Card Statistics
Author
Does anybody here know a general equation for the statistics for drawing a certain card out of a random deck of cards.
ie. I have a deck which has 5 cards of card A, 2 of card B, 3 of card C, 4 of card D, and so on with a total number of cards n. Is there an equation to give me the statictics of drawing card C after drawing 10 cards out of the deck of n cards?
For any general card in any draw the equation is just:
# of particular card
--------------------
total # of cards
If you know the 10 previous draws it is possible to calculate the probability of drawing your card. You would just subtract the # of the certain card that happened to get drawn from the numerator and subtract the # of cards drawn from the denominator. If you don''t know the first 10 draws I can''t think of any way you would acurrately find the answer since all of your cards could have been drawn already. The chances of a particular order of 10 cards being drawn would be very small. You would just use the equation above and multiply together for every card drawn.
I hope this helps some.
Brenden
# of particular card
--------------------
total # of cards
If you know the 10 previous draws it is possible to calculate the probability of drawing your card. You would just subtract the # of the certain card that happened to get drawn from the numerator and subtract the # of cards drawn from the denominator. If you don''t know the first 10 draws I can''t think of any way you would acurrately find the answer since all of your cards could have been drawn already. The chances of a particular order of 10 cards being drawn would be very small. You would just use the equation above and multiply together for every card drawn.
I hope this helps some.
Brenden
Sounds suspiciously like a homework problem....
Anyway, your question is what are the odds of drawing at least 1 of card C in 10 draws? The way to go about that is to calculate the odds of not drawing card C in 10 draws and subtracting that from 1 to get your answer.
So, one the first draw, the odds of not drawing C are
(# non-C cards)/(# of total cards)
The second draw must be adjusted because now there''s one less non-C card, so it''s
(# non-C cards - 1)/(# of total cards - 1)
Follow that out to the 10th draw, you get
(# non-C cards - 9)/(# of total cards - 9)
Take those 10 propabilities and multiply them together. Then subtract the result from 1 for your answer.
Anyway, your question is what are the odds of drawing at least 1 of card C in 10 draws? The way to go about that is to calculate the odds of not drawing card C in 10 draws and subtracting that from 1 to get your answer.
So, one the first draw, the odds of not drawing C are
(# non-C cards)/(# of total cards)
The second draw must be adjusted because now there''s one less non-C card, so it''s
(# non-C cards - 1)/(# of total cards - 1)
Follow that out to the 10th draw, you get
(# non-C cards - 9)/(# of total cards - 9)
Take those 10 propabilities and multiply them together. Then subtract the result from 1 for your answer.
Author
This is not for a homework problem, it is for a stats caculation program that I have been working on and I have already figured out that the odds of drawing one or more cards is 1 - [(Total Number of Cards - Number of Certain Card) Cr (Number of Cards Drawn) / (Total Number of Cards) Cr (Numver of Cards Drawn) ], however what I really need to know is how to figure out what the odds of drawing two of the same card are, and three of the same card, and four of the same card, and so on so forth.
Author
Note: In my post above Cr is the combination, ie. 60 Cr 7 would be Combination of 60 and 7
OK, the last question you actually asked was the probability of drawing two of the same card. If you mean a specific card (A), then it''s just (for two cards, eg):
((# A) / (# cards)) * ((* A - 1) / (# cards - 1))
But I don''t think this is what you want. Could you elaborate a little more on the question from your last post?
-Brian
((# A) / (# cards)) * ((* A - 1) / (# cards - 1))
But I don''t think this is what you want. Could you elaborate a little more on the question from your last post?
-Brian
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