A practical question. Given two decks of cards (52 cards) if i pick 15 cards randomly from one deck and then 5 cards from another deck, what is the probability that exactly 3 cards, between two picked sets, will be same ? What is the probability that at least 3 cards will be the same?
Thank you so much !!
Apendix: added on 31.12.13
let me see if i got this right. if i modify the above example so that i am returning my picked cards back to the deck (sampling with replacement) then once i pick 15 cards i count the number of different cards which gives me the probability for picking a matched card from the second deck. thherefore i have :
$p(pick)=15/52 $
$q(not\_pick) = 1-15/52$
and then :
my probability for picking a matching pair is $\binom {5} {2}p(pick)q(not\_pick) \times \binom {5} {2} p(pick)q(not\_pick) \times \binom {5} {2} p(pick)q(not\_pick)$
or did i messed up again??
and thank you so much for all your help :)
We interpret "the same" as meaning that among our $20$ cards, there are exactly $3$ identical pairs, as in $\spadesuit$ Q, $\spadesuit$ Q, $\diamondsuit$ $4$, $\diamondsuit$ $4$, and $\heartsuit$ $7$, $\heartsuit$ $7$.
Let our sample space consist of all ordered pairs $(A,B)$, where $A$ is a hand of $15$ from the first deck, and $B$ is a hand of $5$ from the second deck. The sample space has $\binom{52}{15}\binom{52}{5}$ equally likely elements.
Call an element of the sample space good if there are exactly $3$ matching cards in the two hands. We count the number of good elements.
Which cards match? They can be chosen in $\binom{52}{3}$ ways. For each such way, the hand from the first deck can be completed to $15$ in $\binom{49}{12}$ ways. And for each such completion, the hand from the second deck can be completed to $5$ in $\binom{37}{2}$ ways.
So our sample space has $\binom{52}{3}\binom{49}{12}\binom{37}{2}$ good elements. For the probability, divide.
For "at least $3$ matches," calculate in a similar way the probability of $4$ matches, and of $5$ matches, and add.