http://www.estiloskateboards.com/deck-decks-complete/
You are dealt 4 cards from a single (initially complete) deck. likelihood that you receive 3 aces?
52 initially
4/52 = 7.69%
3/51 = 5.88%
2/50 = 4%
do i add this? or multiply this? am i even on the right track?
3 aces can be selected from 4 in 4C3 ways and balance one card from balance 48 in 48 ways
=> there are 48C3 x 48 ways of getting 3 aces out of 52C4 total n umber of getting any 4 cards.
=> required probability
= 4C3 * 48/52C4
= (4×48x24)/(52×51x50×49)
= 0.0007092.
I have rechecked my answer and it is correct. To explain it differently,
Suppose the first card selected is ace. Its probability is 4/52 as you rightly thought. Similarly 2nd ace and 3rd ace probabilities are 3/51 and 2/50 Now fourth non-ace probability is 48/49. But one more thing to be kept in mind is that the nonace card may be picked up as first, second, third or fourth attempt which makes the final asnwer as
(4/52)*(3/51)*(2/50)*(48/49)*4
which is the same as what I worked out above.
Complete YuGiOh Zombie Deck [ with a sexy mat ]
