Standard playing card deck without jokers: Deal 13 cards. What is the probability at least 1 suit is not present?

Similar questions:

What is the probability that 13 cards drawn from a standard deck has at least one card from each suit?

Deal 4 cards from a deck. What is the probability that we get one card from each suit?

What I tried:

P(at least 1 suit is not present) = 1 - P(all suits are present)

where

P(all suits are present) =

$$\sum_{a}\sum_{b}\sum_{c}\sum_{d}\frac{\binom{13}{a}\binom{13}{b}\binom{13}{c}\binom{13}{d}}{\binom{52}{13}}$$

where $a+b+c+d=13; a,b,c,d \ge 1$

Is that right? If so, is there a better way to approach this? If not, why?

Following the second question:

$$\frac{\binom{13}{1}^{13}}{\binom{52}{13}}$$

Something else:

$$\frac{\binom{13}{1}^{4} \binom{48}{9}}{\binom{52}{13}}$$

which is ~75 (far greater than 1)

• Hint: use inclusion-exclusion. #hands - #hands missing at least one suit +#hands missing at least two suits - #hands missing three suits. – lulu Jun 16 '16 at 10:00
• @lulu thanks ^-^ Is my 4 sum approach wrong though? What exactly is wrong with the last one? – BCLC Jun 16 '16 at 19:05
• @d_e I don't understand how it doesn't represent this situation though. I want to pick 1 card from each group of 13. From the unpicked 48 cards, i want to pick 9. I really have to use inc-exc for this? – BCLC Jun 16 '16 at 22:43
• @d_e Is there a way to recycle those 48 cards? also the 4 sum approach is wrong? – BCLC Jun 16 '16 at 23:04
• I deleted my previous comments. now I understand what you have tried to do and what is your mistake. give me few minutes to write it down. – d_e Jun 16 '16 at 23:05

2 Answers

Use inclusion/exclusion principle:

• Include the number of combinations containing at most $\color\red3$ suits: $\binom{4}{\color\red3}\cdot\binom{\color\red3\cdot13}{13}$
• Exclude the number of combinations containing at most $\color\red2$ suits: $\binom{4}{\color\red2}\cdot\binom{\color\red2\cdot13}{13}$
• Include the number of combinations containing at most $\color\red1$ suit: $\binom{4}{\color\red1}\cdot\binom{\color\red1\cdot13}{13}$

Finally, divide the result by the total number of combinations:

$$\frac{\binom43\cdot\binom{3\cdot13}{13}-\binom42\cdot\binom{2\cdot13}{13}+\binom41\cdot\binom{1\cdot13}{13}}{\binom44\cdot\binom{4\cdot13}{13}}\approx5.1\%$$

• thanks barak manos ^-^ Is my 4 sum approach wrong though? What exactly is wrong with the last one? – BCLC Jun 16 '16 at 19:05
• However, $\binom42\binom{3\cdot 13}{13}$ is not the number of combinations with a most 3 suits -- some of them will be double-counted, such as "all the spades", which is counted as both {spades,diamonds,hearts} and {spades,clubs,diamonds} and {spades,clubs,hearts}. – hmakholm left over Monica Jun 16 '16 at 20:07

This is probably handled best as an inclusion-exclusion problem, counting the number of hands that omit at least one suit:

Let's take it from the bottom: The number of hands with exactly one suit is of course $$A_1 = \binom{4}{1} \binom{13}{13} = 4$$

The number of hands with exactly two suits is $$A_2 = \binom42 \binom{26}{13} - \binom31 A_1$$ where the second term corrects for the fact that $\binom42\binom{26}{13}$ counts each of the $4$ one-suit hands three times.

The number of hands with exactly three suits is then $$A_3 = \binom43 \binom{39}{13} - \binom21 A_2 - \binom32 A_1$$ where the subtracted terms correct for counting hands that actually contained fewer than three suits.

Now your result would be $$1 - \frac{A_1+A_2+A_3}{\binom{52}{13}}$$

• thanks barak manos ^-^ Is my 4 sum approach wrong though? What exactly is wrong with the last one? – BCLC Jun 16 '16 at 19:06