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I have an exercise as follows: There is a collection of cards consisting of 52 cards (13 types and 4 colours each type). We draw 5 cards from the collection. Then what is the probability of having exactly 1 pair (pair means same colour or same type)? Thanks for any indication.

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What about three of the same kind? Are they counted as 0 pairs, 1 pair or 3 pairs? – azimut Mar 13 '13 at 16:56
When you say colour, do you mean red vs. black (in which case the chance of exactly one pair is zero) or the four suits. At least in French, the suits are referred to as $couleurs$ – Ross Millikan Jun 26 '13 at 23:27
@RossMillikan OP is talking about $4$ 'colours', so about suits I think. – drhab Dec 9 '13 at 9:29

We have $u$ different colors $C=\{c_i, i=1..u: k,l=1..u, k\ne l: c_k\ne c_l\}$
and $v$ different types $T=\{t_i, i=1..v: k,l=1..v,k\ne l: c_k\ne c_l\}$
Next we have $k$($=2$) cards with the properties that we are interested in(the same colors or the same type), then we have these cards add a $m-k$ cards that do not form any other with the properties that we are interested(question is "having exactly 1 pair").

  1. First, assume the option of the same color:
    Here we deck of cards imagine as subsets of types:
    $\{c_1,c_1,...\}\cup\{c_2,c_2,...\}...\cup\{c_u,c_u,...\} $, where each subset contains $v$ elements with the same colour. Now we'll take from the first subset $k$ elements($k$ the same colours). This we can carry out $\binom{v}{k}$ ways(here, naturally, each has different type). Now we have selected a group of appropriate elements(pairs with the same colour) and next we need to add $m-k$ elements with different color and different type. Elements with different colour we can add by $\binom{u-1}{m-k}(m-k)!$ while any two cards must have the same type. This can be done in such a way that the from first unused colors subset($c_2$) and continue with next color($c_2$). Thus we can take $(v-k-1)(v-k-2)...(v-m)=\frac{(v-k-1)!}{(v-m-1)!}$.
    So total $m$ options with the same color:$\binom{v}{k}k!\binom{u-1}{m-k}(m-k)!\frac{(v-k-1)!}{(v-m-1)!}m!$
  2. Next, assume the option of the same type: Here we deck of cards imagine as subset of types:
    $\{t_1,t_1,...\}\cup\{t_2,t_2,...\}...\cup\{t_v,t_v,...\} $, where each subset contains $u$ elements with the same type. And we're similarly as in the first case.
    So total $m$ options with the same type:$\binom{u}{k}k!\binom{v-1}{m-k}(m-k)!\frac{(u-k-1)!}{(u-m-1)!}m!$

    The total number of permutations of $m$ elements from $n$ elements equals to $\binom{n}{m}m!$.

Probability of having exactly 1 pair (pair means same colour or same type) equals to: $$p(n,u,v,m,k)=\frac{\left(\binom{v}{k}\binom{u-1}{m-k}\frac{(v-k-1)!}{(v-m-1)!}+\binom{u}{k}\binom{v-1}{m-k}\frac{(u-k-1)!}{(u-m-1)!}\right)k!(m-k)!}{\binom{n}{m}}$$

Probability for $u=4$ colors, $v=13$ types, $n=u\cdot v=52$, $k=2$ and $m=5$: $$p(52,4,13,5,2)=\frac{\left(\binom{13}{2}\binom{3}{3}\frac{10!}{7!}+\binom{4}{2}\binom{12}{3}\frac{1!}{1}\right)2!3!}{\binom{52}{5}}\doteq 27\%$$
That's quite a lot.
In addition, if any of these cases occurs, it will be much more likely a pair with the same color than a pair with the same type (about 42 times more).

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HINT : As per the pigeonhole principle, if you have $n$ holes and $m$ balls with the condition that $m \gt n$, one hole will receive at least $2$ balls.

When you're trying to insert $5$ balls into $4$ holes, you'll have one hole with at least $2$ balls. Think along these lines.

Is it not guaranteed that you'll have at least one pair when you pick $5$ cards from the given $52$ card set?

If yes, now, how'll you restrict it to exactly one?

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I assume that three of a kind are forbidden.

For the pair there are $13$ different types, and you may select any $2$ out of the $4$ cards. so there are $$13\binom{4}{2} = 13\cdot 6 = 78$$ ways to select the pair. The third card must be of a different type, so $52-4 = 48$ possibilities. Since no second pair is allowed, for the fourth card there are $44$ possibilities, and for the fifth card $40$. The order in which those three single cards are drawn does not matter, so we have to divide by $3!$. We have found that $$ 13\binom{4}{2} \cdot \frac{48\cdot 44\cdot 40}{3!} = 1098240$$ hands out of the $\binom{52}{5} = 2598960$ hands contain exactly $1$ pair. So the probability is $$1098240/2598960 \approx 42.26\%.$$

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please note that the question states "pair means same color or same type". Are you not counting for 2 pairs also then( 1 pair of same type and 1 pair of same color - as per pigeon hole principle - we have only 4 colors and selecting 5 balls )? – Novice Mar 13 '13 at 16:50
Novice: Thanks for spotting this, you are right. I thought that a pair consists of two cards of the same type. – azimut Mar 13 '13 at 16:54

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