Probabilities in pinochle with 3 players

Question

I want to calculate some probabilities of the game Pinochle. But since it's difficult to translate from german into english, I'll simplify it with other words)

Basically, a deck has

• $48$ cards
• $4$ colors (red, blue, green, yellow) to each $12$ cards ($2 \times$ Ace, $2 \times$ Ten, $2 \times$ King, $2 \times$ Upper, $2 \times$ Lower, $2 \times$ Seven)

There are $3$ players. Player A, B and C. Each player receives $14$ cards and the last $6$ cards are for the tap.

What are the probabilities that

• $a)$ Player A has both red upper and blue lower in his hands
• $b)$ Player B has at least one green seven in his hand
• $c)$ Player C has exactly 5 yellow cards in his hand
• $d)$ The tab only contains Tens

This is what I have, but I don't think they are correct:

$a)$ There are 8 "upper" and 8 "lower" cards in total and there are exactly $2$ red uppers and $2$ blue lowers in the card. $\frac{2 \cdot \binom{12}{2}}{\binom{48}{12}} = 0.000000002 $

$b)$ There are exactly $2$ green sevens in the deck. $\frac{1}{48}$

$c)$ There are 12 yellow cards in total. $\frac{12}{48} \cdot \frac{11}{47} \cdot \frac{10}{46} \cdot \frac{9}{45} \cdot \frac{8}{44} = 0.00046253469$

$d)$ The tap consists of only 6 cards. There are 8 "Tens" in total. $\frac{8}{48} \cdot \frac{7}{47} \cdot \frac{6}{46} \cdot \frac{5}{45} \cdot \frac{4}{44} \cdot \frac{3}{43} = 0.0000022817$

Any help is appreciated!

Answer 1

$(d)$ is correct.

a) player $A$ has both red upper and blue lower in his hands. So player $A$ has at least one of both red upper and blue lower cards. There are $2 \cdot 2$ ways to choose one of each, $2 \cdot 2$ ways to choose two cards of one type and one card of another and just one way to choose two cards of both.

$\displaystyle 2 \cdot 2 \cdot {44 \choose 12} + 2 \cdot 2 \cdot {44 \choose 11} + {44 \choose 10}$

Divide by $\displaystyle {48 \choose 14}$ to find desired probability which comes to $\displaystyle \frac{47411}{194580} \approx 0.2437$.

b) Player B has at least one green seven in his hand

Number of ways in which player $B$ has no green seven is $\displaystyle{46 \choose 14}$

So number of ways in which $B$ has at least one green is $\displaystyle{48 \choose 14} - \displaystyle{46 \choose 14}$

Dividing by $\displaystyle {48 \choose 14}$, the desired probability is $\displaystyle \frac{189}{376}$.

c) Player $C$ has exactly $5$ yellow cards

Number of ways are $\displaystyle {12 \choose 5}{36 \choose 9}$. Dividing by $\displaystyle {48 \choose 14}$, the desired probability is $\approx 0.1546$

Answer 2

You seem to be quite confused. I'll just talk about part (a). First of all, each player is dealt $14$ cards, so the denominator should be $\binom{48}{14}$ not $\binom{48}{12}$. As to the numerator there are $2$ blue uppers, $2$ red uppers, and $44$ other cards. There are several possibilities. The player could get

• $1$ blue upper and $1$ red upper in $\binom{2}{1}\binom{2}{1}\binom{44}{12}$ ways

• both blue uppers and $1$ red upper in $\binom{2}{2}\binom{2}{1}\binom{44}{11}$ ways

• $1$ blue upper and both red uppera in $\binom{2}{1}\binom{2}{2}\binom{44}{11}$ ways

• both blue uppers and both red uppers in $\binom{2}{2}\binom{2}{1}\binom{44}{10}$ ways

We have to add all these up and divide by $\binom{48}{14}$.

We have to take into account all the ways there are of choosing cards that aren't red or blue uppers.