3
$\begingroup$

A standard deck of cards has 52 cards. If order does not matter, what are the chances of getting 3 cards of different suits, an 8 and a face card?

I thought like this:

  • P(first card being a different suit) = $1/1 $
  • P(second card being a different suit) = $39/51 $
  • P(third card being a different suit) = $26/50$
  • P(fourth card being an 8) = $4/49 $
  • P(fifth card being a face card) = $12/48$

My answer : $\frac{1}{1} \cdot \frac{39}{51} \cdot \frac{26}{50} \cdot \frac{4}{49} \cdot \frac{12}{48} = 0.81\%$

$\endgroup$
5
  • $\begingroup$ You need to consider the cases where any of your first 3 card fulfill any/all the other 2 condition (8 or face). $\endgroup$ – sentheta Feb 1 '20 at 14:42
  • $\begingroup$ You are calculating the probability of drawing those cards in that particular order. Also, you have not accounted for the fact that a card could be both an 8 and a member of one of those suits or both a face card and a member of one of those suits. A point of clarification: Do you mean exactly or at least three different suits? $\endgroup$ – N. F. Taussig Feb 1 '20 at 14:43
  • $\begingroup$ @Vanwij that was not considered and it's a good point. i'm assuming that my solution is rendered invalid... $\endgroup$ – Stephen Reeves Feb 1 '20 at 14:44
  • $\begingroup$ @N.F.Taussig again , good point. this exercised was carved out from an elementary combinatorics exercise book. $\endgroup$ – Stephen Reeves Feb 1 '20 at 14:48
  • $\begingroup$ @N.F.Taussig i believe the order doesnt matter. $\endgroup$ – Stephen Reeves Feb 1 '20 at 15:08
3
$\begingroup$

Total no of cases are ${52 \choose 6}$ , i am gonna write ${n \choose r }$ as nCr. (I am assuming that we have to take in consideration that which suit are of the 8 and face cards are of).


Now as we need 6 cards of three 3 different suites, the possible sums can be:- $a)1+1+4\ \ b)1+2+3\ \ c)1+1+1+3\ \ d)1+1+2+2\ \ e)2+2+2$


a)1+1+4 : rather than counting an 8 appears and an face cars appers it's easier to count total - either one of them does not appears

= Total - 8 does not appear - face card does not appear + both 8 and face card does not appear.

(13C1.13C1.13C4 - 12C1.12C1.12C4 - 10C1.10C1.10C4 + 9C1.9C19C4)$\times$(number of ways of selection of suits)

Selection of suits -> 4C2(for selecting 2 suits from 1 cards come in) .2C(1) (from where 4 cards come)=12

So it comes out 465132


Same way b)->2135520 ,c)->757672 ,d)->1820448, e)->570348


Final answer $\frac{465132+2135520+757672+1820448 +570348}{{52 \choose 6}}$

=$\frac{5749120}{20358520}$=0.28239380858

$\endgroup$
0
2
+50
$\begingroup$

The chance of getting no $8$s is $\frac {48\cdot47\cdot46\cdot45\cdot44\cdot43}{52\cdot51\cdot50\cdot49\cdot48\cdot47}=\frac{48!46!}{52!42!}$

The chance of getting no face cards is $\frac{40!46!}{52!34!}$

The chance of getting neither an $8$ nor a face card is $\frac{36!46!}{52!30!}$

By inclusion-exclusion the chance of getting at least one $8$ and at least one face card is then $$1-\frac{48!46!}{52!42!}-\frac{40!46!}{52!34!}+\frac{36!46!}{52!30!}=\frac{309821}{1017926}\approx 0.304365$$

It is not quite fair to say the chance of getting three suits is independent of the $8$ and face card probability because getting a lot of the same rank improves your chance for multiple suits and asking for an $8$ and a face card (very slightly) decreases the chance of lots of the same rank. I am going to consider them independent.

The chance of getting all six cards of one suit is $4\cdot \frac{13!46!}{52!7!}$ The chance of getting all six cards in two suits would be $6\cdot \frac {26!46!}{52!20!}$ but we have counted the all one suit cases twice each, so the chance of single suits three times, so we have to subtract it twice. The chance of at least three suits is then $$1-6\cdot \frac {26!46!}{52!20!}+8\cdot \frac{13!46!}{52!7!}=\frac {365209}{391510}\approx 0.932822$$

The total probability is then $$\frac{113149417589}{398528208260}\approx 0.283918$$ Most of the risk of failure comes from not getting an $8$.

$\endgroup$
6
  • $\begingroup$ Enumeration via code gives $5749120/20358520\approx 0.2824$ $\endgroup$ – Daniel Mathias Feb 4 '20 at 0:17
  • $\begingroup$ @DanielMathias: I am surprised that it is so far from my number, suggesting that at least one of us has an error. Your denominator is correct. Could you check separately the ranks and the suits? $\endgroup$ – Ross Millikan Feb 4 '20 at 0:27
  • $\begingroup$ Second term in your 8+face probability is incorrect. Also, your suits probability is slightly off, as your two-suit probability counts all one-suit cases three times each. $\endgroup$ – Daniel Mathias Feb 4 '20 at 2:34
  • $\begingroup$ @DanielMathias: Thank you very much. The first was confusion in copying. Now we are very close, well within a reasonable error for the independence assumption. $\endgroup$ – Ross Millikan Feb 4 '20 at 3:02
  • $\begingroup$ Final guess as to why this answer is wrong: only 3 suits means you can't get 4 8's. $\endgroup$ – JMP Feb 4 '20 at 12:57
-1
$\begingroup$

When order doesn't matter (i.e., the number of unordered samples/$\textit{combinations}$ is what we are interested in), then $n\choose{k}$ methods are usually the safest way to go.

With that said, when all outcomes of an experiment are equally likely, then the probability of an event $A$ is

$$ \mathbb{P}(A) = \frac{|A|}{|\Omega|}, $$ where $|\cdot|$ denotes the number of elements in the set, and $\Omega$ denotes the sample space.

For example, the probability of obtaining 2 aces, 1 King and a card not ace or King from a deal of 4 cards is $$ \frac{{4\choose{2}} {4\choose{1}} {44\choose{1}}}{52\choose 4}. $$ Think of it as "from the 4 aces available, we want 2" and so on. The handy thing about using $n\choose{k}$ approach is that once you've worked out what the correct numbers are, you don't have to worry about extra combinations on top of that.

So back to our example (I assumed from your working that we do not care what suit the 8 and the face card belong to). The number of ways 3 cards of different suits can be dealt from a pack is

$$ \text{(Number of ways of choosing 3 suits from 4)}\times\text{(Number of ways of choosing 1 card from each suit 3 times)}\\ = {4\choose{3}}{13\choose{1}}^3. $$ With these 3 cards chosen, the number of ways of choosing one 8 from the four available is ${4\choose{1}}$; same for the King. This leaves us with $$ \frac{|A|}{|\Omega|} = \frac{{4\choose{3}}{13\choose{1}}^3{4\choose{1}}{4\choose{1}}}{{52\choose{5}}} \approx 0.054102 = 5.4102\%. $$ I was concerned about the assumptions of the question, since it's not clear that the suit draws have to account for drawing an 8 or a King, but it may be the case that a priori we don't take this into account, since the cards are all being drawn simultaneously. I'm not sure. In any case, this would be my approach and I hope it at least sheds some light for you on how these questions can be tackled.

$\endgroup$
3
  • $\begingroup$ Your answer is incorrect for a number of reasons. Firstly, six cards are drawn, not five. Second, there may be more than one 8 or more than one face card in the draw. You count those cases multiple times. $\endgroup$ – N. F. Taussig Feb 1 '20 at 19:14
  • $\begingroup$ Ok, then the answer needs to be adapted to 6 draws, which is a simple generalisation. And I felt I made it clear the assumptions that I was working under (3 cards of different suits, one King (any suit) one eight (any suit)). I believe that under those assumptions my answer is correct, but please let me know if not. For one reason or another the question does leave substantial details missing. I provided the best answer I could with the details I was given. $\endgroup$ – Dan Burrows Feb 1 '20 at 19:40
  • $\begingroup$ @DanBurrows very thorough and well explained! so for the last card being drawn, how do i calculate its probability? $\endgroup$ – Stephen Reeves Feb 2 '20 at 13:48

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.