Probability of a deck of cards?

Question

A standard deck of 52 cards contains 4 suits of 13 cards each. What is the probability of having cards of all one suit if 13 cards are drawn and no replacement occur?

Answer 1

Okay so you start out like this
the chances of picking a card that is a suit is 100% so you start 52/52 then there are 51 cards remaining and only 12 of them would be part of that suit so it's 12/51. Continue doing this until you reach the last card. You multiply all these numbers together:

(52/52) (12/51) (11/50) (10/49) (9/48) (8/47) (7/46) (6/45) (5/44) (4/43) (3/42) (2/41) (1/40)

The denominators multiply to 3954242643911239680000
The numerators multiply to 24 908 083 200
If you simplify the two you get 1 over 158753389900


Therefore the answer is

1
----------------------
158,753,389,900

or 1 / 158,753,389,900

Answer 2

the first card can be any darn card in the deck, right? So its 52 out of 52 --> 52/52 = 1

But once you pick that first card, the remaining 12 cards must also be members of the suit that appeared in the first pick. For the second pick, there are 12 of that suit left out of a total of 51 cards. So this probability is 12/51. For the third pick, there are now 11 of that suit left out of a total of 50 cards. So this card's probability is 11/50.

Continue this until you get to the 13th pick.

Multiply these 13 probabilities together:

(1)*(12/51)*(11/50)*(10/49)*(9/48)*. . . *(1/40)
= 2.52x10^-10. This is an almost impossible chance, you have almost 200 times more chance of winning a pick six lottery!!!

Answer 3

Can sequence this easily... Probability that the first card will "fit the pattern" is 1. The next card must match: 12/51. The next card must match those: 11/50. Etc. Multiply the thirteen numbers.

Alternatively, can solve it by a "counting" argument. There are 52 choose 13 different "hands" of 13 cards. Four of them are single-suited. Thus 4/(52 choose 13).

Answer 4

52/52*12/51*11/50*10/49*9/48*8/47*7/46*6/45*5/44*4/43*3/42*2/41*1/40

=1/1*4/17*11/50*10/49*3/16*8/47*7/46*2/15*5/44*4/43*1/14*2/41*1/40

=1/158753389900

at least.. that's what my trusty spreadsheet tells me :P


this is assuming that the cards from the same suit do not have to be pulled out in order, rather, if the first card pulled out is a clubs, then the next twelve must merely be clubs also

Answer 5

not 100% sure, but i think it should be:
13/52 * 12/51 * 11/50 * 10/49 * ...

Answer 6

Probability = (4C1) / (52C13) = 4 / {52! / [(13!)(39!)]}

Probability = 4*(13!)(39!) / 52!

Probability = 1 / 158,753,389,900

Here is a link.

http://mathworld.wolfram.com/Bridge.html

Answer 7

i use to be really good at this stuff but then my brain decided to get rid of my awsum math powers. so now all i no is that it's 1 out of a really BIG number