Similar to the standard 52-card deck questions, I have a question about choosing cards; however, I am working with a 60 card deck and I have 4 players. At the end of each game, 1 player will have 15 cards. In this game, there are 10 different "suits" of cards. What are the odds that 1 player will have at least 4 cards from the same suit (each suit has 6 cards)? If possible, express your answer in terms of combinations as well as probability. Edit 1 So far, I have determined that 15 C 60 (15 choose 60) will get the 15 cards that the player ends up with. Then, I attempted to figure out how many ways I can have 4 cards from the same suit of 6. Think of this as getting 4 hearts or spades in a standard deck. Perhaps the formula is something similar to (15 choose 4)/(60 choose 15)?

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    $\begingroup$ Given that you imply this is the result of a game, rather than a purely random assignment of 15 cards, it is hard to figure out the probability without knowing the game. $\endgroup$ – Thomas Andrews Jan 12 '16 at 2:48

If you are wanting to know the probability that a hand of $15$ from this deck of $60$ has at least $4$ cards from the same suit (i.e. the "game" has nothing to do with it):

Suppose the $10$ suits are $A, B, C, .... J$ There will be many combinations of cards that would meet the requirements, e.g. one of them could be $6A-3C-3F-3H$. Then

$$Pr = {{\binom66\binom63\binom63\binom63/3!}\over\binom{60}{15}}$$

The division by 3! is because there are the same number of cards from 3 suits.

You will need to compute similarly for other combos and add up in the numerator.


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