Probability of having exactly 1 pair from drawing 5 cards

Question

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.

Answer 1

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?

Answer 2

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).

Answer 3

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 $$1098240/2598960 \approx 42.26\%.$$