drtdrums Posted March 19, 2010 Share Posted March 19, 2010 Helping a buddy with his prob and stats homework. He has a problem that's driving me nuts because I think I've done something wrong. The problem: You are playing a collectible card game (ed: like MTG or Pokemon or some other **** game). You have a deck of 120 cards. You draw 14 cards at the beginning of the game. To have a good chance at winning, you need to draw one of a particular type of card, Card X. You have 16 Card X's in your deck. What are the chances that you draw at least one Card X in the initial 14-card draw? .. I'll hold my answer for now so that I don't influence the thought of anyone. Thanks guys! Link to comment Share on other sites More sharing options...
DjTj Posted March 19, 2010 Share Posted March 19, 2010 There are 104 cards in your deck that are not card X. The odds of drawing all 14 of your cards from that 104 is: 104/120 * 103/119 * 102/118 * ... * 91/107 = 11.9% Thus the odds of drawing at least one Card X is 100% - 11.9% = 88.1% Link to comment Share on other sites More sharing options...
Corcaigh Posted March 19, 2010 Share Posted March 19, 2010 It's 1 - the probability of drawing no cards with an X Link to comment Share on other sites More sharing options...
drtdrums Posted March 19, 2010 Author Share Posted March 19, 2010 There are 104 cards in your deck that are not card X.The odds of drawing all 14 of your cards from that 104 is: 104/120 * 103/119 * 102/118 * ... * 91/107 = 11.9% Thus the odds of drawing at least one Card X is 100% - 11.9% = 88.1% Beautiful. Thank you. I was off... Added the probabilities of drawing a card with an X together. Kinda crazy how it doesn't work...lol. Link to comment Share on other sites More sharing options...
PokerPacker Posted March 19, 2010 Share Posted March 19, 2010 I believe this is hypergeometric in nature, and you want to find the compliment. the compliment: (16) (104) (0) *(14) (120) (14) = 1*(104!/(104-14)!) (120!/(120-14)!) = 0.118783757 compliment of compliment = 0.881217243 Link to comment Share on other sites More sharing options...
Corcaigh Posted March 19, 2010 Share Posted March 19, 2010 You can do it either way, but it amounts to the same. Link to comment Share on other sites More sharing options...
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