Let there be four cards (2 queens and 2 kings). You draw two cards. What is the probability that the cards are a pair? Is there a difference in probability if you draw two cards together (in which case probability is 0.5 if i'm not wrong) or draw one card at a time (probability of drawing a pair is 1/3)?


Let the Cards be $Q_1,Q_2,K_1,K_2$. Let $AB$ be the event that denotes drawing of $A$ followed by drawing of $B$. Then probability $p$ of getting a pair when drawing cards one after another is nothing but

$$\begin{align} p & = P(K_1 K_2) + P(K_2 K_1) + P (Q_1 Q_2) + P(Q_2Q_1). \\ & = P(K_2 \mid K_1) P(K_1) + P(K_1 \mid K_2) P(K_2) + P(Q_2 \mid Q_1) P(Q_1) + P(Q_1 \mid Q_2) P(Q _2). \\ & = \frac 1 4 \cdot \frac 1 3 + \frac 1 4 \cdot \frac 1 3 + \frac 1 4 \cdot \frac 1 3 + \frac 1 4 \cdot \frac 1 3. \\ & = 4 \cdot \frac {1} {12}. \\ & = \frac 1 3. \end{align}$$

Which is same as the probability of getting a pair taking two cards at a time.

  • $\begingroup$ Hmm. Thanks. But you are conditioning on the fact that first card was a king or queen or second is king or queen. But if you draw two cards together then there is no first or second draw. Either it is a pair or it's not. $\endgroup$
    – pb10
    Feb 11 '19 at 8:37
  • $\begingroup$ Yeah that's correct. But in both of the cases you will end up with same probability. $\endgroup$
    – little o
    Feb 11 '19 at 8:38
  • $\begingroup$ Thanks. Unfortunately I just can't make a leap from state space of {KK,KQ,QQ,QK} to what you are proposing. $\endgroup$
    – pb10
    Feb 11 '19 at 8:43
  • $\begingroup$ @pb10 Maybe think of it this way. Consider the left card out of the two you took. Either it is a king or a queen, it doesn't matter. Now, in order to form a pair, the right card needs to be the other king or queen. Since there are three cards left, and only one of them completes your pair, the probability is $\frac{1}{3}$. $\endgroup$
    – jvdhooft
    Feb 11 '19 at 8:44
  • $\begingroup$ Hmm. So we treat the process of drawing a pair as two events no matter what? My thought process had been : Number of ways to draw a pair is 2 (QQ,KK). Number of ways to draw two cards is 4. Hence probability is 0.5. $\endgroup$
    – pb10
    Feb 11 '19 at 8:51

After drawing the first card there are $3$ cards left, and exactly one of them is the "mate" of the card that has been drawn allready.

So the probability that this card will be drawn (resulting in a pair) is $\frac13$.

In this situation (no replacement) there is no essential difference between drawing cards one by one or drawing two cards together.

(To understand implant some order like: "the first card touched by my fingers is the first" or "after drawing I select a card and label it as first card". The only difference is that you did not had chance to see what card it is, but that on its own is not relevant for the probabilities.)

  • $\begingroup$ Please let me know how this is flawed: Number of ways to draw a pair is 2 (QQ,KK). Number of ways to draw two cards is 4. Hence probability is 0.5 $\endgroup$
    – pb10
    Feb 11 '19 at 8:53
  • $\begingroup$ $P(Q=2)=P(K=2)=\frac12\frac13=\frac16$ and $P(Q=1,K=1)=P(\text{first a queen, then a king})+P(\text{first a king then a queen})=2\cdot\frac12\frac23=\frac23$. Here $Q$ denotes the number of queens that are drawn and $K$ denotes the number of kings that are drawn. Note that $P(\text{a pair})=P(Q=2)+P(K=2)=\frac13$. $\endgroup$
    – drhab
    Feb 11 '19 at 9:01
  • $\begingroup$ "number of ways to draw two cards is 4". I guess you think of KK,KQ,QK,QQ. Yes, $4$ ways, but they do no have equal probabilities. The probability on KK is $\frac16$ (equals probability on QQ) and on KQ is $\frac13$ (equals probability on QK). So finding the probability is not just a matter of dividing number of specific ways by total number of ways. $\endgroup$
    – drhab
    Feb 11 '19 at 9:10
  • $\begingroup$ Perfect. Thank you so much. My answer would make sense if we are for example tossing 2 coins where all probabilities are same for HH,HT,TH,TT. $\endgroup$
    – pb10
    Feb 11 '19 at 9:13
  • $\begingroup$ Indeed. That is the same as the card problem under the condition that the cards are replaced after each drawing. $\endgroup$
    – drhab
    Feb 11 '19 at 9:14

Simpler: You draw a card. No matter what it is, there is only 1 of three remaining cards that will match it. Hence $P = 1/3$.

It makes no difference if you draw two at a time. You can always say: one of the cards is on the top of the two I chose....


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