You deal 5 cards from a well-shuffled deck of playing cards. What is the probability that the 5th card is the queen of spades?

Just from analysis, P(5th queen spade) = (51*50*49*48*1)/(52*51*50*49*48) = 1/52

However why wont this method of logic thinking incorrect?

P(5th queen spade) = (51Cr4) / (52Cr5) = 5/52. Reasoning: choose any first 4 cards and last card is queen spade, divide by all possible choice

  • $\begingroup$ Isn't that fifth card as likely to be the queen of spades as the three of clubs? $\endgroup$ – Lord Shark the Unknown Sep 14 '18 at 6:46
  • $\begingroup$ But what if the three of clubs was selected before the 5th card? Is it the order of when cards are drawn does not matter? $\endgroup$ – userName Sep 14 '18 at 6:49
  • $\begingroup$ But I am confuse because the question specifically state 5th card, hinting that order matters? I am not sure about this part. Please advice $\endgroup$ – userName Sep 14 '18 at 6:50
  • $\begingroup$ You are drawing four cards and then a fifth card. You don't care about order of the first four cards, but pay special attention to the fifth. Count ways to do so in both numerator and denominator (that is: favoured event and total space). $\endgroup$ – Graham Kemp Sep 14 '18 at 8:28
  • $\begingroup$ @graham So since first four cards order does not matter, the probability of that cancel each other in numerator and denominator is that what it is? $\endgroup$ – userName Sep 14 '18 at 8:39

Your (51Cr4) / (52Cr5), or as I prefer to write it $\dfrac{51 \choose 4}{52\choose 5}$, is the probability that the first five cards contain the Queen of Spades.

If that happens, there then is a $1$ in $5$ chance that the Queen of Spades is the fifth of these five cards,

making the result to the original question $\dfrac5{52}\times \dfrac{1}{5} = \dfrac{1}{52}$

  • $\begingroup$ Oh okay this makes sense... So does it mean that if I use the formula P(number of sample points in an event) / P(number of sample points in sample space) it will always be like the explanation that you have given? $\endgroup$ – userName Sep 14 '18 at 6:55
  • $\begingroup$ @userName - it depends on the precise question, especially on whether each element of the sample space is equally likely $\endgroup$ – Henry Sep 14 '18 at 7:06
  • $\begingroup$ What about this question? Find probability that the 5th card is the queen of spades, given that the first 4 cards are hearts? I consider P(first 4 cards hearts ^ last card queen of spade) / P(first 4 cards hearts) = (13Cr4 * 1 / 5) / (13Cr4 x 39Cr1 x 1 / 5) why is this reasoning incorrect? [multiplied by 1/5 due to the order, like in the previous example] correct ans: 1/48 $\endgroup$ – userName Sep 14 '18 at 7:16
  • $\begingroup$ @userName After you have chosen four particular hearts, there would be $52-4=48$ cards remaining. So you could answer that question with $\dfrac{{13 \choose 4}{1 \choose 1}}{ {13 \choose 4}{48 \choose 1}} =\dfrac{1}{48}$ $\endgroup$ – Henry Sep 14 '18 at 7:35
  • $\begingroup$ Alright thanks Henry. But just like to clarify why do I not need to multiply by 1/5 in this situation? Isn't P(first 4 cards hearts and last card queen spade) also mean that we have to be careful and consider that the queen spade is the last card? $\endgroup$ – userName Sep 14 '18 at 8:11

Since you are asking for specifically the fifth card to be $\spadesuit Q$, this is a problem in which order is important.

The number of ways to select $5$ different cards with order important is $P(52,5)$. The number of ways in which the fifth card is $\spadesuit Q$ is $P(51,4)$. So the probability is $$\frac{P(51,4)}{P(52,5)}=\frac{51\times50\times49\times48}{52\times51\times50\times49\times48}=\frac1{52}\ .$$ Your mistake was to use $C$s instead of $P$s.

  • $\begingroup$ Thanks David. Can I check how can the use of Permutation solve this qn other qn? Find probability that the 5th card is the queen of spades, given that the first 4 cards are hearts? If I use that formula I got stuck when (13Pr4) / (13Pr4 x 39Cr1) which gives 1/39... The correct ans is 1/48 instead $\endgroup$ – userName Sep 14 '18 at 7:27
  • $\begingroup$ Is it suppose to be (13Pr4) / (13Pr4 x 48Cr1) since I only select 4 cards and now left with 48 cards to choose from? $\endgroup$ – userName Sep 14 '18 at 7:30
  • $\begingroup$ You seem to have got an answer from someone else so I trust that is OK. But can I point out that on this site it is preferred that new questions should not be asked in comments. This is because it makes it harder for other people to find the question later. It would be better if you asked a new question. You can always cross-reference your present question if relevant. Thanks. $\endgroup$ – David Sep 17 '18 at 0:09

$\def\cbinom#1#2{{^{#1}\mathsf C_{#2}}}\frac{\cbinom{51}4\cbinom 11}{\cbinom{52}5}$ is the probability for selecting the queen of spades and four from the fifty-one other cards when selecting any five from all fifty two cards. It does not consider the fifth place as a seperate group from the first four.

You require the probability for selecting any four from the fifty-one non-queen of spaces then the queen of spades when selecting any four from fifty-two cards then any one from the fourty-eight other card.

That is $\frac{\cbinom {51}4\cbinom 11}{\cbinom {52}4\cbinom{48}{1}}$, which equals $1/52$.


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.