Actual Hand in 7-card stud

[jakebrandon]

I was in a 7-card stud game and one player ended up with 4 Jacks, another player with 4 Queens, and another with 4 Kings. As a side note, the guy with 4 Jacks lost all his money and basically felt he was cursed forever.
Can anyone calculate the odds of this hand happening? I know it has to be astronomical and I've always wondered about the probability of it happening.
Thanks!

 

[soroban]

Hello, jakebrandon!

I was in a 7-card stud game and one player ended up with 4 Jacks, another player with 4 Queens, and another with 4 Kings.
As a side note, the guy with 4 Jacks lost all his money and basically felt he was cursed forever.
Can anyone calculate the odds of this hand happening?
I know it has to be astronomical and I've always wondered about the probability of it happening.
Thanks!

You said you were in the game; I'll assume there are four players.

The deck has 12 Face cards and 40 Others.

A player has 4 Kings and 3 Others. .There are: .\(\displaystyle {4\choose4}{40\choose3} \) ways.
There are: .\(\displaystyle {52\choose7}\) possible outcomes.
\(\displaystyle P(\text{4 Kings}) \:=\:\frac{{40\choose3}}{{52\choose7}} \)

A player has 4 Quenns and 3 Others. .There are: .\(\displaystyle {4\choose4}{37\choose3} \) ways.
There are: .\(\displaystyle {45\choose7}\) possible outcomes.
\(\displaystyle P(\text{4 Queens}) \:=\:\frac{{37\choose3}}{{45\choose7}} \)

A player has 4 Jacks and 3 Others. .There are: .\(\displaystyle {4\choose4}{34\choose3} \) ways.
There are: .\(\displaystyle {38\choose7}\) possible outcomes.
\(\displaystyle P(\text{4 Jacks}) \:=\:\frac{{34\choose3}}{{38\choose7}} \)

You have 7 Others. .There are: .\(\displaystyle {31\choose7}\) ways.
There are: .,\(\displaystyle {31\choose7}\) possible outcomes.
\(\displaystyle P(\text{7 Others}) \:=\:1\)


\(\displaystyle \displaystyle P(\text{4 Kings, 4 Queens, 4 Jacks, 7 Others}) \:=\:\frac{{40\choose3}}{{52\choose7}} \cdot \frac{{37\choose3}}{{45\choose7}} \cdot \frac{{34\choose3}}{{38\choose7}} \cdot 1 \)