and 2 B's, the probability can be expressed as the ratio of the specific number of combinations and the the total number of combinations above.

#### Orjanbd

##### New Member
So I understand how you get to 230230 by calculating 165765600/720. However, I can’t figure out this next part:

"I realize that the magnitude of the numbers seems big, so it might be useful to reduce it down to something that you can manage and sketch out a probability tree and generalize (e.g., a total of 2 A's and 3 B's and try to find the chance of 1 A and 2 B's).

Once you know the number of combinations of 4 A's and 2 B's, the probability can be expressed as the ratio of the specific number of combinations and the the total number of combinations above."

Could you please be more specific? Am I closer by reducing 230230 to 1001?

#### pbl24

##### New Member
I failed to mention this in my initial post. I made a structural assumption in that the elements in groups A and B can be distinguished in some way. For example, if there are 2 groups of male and female that we can call A and B, we can assign 4 separate males into group A and 2 separate females into group B. If that's not the case, then what I've mentioned above is incorrect. With this assumption, your elements in groups A and B can essentially be numbered:

A_1, A_2, A_3, ..., A_10
B_1, B_2, B_3, ... B_16

In the reduced example I provided, there are 2 elements in A and 3 in B:

A_1, A_2
B_1, B_2, B_3

C(5, 3) states there are 10 possible combinations:

A_1, A_2, B_1
A_1, A_2, B_2
A_1, A_2, B_3
A_1, B_1, B_2
A_1, B_1, B_3
A_1, B_2, B_3
A_2, B_1, B_2
A_2, B_1, B_3
A_2, B_2, B_3
B_1, B_2, B_3

Of these 10, only 6 match the pattern of 1 A and 2 B's. We can then conclude that the probability of getting 1 A and 2 B's is 6/10 = 0.6. You should be able to generalize this to your problem.

#### mvernengo

##### New Member
Combinat(10;A) *combinat(16:B) gives total per case then divide by total cases

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