Probability of $3+3$ cards, out of $6$ cards drawn from a solitaire

Question

A solitaire consists of $52$ cards. We take out $6$ out of them (wihout repetition). Find the probability there are $3+3$ cards of the same type (for example, $3$ "1" and $3$ "5").

Attempt.

First approach. There are $\binom{13}{2}$ ways to choose $2$ out of the $13$ types and by the multiplication law of probability, the desired probability is $$\binom{13}{2}\frac{4}{52}\,\frac{3}{51}\,\frac{2}{50}\, \frac{4}{49}\,\frac{3}{48}\,\frac{2}{47}.$$

Second approach. There are $\binom{13}{2}$ ways to choose $2$ out of the $13$ types and the desired probability is $$\binom{13}{2}\frac{\binom{4}{3}\binom{4}{3}\binom{4}{0}\ldots\binom{4}{0}}{\binom{52}{6}}.$$

These numbers don't coincide, so I guess (at least) one of them is not correct.

Thanks in advance for the help.

Answer 1

In order to fully clear your confusion, let us tackle a simpler problem first.

We are dealing with drawing w/o replacement, (hypergeometric distribution)
If asked to find the Pr of drawing $2$ red and $3$ blue balls from a pool of $5$ red and $4$ blue balls,

Using the multiplication rule, $P(RRBBB)$ in that particular order$\;= \dfrac59\dfrac48\dfrac47\dfrac36\dfrac25$,
but we would need to multiply it by $\dfrac{5!}{2!3!}$ to take care of all possible orders.
[ But this multiplication factor is all too often forgotten by students]

By the combination approach, we would simply use $\dfrac{\binom52\binom43}{\binom95}$

I would advise that you use direct multiplication of probabilities when a specific order is given, and combinations otherwise.

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To come back to your problem, you should be able to see that in your first approach, since there are $3$ each of the two types, you need a multiplier of $\dfrac{6!}{3!3!}$,

thus $\dbinom{13}2\dfrac4{52}\dfrac3{51}\dfrac2{50}\dfrac4{49}\dfrac3{48}\dfrac2{47}\times \dfrac{6!}{3!3!}$

whereas the second approach directly gives the correct answer,

in fact you should simplify it to $\dbinom{13}2 \frac{\binom43\binom43}{\binom{52}6}$

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For a variety of problems on drawing colored balls from an urn without replacement, you could have a look here

Answer 2

Start with a simple test case, and build up:

Say we are to find $A\heartsuit$, $A\clubsuit$, $A\diamondsuit$, $J\heartsuit$, $J\clubsuit$, $J\spadesuit$, in any order, in the first six cards dealt.

There are $52!$ shuffles, of which $6!\times 46!$ have our cards in the required position.

Now we want to know how many ways we can produce $3$ Aces and $3$ Jack's from the $4$ available in each suit, and this is obviously $\binom43^2=16$.

So far then we have $6!\times 46!\times16$ shuffles that give us what we want.

Now we have $\binom{13}{2}$ ways to pick our two card ranks, so the final answer is:

$$\binom{13}{2}\frac{6!46!16}{52!}\approx0.0000613$$

Answer 3

There are $13$ possible numbers/people. We need to pick $3$, so the probability is $\binom{4}{3} \cdot \binom{13}{2} \cdot \binom{4}{3}$. The total probability of choosing 6 cards is $\binom{13}{2}$. The total probability is $\frac{4 \cdot 78 \cdot 4}{20358520}$ , which equals $\frac{1248}{20358520} \Rightarrow \approx 6.13011162*10^{-5}$