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

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.

• You need to be very clear about what it is you want, not only to us but to yourself. When you take out the cards, if they happen in sequence, would $1\heartsuit~5\spadesuit~1\diamondsuit~5\diamondsuit~5\clubsuit~1\spadesuit$ have counted? The way you have phrased it, it sounds like it should. Does your first approach take this into consideration at all? – JMoravitz Feb 13 '18 at 0:36
• Examine your calculations in the simpler case where you draw two cards and ask for the probability that they have different ranks. Would your first method yield $\binom {13}2\times \frac 4{52}\times \frac 4{52}$? Does that make sense? – lulu Feb 13 '18 at 0:38
• The probability given in your first approach is actually the probability that the very specific order in which the top six cards are drawn in sequence is three cards all of the same rank followed by three more cards all of some different higher rank. (That the second rank must be higher than the first, or some equivalent alteration to the problem, is caused by your use of $\binom{13}{2}$ rather than $13\cdot 12$) – JMoravitz Feb 13 '18 at 0:39
• @lulu his first approach would have yielded $\binom{13}{2}\times \frac{4}{52}\times \frac{4}{\color{red}{51}}$, note the decreasing denominators (which is still wrong, but not by as much as you seem to imply) – JMoravitz Feb 13 '18 at 0:41
• @JMoravitz Yes, thank you. That is what I meant to write. – lulu Feb 13 '18 at 0:49

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.

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 $\dbinom{13}2 \frac{\binom43\binom43}{\binom{52}6}$

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

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 $\frac{1248}{20358520} \Rightarrow \approx 6.13011162*10^{-5}$

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