correctchendawg wrote: So with this in mind, am I doing these scenarios and applying this principle correctly?
If I want to pull a King and Queen for hole cards (suits do not matter, nor does order), then the probability would be 8/663(4/52*4/51*2). I would be multiplying by two since the cards are from the same pool.
no, this is wrong.If I want to pull a King and Queen(suits don't matter, nor order), with a pull each from TWO standard decks, then it would be 4/169(8/52*8/52). I'm not multiplying by two this time because they are from pulls from separate decks each time.
imagine that the decks are green and blue.
in this case, you must actually add probabilities for both events, because they are qualitatively different events: i.e., drawing a king from the green deck and a queen from the blue deck is not the same thing as drawing a queen from the green deck and a king from the blue one.
so this would be (4/52)(4/52) + (4/52)(4/52), i.e., (green king)(blue queen) + (blue king)(green queen).
what you don't switch is the order of the pools themselves -- in other words, once you've settled on selecting from the green deck first and from the blue deck second, you calculate ALL probabilities using that same selection order.
so, for instance, the probability of getting a king from both decks would just be (4/52)(4/52).
two errors.If I want to pull a King and Queen(suits don't matter, nor order), with a pull from TWO standard decks, but this time we combine the decks and pull the cards, then it would be 60/1339(16/104*15/103*2). I would be multiplying by 2 this time since it's from the same pool.
first, you appear to be operating under the assumption that each deck contains 8 kings and 8 queens. (weird, because you had the right numbers of kings and queens in the first part.)
second, i don't know why you decreased the numerator; that is what you'd do if you were picking two of the same card.
the numerators should be 8 and 8, not 16 and 15. otherwise, correct.
