$5$-card draw: drawing $3$ cards with a Pair

Question

A $5$-card poker hand is dealt to us of standard $52$-card deck. We hold a pair, and exchange $3$ cards (without replacement) in order to improve the One pair hand we hold.

When enumerating the second pair, I stumbled upon a certain obstacle - need to calculate the free and discarded ranks separately, I guess and add them up:

We discarded $3$ cards of $3$ different ranks (call them "discarded" ranks) and we have other $9$ "free" ranks with which we can enhance our hand.

Number of obtaining of a "free" pair:

${9\choose1}\times{4\choose2}$, and the $5^{th}$ card, which also could be of either of $8$ "free" ranks or the one of $3$ "discarded" ones.

So the total no of "free" pair would be:

$${9\choose1}\times{4\choose2}\times\left[{8\choose1}\times{4\choose1}+{3\choose1}\times{3\choose1}\right].$$

Is this correct?

We must also calculate number of "discarded" pairs, which would be:

$${3\choose1}\times{3\choose2}\left[{9\choose1}\times{4\choose1}+{2\choose1}\times{3\choose1}\right].$$

And the sum of these two numbers would be chances of improving a one pair hand to the two pair?

Of course, we divide the number by $47\choose3$ to get the probability.

Verification required, please.

Answer 1

Without loss of generality, assume that the original hand was AAKQJ, and we traded in the KQJ.

The possible outcomes are as follows:

• Draw two aces and another rank to obtain four of a kind (AAAAx).
• Draw three of another rank to obtain a full house (AAxxx).
• Draw an ace and another pair to obtain a full house (AAAxx).
• Draw an ace and two other ranks to obtain three of a kind (AAAxy).
• Draw a pair and another rank to obtain two pairs (AAxxy).
• Draw three other ranks to retain one pair (AAxyz).

There are only two aces left in the remainder of the deck. These can be combined with any of the other $45$ cards to obtain a four of a kind, so there are $45$ such draws.

There is only one way each of drawing KKK, QQQ, or JJJ, plus four ways each of drawing TTT, 999, down to 222, so there are $4 \times 9 + 3 = 39$ such draws to obtain a full house AAxxx.

There are three ways each of drawing KK, QQ, or JJ, plus six ways each of drawing TT, 99, down to 22, so there are $6 \times 9 + 3 \times 3 = 63$ such pairs; each of these can be combined with either of the two remaining aces in the $2 \times 63 = 126$ draws to obtain a full house AAAxx.

There are $36 \times 32 \div 2 = 576$ non-matching pairs containing no K, Q, or J; there are $36 \times 9 = 324$ non-matching pairs containing exactly one K, Q, or J; and there are $9 \times 6 \div 2 = 27$ non-matching pairs containing only K, Q, and J. There are thus $576+324+27 = 927$ such pairs, each of which can be combined with either of the two remaining aces in the $2 \times 927 = 1854$ draws to obtain three of a kind AAAxy.

There are three ways each of drawing KK, QQ, or JJ; each of these can be combined with any of $42$ non-matching, non-ace cards for $9 \times 42 = 378$ draws. To this must be added the six ways each of drawing TT, 99, down to 22; each of these can be combined with any of $41$ non-matching, non-ace cards for $54 \times 41 = 2214$ draws. Thus, there are $378+2214 = 2592$ draws to obtain two pairs AAxxy.

Finally, there are $36 \times 32 \times 28 \div 6 = 5376$ non-matching triples xyz not containing K, Q, or J; there are $9 \times 36 \times 32 \div 2 = 5184$ non-matching triples xyz containing exactly one K, Q, or J; there are $9 \times 6 \times 36 \div 2 = 972$ non-matching triples containing exactly two K, Q, or J; and there are $9 \times 6 \times 3 \div 6 = 27$ non-matching triples KQJ. This totals $5376+4032+864+27 = 11559$ non-matching triples to retain one pair AAxyz.

As a double check, we have accounted for $11559+2592+1854+126+39+45 = 16215$ hands, and $\binom{47}{3} = 16215$.