By a hand containing 5 different kinds, is this including face value, suit and sequence or just face value?
I think it's just face value, because 5 different suits wouldn't make sense. -Zoe
13 different kinds and 4 different suits
i did (13 choose 5)*4 on top (to represent there are 13 numbers, pick one ad then taking the suits into account - ace of hearts is different from ace of spades) and (52 choose 5) on the bottom to show all the possible hands with 5 cards.
I think that is mostly right but I believe it should be (13 choose 5)*4^5, in order to account for there being 4 choices for each of the kinds.--Spfeifer 16:55, 18 February 2009 (UTC)
I haven't confirmed this solution with anyone, but I found it easier to get the probability of the complement, having a hand with at least two cards of the same type. For the top: Choose any card (52C1), choose another card of that type (3C1) and choose 3 arbitrary cards (50C3)-jdrummon
Same kind could also mean all the cards have the same suit. So you have to eliminate all the flush hands. so subtract $ \frac{{4 \choose 1}{13 \choose 5}}{{52 \choose 5}} $ from the above statement. ---Aaron 00:35, 19 February 2009 (UTC)
