There are 20 cards and inside them there are 4 specials (i.e. hearts) and 16 "others" (i.e. other colors).
We need to split it to 2 players, therefore each of them have 10 card.
What is the probability that one of them holds all 4 of the "special" cards (and the other one have therefore 0)?
I am trying to find out the probability in game https://en.wikipedia.org/wiki/Mari%C3%A1%C5%A1
If you know it, its the part of the game where you play 2x7 and you have only 4 cards of "supporting" color. In such case, if any player you play against have also 4, the game is already lost.
My approach so far:
As we know nothing about 20 cards and the order does not matter, we can simulate it in a way that 1 player takes 10 cards from pack of 20 cards where 4 cards are "special". Then add two chances: What is the chance that player gets 0 of them? And then add the "whats the chance the player gets all 4 of them"?
The "0" is quite straightforward. First card I select from pack of 20 cards where 16 of them are "fine". If I succeed the next I have pack of 19 cards where 15 of them are "fine", therefore the overall chance should be: 16/20 * 15/19 * 14/18 * ... * 7/11
But to get all 4 is not that easy with this approach. I can get Binomial coefficient (as orders do not matter) as (20 on top of 10) = 20!/(10!*10!). Which finds me all possible combinations. But to find probability I need to know how much of combinations contains all 4 "special" cards - which I cant figure out.
The number of $10$-card hands that contain all $4$ specials is just ${{16}\choose{6}}$. So the probability that a particular player has all $4$ special cards is $$ \frac{{16}\choose{6}}{{20}\choose{10}}=\frac{10! \cdot 16!}{20! \cdot 6!}=\frac{10\cdot 9 \cdot 8 \cdot 7}{20\cdot 19 \cdot 18\cdot 17}=\frac{14}{323} \approx 0.04334, $$ and the probability that either player does is twice that.