1
$\begingroup$

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)

$\endgroup$
9
  • 2
    $\begingroup$ Hint: use inclusion-exclusion. #hands - #hands missing at least one suit +#hands missing at least two suits - #hands missing three suits. $\endgroup$ – lulu Jun 16 '16 at 10:00
  • $\begingroup$ @lulu thanks ^-^ Is my 4 sum approach wrong though? What exactly is wrong with the last one? $\endgroup$ – BCLC Jun 16 '16 at 19:05
  • $\begingroup$ @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? $\endgroup$ – BCLC Jun 16 '16 at 22:43
  • $\begingroup$ @d_e Is there a way to recycle those 48 cards? also the 4 sum approach is wrong? $\endgroup$ – BCLC Jun 16 '16 at 23:04
  • 1
    $\begingroup$ 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. $\endgroup$ – d_e Jun 16 '16 at 23:05
3
$\begingroup$

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:

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

$\endgroup$
2
  • $\begingroup$ thanks barak manos ^-^ Is my 4 sum approach wrong though? What exactly is wrong with the last one? $\endgroup$ – BCLC Jun 16 '16 at 19:05
  • $\begingroup$ 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}. $\endgroup$ – hmakholm left over Monica Jun 16 '16 at 20:07
2
$\begingroup$

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 $$ 1 - \frac{A_1+A_2+A_3}{\binom{52}{13}} $$

$\endgroup$
1
  • $\begingroup$ thanks barak manos ^-^ Is my 4 sum approach wrong though? What exactly is wrong with the last one? $\endgroup$ – BCLC Jun 16 '16 at 19:06

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.