A deck of cards has 54 cards…

probability

thanks for checking my work!

A deck of cards has 54 cards: 2,3, . . . ,10, Jacks, Queens, Kings, Aces of all four suits, plustwo Jokers. Jokers are assumed to have no suits. We’ll say that a combination of 5 cards is a flush if one ofthe following three conditions hold:

  • All five cards are of the same suit, OR
  • Four cards are of the same suit and the remaining card is a Joker, OR
  • Three cards are of the same suit and the remaining 2 cards are Jokers.

Compute the probability that 5 randomly chosen cards from the deck give the flush combination.

does my solution make sense?

all combinations = $54\cdot53\cdot52\cdot51\cdot50$

5 same suit = $\dbinom{4}{1} \cdot\dbinom{13}{5}$

4 same suit 1 joker = $\dbinom{4}{1}\cdot\dbinom{13}{4}\cdot\dbinom{2}{1}$

3 same suit 2 joker = $\dbinom{4}{1}\cdot\dbinom{13}{3}$

so…

$\dfrac{\dbinom{4}{1}\cdot\dbinom{13}{5} + \dbinom{4}{1} \cdot\dbinom{13}{4} \cdot\dbinom{2}{1}+\dbinom{4}{1}\cdot\dbinom{13}{3}}{54\cdot53\cdot52\cdot51\cdot50}$

Best Answer

You have correctly counted combinations for the three favoured cases.

Since you are counting combinations in the numerator, you should do so in the denominator too.

$$\dfrac{\dbinom{4}{1}\left(\dbinom{13}5\dbinom{2}0+\dbinom{13}4\dbinom{2}1+\dbinom{13}3\dbinom{2}2\right)}{\dbinom{54}{5}}$$


Another way to count the numerator: a flush is formed when we select a suit, then select five cards from the fifteen cards that are either of that suit or of the jokers.$$\dfrac{\dbinom 41\dbinom{15}5}{\dbinom{54}5}$$

See also Vandermonde's Identity.