Probability problem with cards

[natugnaro]

Homework Statement

A deck of 32 Skat cards of the four colors clubs, spades, hearts, diamonds in each case ace,
king, queen, jack, ten, nine, eight, seven is randomly distributed among three players A, B, C,
each player receiving 10 cards and the remaining to cards go to the so-called Skat. Player A
received exactly 2 jacks. How large is the probability p that player B or C holds the other two
jacks in his hand?
Note: Regard most simply the relationship of the favorable (= B has exactly two jacks) to the possible
hands for players B.

The Attempt at a Solution

Two jacks can be distributed between Skat, B and C in six different ways, but only two are favorable (B hold two jacks or C holds two jacks).
Then the probability of B or C holding other two jacks is:

P=2/6=1/3

 

[Hurkyl]

The Attempt at a Solution

Two jacks can be distributed between Skat, B and C in six different ways, but only two are favorable (B hold two jacks or C holds two jacks).
Then the probability of B or C holding other two jacks is:

P=2/6=1/3

It's clear that that's the proportion of splits that are favorable, but less so that that's the probability.

 

[natugnaro]

Two jacks can be distributed between Skat S, player B and player C.
Therefore, there are 6 possible options:
S(11),B(0),C(0) {two jacks on Skat, player B and C don't have jacks, etc...}
S(1),B(1),C(0)
S(0),B(11),C(0)
S(0),B(1),C(1)
S(0),B(0),C(11)
S(1),B(0),C(1)

this is my sample space, and there are two favorable points (S(0),B(11),C(0)) and (S(0),B(0),C(11)),
so probability for player B or C to hold other two jacks in his hand is P=2/6=1/3.
I do not see where I'm wrong, can you explain a little more ?

 

[Hurkyl]

I do not see where I'm wrong, can you explain a little more ?

The ratio of the number of things with a property over the total number of things is called a proportion.

Proportions have pretty much nothing to do directly with probability, except in the special case that "things" are disjoint events in a finite probability space, and each of these events has equal probability measure.


While you've correctly computed the proportion of favorable ways to assign jacks to players, you have made no attempt to argue that each such assignment occurs with equal probability in your probability space.

In fact, it turns out that they don't. (Assuming when you said "is randomly distributed" you meant that each possible deal occurs with equal probability)

 

[natugnaro]

Ok, a different approach then.
I'll first try to solve the simplified problem, let's say there is no player C, and the rest of the problem is the same.
After a player A has received 10 cards there are 22 cards left.
From these 22 cards, 10 go to player B, the rest goes to the Skat. The probability that player B has received 2 jacks is:

(Binomial[2, 2]*Binomial[20, 8])/Binomial[22, 10] = 15/77

Binomial[2,2] - the number of ways to choose tow jacks from two jacks
Binamial[20,8] - the number of ways to choose the rest 8 cards
Binomial[22,10] - the number of ways a player B can receive 10 cards

did I solve this simplified case good ?