Probability for Magic Trick I had a probability problem to solve , but could not proceed further , we have a m identical decks having n cards , where each deck has n different cards . Now shuffle them and select n cards . Now a someone chooses a card from the new deck , I do not know the card but I also take card after he has put it back , what would be the probability I would pick the same card. So any hint or method to solve would be appreciated.
 A: One suggestion would be to use conditional probability / the law of total probability. 
If I understand the question correctly, we have m identical decks, each with n cards. We shuffle all m * n cards together, and now have a pile with n distinct card types, each with m copies. We then draw n cards randomly from these shuffled cards, and we want to compute the probability that two cards selected from this new deck are of the same type (call it type i). Note: My apologies for the lack of mathematical notation in the following explanation -- not super quick with LaTeX yet, and I figured you'd get the gist anyways.
Now, what is the probability of selecting two cards of type i if we selected 0 of type i in our initial draw of the new deck? What about if we had selected 1 of type i in our initial draw? What about two, and so on? Once you've calculated these conditional probabilities (there will be m of them, unless m > n, in which case there will be n), you can multiply each of them by the probability of actually drawing the specified number of cards of type i when drawing the new deck (can be calculated with some basic combinatorics ... or remembering a handy discrete distribution). You now have the joint probability of selecting two cards of the same type after drawing k cards of that type, where k runs from 0 to m for m < or = n, and 0 to n for m > n. Since each of the "draw" events are disjoint and their union covers the event space, we can use the law of total probability to calculate the desired probability (just add them up). For more information, take a look at the wikipedia article on the law of total probability.
A: Let $x$ denote the card drawn from the new deck. The probability that the new deck contains (exactly) $k$ copies of $x$ will be
$$
P(k)=\frac{\binom{m}{k}\binom{mn-m}{n-k}}{\binom{mn}{n}}
$$
The probability you seek will then be
$$
P=\tfrac{1}{n}\cdot P(1)+\tfrac{2}{n}\cdot P(2)+...+\tfrac{m}{n}\cdot P(m)
$$
Unless of course $m>n$ then you must truncate the above sum at $\frac{n}{n}\cdot P(n)$.
maybe the parameter mn-n should be mn-m          
