2
$\begingroup$

I was thinking that you first pick 1 of the 13 card types, and in that choose any two for the first pair. Then pick 1 of the 12 remaining card types and in that, choose any two for the second pair. Finally, pick 1 of the 11 remaining cards and in that, choose 1 card. Hence, the probability is:

$$ \frac{13\binom{4}{2} \cdot 12\binom{4}{2} \cdot 11\binom{4}{1}}{\binom{52}{5}} = 0.0951 $$

However, the solution is 0.0475 and it turns out I'm off by a factor of 2. Does anyone know why this method is not getting the right answer? I think it's something to do with double counting... But I can't seem to wrap my head around it.

$\endgroup$
1
  • $\begingroup$ I think you meant to write $\frac{13\binom{4}{2} \cdot 12\binom{4}{2} \cdot 11}{\binom{52}{5}}$? $\endgroup$
    – nivekgnay
    Commented May 4, 2016 at 7:12

2 Answers 2

Reset to default
2
$\begingroup$

Pick two values (eg 2s and Kings): ${13\choose 2}$

Pick two suits (eg for each): ${4\choose 2},{4\choose 2}$

Pick value for the 5th card (not 2 or King): ${11\choose1}$

Pick suit for 5th card (any): ${4\choose1}$.

Hence ${13\choose2}{4\choose 2}{4\choose 2}{11\choose1}{4\choose1}=123552$. So probability $=123552/{52\choose 5}= 4.75\%$

You double-counted on the first step.

$\endgroup$
2
  • $\begingroup$ Using this method why can't I pick three values first and then from each of these ranks pick the pairs and the single card. Ie. (13C3)(4C2)(4C2)(4C1)? This gives me an answer that is off by 3. $\endgroup$
    – jlcv
    Commented May 4, 2016 at 16:00
  • $\begingroup$ @ZoomBee Suppose you picked the values 1,2,3. There are still 3 possibilities: the pairs are 1s and 2s; or the pairs are 1s and 3s; or the pairs are 2s and 3s. You have missed that factor 3. $\endgroup$
    – almagest
    Commented May 4, 2016 at 20:27
1
$\begingroup$

Since you pick the pairs first from the 13 possible values, then from the 12 remaining values, you are indeed double counting - the hand AA335 is the same as the hand 33AA5, but you consider them to be distinct.

$\endgroup$
2
  • $\begingroup$ Can't that also be said for the last card (ie. 5 in your example)? As in 5AA33 is the same as AA335 so I will need to divide by 3!? $\endgroup$
    – jlcv
    Commented May 4, 2016 at 7:23
  • 1
    $\begingroup$ @ZoomBee When you selected one of the thirteen ranks from which to select a pair, then selected one of the twelve remaining ranks from which to select a second pair, you counted the hand consisting of two aces, two threes, and a five twice, once when you selected aces, then threes and once when you selected threes then aces. The order in which the pairs are picked does not matter. Notice that you only count the five once, namely when you pick one of the other $11$ suits, then choose one card from that suit. $\endgroup$
    – N. F. Taussig
    Commented May 4, 2016 at 10:18

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .