Probability: Cards

Question

A 52-card deck contains 13 cards from each of the four suits: clubs , diamonds , hearts , and spades . You deal 4 cards without replacement from a well shuffled deck, so that you are equally likely to deal any 4 cards. The probability that you deal no clubs is?

Answer 1

$$\left(\frac{39}{52}\right).\left(\frac{38}{51}\right).\left(\frac{37}{50}\right).\left(\frac{36}{49}\right) = 0.303818 = 30.38\%$$

Answer 2

Hints:

Combinatorial way 1: There are $\binom{52}{4}$ equally likely hands of $4$. There are $\binom{39}{4}$ no club hands.

Combinatorial way 2: Imagine dealing the cards one at a time. There are $(52)(51)(50)(49)$ equally likely strings of four distinct cards. How many strings of $4$ cards have no club?

Conditional probability way: Imagine that the cards are dealt one at a time. The probability the first card is not a club is $\frac{39}{52}$.

Given that the first card is not a club, the probability the second card is not a club is $\frac{38}{51}$. So the probability that neither of the first two cards is a club is $\frac{39}{52}\cdot \frac{38}{51}$.

Given that neither of the first two card is a club, the probability the third card is not a club is $\frac{37}{50}$. Continue.

Answer 3

Calculate the number of combinations with no clubs:

$$\binom{39}{4}=\frac{39!}{4!\cdot35!}=\frac{36\cdot37\cdot38\cdot39}{1\cdot2\cdot3\cdot4}$$

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Calculate the total number of combinations:

$$\binom{52}{4}=\frac{52!}{4!\cdot48!}=\frac{49\cdot50\cdot51\cdot52}{1\cdot2\cdot3\cdot4}$$

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Divide the former by the latter:

$$\frac{\frac{36\cdot37\cdot38\cdot39}{1\cdot2\cdot3\cdot4}}{\frac{49\cdot50\cdot51\cdot52}{1\cdot2\cdot3\cdot4}}=\frac{36\cdot37\cdot38\cdot39}{49\cdot50\cdot51\cdot52}=0.303$$