# PIE question — probability of a void in a tarot deck

The question:

A tarot deck consists of 78 cards. There are 14 cards in each of 4 suits (comprising the usual 13 ranks and a face card called a knight), and there are 21 picture cards and a joker. The picture cards and the joker do not belong to any suit.

A tarot hand consists of 18 cards drawn at random from the deck. What is the probability that a tarot hand has a void, meaning that at least one suit is not present among the 18 cards? Once you've computed the answer in terms of binomial coefficients, use a calculator or computer to determine the answer to the nearest tenth of a percent, and enter that as your answer.

I know I have to use PIE based on where they say "at least one suit", but how would I set it up in terms of PIE?

• Is a $5$ (for example) in one suit distinguishable from the $5$ in each of the other three suits? – barak manos Dec 2 '14 at 21:18
• @barakmanos Not sure...I don't play tarot, so I'm not sure about the rules regarding the game. The only things specified are in the original problem statement--I wasn't provided with any additional hints/disclaimers. – Mathy Person Dec 2 '14 at 21:22
• Well, I've added an answer assuming that the suits are distinguishable from each other (something like, diamond, heart, clover and spade). – barak manos Dec 2 '14 at 21:47

Include the number of ways to choose from all the cards except for $1$ out of $4$ suits:

$$\binom{4}{1}\cdot\binom{64}{18}$$

Exclude the number of ways to choose from all the cards except for $2$ out of $4$ suits:

$$\binom{4}{2}\cdot\binom{50}{18}$$

Include the number of ways to choose from all the cards except for $3$ out of $4$ suits:

$$\binom{4}{3}\cdot\binom{36}{18}$$

Exclude the number of ways to choose from all the cards except for $4$ out of $4$ suits:

$$\binom{4}{4}\cdot\binom{22}{18}$$

Divide the result by the number of ways to choose from all the cards:

$$\frac{\binom{4}{1}\cdot\binom{64}{18}-\binom{4}{2}\cdot\binom{50}{18}+\binom{4}{3}\cdot\binom{36}{18}-\binom{4}{4}\cdot\binom{22}{18}}{\binom{78}{18}}\approx6.72\%$$