magley64 Posted October 27, 2011 Report Share Posted October 27, 2011 What are the odds of pulling either an ace or a diamond from a standard deck of 52 cards.one draw, an ace is considered success, a diamond is considered success.all other cards are considered failure, what are the odds of success? Quote Link to comment Share on other sites More sharing options...
Mykill Posted October 27, 2011 Report Share Posted October 27, 2011 13 diamonds3 aces 16/52 ???? 1 Quote Link to comment Share on other sites More sharing options...
magley64 Posted October 27, 2011 Author Report Share Posted October 27, 2011 my sister's professor has a different answer than my sister, and i agree with my sister, just looking to see where everyone is at. Quote Link to comment Share on other sites More sharing options...
C-bus Posted October 27, 2011 Report Share Posted October 27, 2011 1:3.25 would be the easy answer. But I'm sure I'm missing something. 1 Quote Link to comment Share on other sites More sharing options...
scottb Posted October 27, 2011 Report Share Posted October 27, 2011 $1.60?.....lol Quote Link to comment Share on other sites More sharing options...
Mykill Posted October 27, 2011 Report Share Posted October 27, 2011 .308 is the actual number but its all daimonds plus the 4 aces but since one of the aces is a diamond you dont count it twice so 3 aces plus the 13 diamonds. Quote Link to comment Share on other sites More sharing options...
imprez55 Posted October 27, 2011 Report Share Posted October 27, 2011 Odds? Well, there are 13 diamonds, and 3 more unaccounted for aces so 16 total. So the odds are 16:52, or 16/52=0.30769 or 31% percent 1 Quote Link to comment Share on other sites More sharing options...
cmh_sprint Posted October 27, 2011 Report Share Posted October 27, 2011 4/13. Based on what i just read. So a simple probability problem that is very common would be to ask what is the probability of pulling out a 6. This probability would be 4/52 which could then be simplified to 1/13. Notice there are 4 six’s in a deck out of a total of 52 cards which is how we get 4/52. You should always simplify your final answer when expressing a probability. 1 Quote Link to comment Share on other sites More sharing options...
C-bus Posted October 27, 2011 Report Share Posted October 27, 2011 4/13. Based on what i just read. So a simple probability problem that is very common would be to ask what is the probability of pulling out a 6. This probability would be 4/52 which could then be simplified to 1/13. Notice there are 4 six’s in a deck out of a total of 52 cards which is how we get 4/52. You should always simplify your final answer when expressing a probability.That's how I came up with 1 in 3.25. What does the Prof say? Quote Link to comment Share on other sites More sharing options...
magley64 Posted October 27, 2011 Author Report Share Posted October 27, 2011 That's how I came up with 1 in 3.25. What does the Prof say?he's defending 17/52 because there are 4 aces "plus" 13 diamonds....doesn't sound to me like he should be teaching anyone statistics... Quote Link to comment Share on other sites More sharing options...
Goldie Posted October 27, 2011 Report Share Posted October 27, 2011 Yeah, there isn't an ace of diamonds so the professor would be correct... Quote Link to comment Share on other sites More sharing options...
C-bus Posted October 27, 2011 Report Share Posted October 27, 2011 he's defending 17/52 because there are 4 aces "plus" 13 diamonds....doesn't sound to me like he should be teaching anyone statistics...That man needs a deck of cards. Quote Link to comment Share on other sites More sharing options...
mello dude Posted October 27, 2011 Report Share Posted October 27, 2011 I'm at 16/52 or 30.7 % -- Texas hold 'em anyone? Quote Link to comment Share on other sites More sharing options...
TwiztedRabbit Posted October 27, 2011 Report Share Posted October 27, 2011 count em.... then do the math nuff said Quote Link to comment Share on other sites More sharing options...
Mykill Posted October 27, 2011 Report Share Posted October 27, 2011 count em.... then do the math nuff saidTheres no math in statistics Quote Link to comment Share on other sites More sharing options...
cmh_sprint Posted October 27, 2011 Report Share Posted October 27, 2011 Our education system at work. Quote Link to comment Share on other sites More sharing options...
imprez55 Posted October 27, 2011 Report Share Posted October 27, 2011 he's defending 17/52 because there are 4 aces "plus" 13 diamonds....doesn't sound to me like he should be teaching anyone statistics...I would like to play cards with this man on the weekend Quote Link to comment Share on other sites More sharing options...
Vulcan_Rider Posted October 27, 2011 Report Share Posted October 27, 2011 What a jackass its 4/13 but then again I still think the answer is 2 so what do I know Quote Link to comment Share on other sites More sharing options...
NinjaDoc Posted October 27, 2011 Report Share Posted October 27, 2011 he's defending 17/52 because there are 4 aces "plus" 13 diamonds....doesn't sound to me like he should be teaching anyone statistics...he might be right, since in math term Ace diamond is = two probability ? even thou we end up with one card only? i mean if what is the probability we will pick a diamond = 13 .what is the probability we will pick an ace = 4 then its a combination of these two and never actually a probability of picking back to back 16 cards,, but 17 cards.. remember in this question we might be replacing card after each pick right.? then again maths is not my strong subject.. Quote Link to comment Share on other sites More sharing options...
Vulcan_Rider Posted October 27, 2011 Report Share Posted October 27, 2011 he might be right, since in math term Ace diamond is = two probability ? even thou we end up with one card only? i mean if what is the probability we will pick a diamond = 13 .what is the probability we will pick an ace = 4 then its a combination of these two and never actually a probability of picking back to back 16 cards,, but 17 cards.. remember in this question we might be replacing card after each pick right.? then again maths is not my strong subject..52 cards in a deck, only 16 cards in that deck are a successful pick, odds are 16/52 broken down to 4/13. Replace it or not there are still only 16 cards out of 52 that are successful picks. Not to mention not putting the card back after a draw would change the odds with every draw. Quote Link to comment Share on other sites More sharing options...
NinjaDoc Posted October 27, 2011 Report Share Posted October 27, 2011 i mean what if the argument is like this, first time i picked a diamond ACE i am going to count it as diamond, then i replace the card and then next time i pick it again and i count it as ACE. until the card reaches our hand its not counted, so in the deck of 52 cards this one card hold value of two probability ? Quote Link to comment Share on other sites More sharing options...
magley64 Posted October 27, 2011 Author Report Share Posted October 27, 2011 he might be right, since in math term Ace diamond is = two probability ? even thou we end up with one card only? i mean if what is the probability we will pick a diamond = 13 .what is the probability we will pick an ace = 4 then its a combination of these two and never actually a probability of picking back to back 16 cards,, but 17 cards.. remember in this question we might be replacing card after each pick right.? then again maths is not my strong subject..I think this is the professor's line of thinking...Try this: put an X on 16 cards, and then ask "what are the odds of drawing a card with an X?"your answer would be 16/52, right?now draw an X on the 13 diamonds, and the 3 remaining aces, the cards with the X, are the same cards referenced in the original problem, but now there is only 1 classification (whether or not the card has "X" drawn on it)I gotta believe 16/52 is the only correct answer Quote Link to comment Share on other sites More sharing options...
Mykill Posted October 27, 2011 Report Share Posted October 27, 2011 I gotta believe 16/52 is the only correct answerIt is. Unless in the problem there is an "or" or "and" which would change things. But this is a basic stats problem. Im surprised the prof said that. Quote Link to comment Share on other sites More sharing options...
NinjaDoc Posted October 27, 2011 Report Share Posted October 27, 2011 now ur suggesting the probability of picking a card marked X which becomes 16/52now lets do this, put an x on all diamonds, what is the probability of picking card marked x = 13/52 now put a Y on all aces probability is 4/52 what is the probability that of picking either x or y ... i pick back to back 16 x put it back then back to back 4 y then it is 17, the condition never excluded a thing with both x and y. Quote Link to comment Share on other sites More sharing options...
NinjaDoc Posted October 27, 2011 Report Share Posted October 27, 2011 It is. Unless in the problem there is an "or" or "and" which would change things. But this is a basic stats problem. Im surprised the prof said that.true.. was it the english prof who gave this question ? Quote Link to comment Share on other sites More sharing options...
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